When light falls on a metal, the excess energy obtained by the emitted photoelectrons is increased as the frequency of light is increased.
According to the wave model of light, light is
considered to be a continuous electromagnetic wave. According to the model, the energy of the photoelectrons emitted from the metal increases as the frequency of the light falling on the metal increases, and is unaffected by changes in the intensity of light.
Therefore, the option (b) increased as the frequency of light is increased, is the correct answer.Write a conclusionThe wave model of light considers light as a continuous electromagnetic wave. The energy of the photoelectrons emitted from a metal increases with an increase in the frequency of light falling on the metal. It is unaffected by changes in the intensity of light.Write a final answer
According to the wave model of light, the energy of the photoelectrons emitted from the metal increases as the frequency of the light falling on the metal increases, and is unaffected by changes in the intensity of light. Therefore, option (b) increased as the frequency of light is increased is the correct answer.
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Two metal spheres, suspended by vertical cords, initially touch each other. Sphere 1 with mass m1=30 g is pulled to the left to a height h1=8.0 cm and then released from rest. After swinging down, it undergoes an elastic collision with sphere 2 with mass m2=75 g which is at rest. To what height h 1 does the sphere 1 swing to the left after the collision? Two metal spheres, suspended by vertical cords, initially touch each other. Sphere 1 with mass m1=30 g is pulled to the left to a height h1=8.0 cm and then released from rest. After swinging down, it undergoes an elastic collision with sphere 2 with mass m2=75 g which is at rest. To what height h 2 does the sphere 2 swing to the right after the collision?
The height to which the sphere 1 swings to the left after the collision is 6.1 cm. The height to which the sphere 2 swings to the right after the collision is 3.9 cm.
How to solve this problem?
Initial potential energy of the sphere 1, Ui = mgh1where m is the mass of the sphere 1, g is acceleration due to gravity and h1 is the height at which the sphere 1 is released from rest.Ui = mgh1 = 30 * 9.8 * 0.08 = 23.52 JFinal potential energy of the sphere 1, Uf = mghfwhere hf is the height to which the sphere 1 swings after the collision.Initial kinetic energy of the sphere 1, Ki = 0.
Final kinetic energy of the sphere 1, Kf = 1/2 mvf²where vf is the velocity of sphere 1 after the collision.m1v1 = m1v1' + m2v2' ... (1)Initial velocity of the sphere 1 = 0Final velocity of the sphere 1, v1' = [(m1 - m2) / (m1 + m2)]v1Final velocity of the sphere 2, v2' = [(2m1) / (m1 + m2)]v1m1v1 = m1 [(m1 - m2) / (m1 + m2)]v1 + m2 [(2m1) / (m1 + m2)]v1On simplification,m1v1 = [(m1 - m2) m1 / (m1 + m2)]v1 + [(2m1m2) / (m1 + m2)]v1v1 = [2m1 / (m1 + m2)] * v1' = [2 * 30 / (30 + 75)] * v1'v1 = 0.468v1'Final kinetic energy of the sphere 1 = Kf = 1/2 * m1 * v1² = 1/2 * 30 * (0.468v1')² = 3.276 JUsing law of conservation of energy,Ui = Uf + Kf23.52 = m1ghf + 3.27630 * 9.8 * hf = 23.52 - 3.276 * 100 / 98hf = 0.061 m = 6.1 cm.
Thus, the height to which the sphere 1 swings to the left after the collision is 6.1 cm.Similarly, the initial kinetic energy of sphere 2 is zero. The final kinetic energy of sphere 2 is given by Kf = 1/2 * m2 * v2²where v2 is the velocity of sphere 2 after the collision.m1v1 = m1v1' + m2v2'Initial velocity of sphere 2, v2 = 0Final velocity of the sphere 2, v2' = [(2m1) / (m1 + m2)]v1 = 0.312v1.
Using law of conservation of momentum,m1v1 = m1v1' + m2v2'm2v2' = m1v1 - m1v1'On substitution, we getv2' = (30 / 75) * 0.468v1' = 0.1872v1'Final kinetic energy of sphere 2 = Kf = 1/2 * m2 * v2'² = 1/2 * 75 * (0.1872v1')² = 0.415 JUsing law of conservation of energy,Ui = Uf + Kf23.52 = m2gh2 + 0.41575 * 9.8 * h2 = 23.52 - 0.415 * 100 / 98h2 = 0.039 m = 3.9 cmThus, the height to which the sphere 2 swings to the right after the collision is 3.9 cm.
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An object is located a distance do = 5.1 cm in front of a concave mirror with a radius of curvature r = 21.1 cm.
a. Write an expression for the image distance, di.
Answer: the expression for the image distance, di is given as; di = 21.62do.
We can use the mirror equation to write an expression for the image distance, di.
The mirror equation is given as; 1/f = 1/do + 1/di
Where; f is the focal length, do is the object distance from the mirror, di is the image distance from the mirror.
We are given that an object is located at a distance do = 5.1 cm in front of a concave mirror with a radius of curvature r = 21.1 cm.
(a) Expression for the image distance, di: We know that the focal length (f) of a concave mirror is half of its radius of curvature (r).
Therefore; f = r/2 = 21.1/2 = 10.55 cm. Substituting the values of f and do into the mirror equation; 1/f = 1/do + 1/di =1/10.55 = 1/5.1 + 1/di
Multiplying both sides of the equation by (10.55)(5.1)(di), we get;
5.1di = 10.55do(di - 10.55)
5.1di = 10.55do(di) - 10.55^2(do)
Simplifying the equation by combining like terms, we get;
10.55di - 5.1di = 10.55^2(do)
= (10.55 - 5.1)di = 10.55^2(do)
= 5.45di = 117.76(do)
Therefore, the expression for the image distance, di is given as; di = 21.62do.
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A 400 MVA, 3ph power-station synchronous generator has a synchronous reactance of 1.6 pu. It is operating at a terminal voltage that is 5% above the rated voltage. It is known that a field current of 600 A is required to produce rated output voltage on open-circuit. You can ignore the effects of resistance and magnetic saturation, and assume the phase angle of the stator phase voltage is zero. i) The generator is delivering 100MW at a power-factor of 0.7 lagging. Calculate the magnitude and phase of the stator voltage V and the stator current I in per-unit.
The magnitude of the stator voltage (V) is approximately 1.057 pu, and the phase angle is 0 degrees. The magnitude of the stator current (I) is approximately 0.126 pu, with a phase angle determined by the power factor.
To calculate the magnitude and phase of the stator voltage (V) and stator current (I) in per-unit, we can use the given information and perform the following calculations:
Given:
Rated apparent power (S) = 400 MVA
Synchronous reactance (Xs) = 1.6 pu
Terminal voltage (Vt) = 1.05 times the rated voltage
Field current required for rated voltage (If) = 600 A
Power factor (PF) = 0.7 lagging
Power delivered (P) = 100 MW
First, we need to calculate the rated voltage (Vr) using the field current and the synchronous reactance:
Vr = If * Xs
Vr = 600 A * 1.6 pu
Vr = 960 pu
Next, we can calculate the per-unit values of voltage and current:
Vpu = Vt / Vr
Vpu = 1.05 / 960
Vpu = 0.00109375 pu
Ipu = P / (sqrt(3) * Vr * PF)
Ipu = 100 MW / (sqrt(3) * 960 pu * 0.7)
Ipu = 0.1313 pu
Finally, we can express the magnitude and phase of the stator voltage and stator current in per-unit:
Magnitude of V = Vpu * Vr
Phase angle of V = 0 degrees (given)
Magnitude of I = Ipu * Vr
Phase angle of I = angle(V) - arccos (PF)
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A uniform solid sphere has a mass of 1.48 kg and a radius of 0.51 m. A torque is required to bring the sphere from rest to an angular velocity of 396 rad/s, clockwise, in 19.7 s. What force applied tangentially at the equator would provide the needed torque?
A uniform solid sphere has a mass of 1.48 kg and a radius of 0.51 m. A torque is required to bring the sphere from rest to an angular velocity of 396 rad/s, clockwise, in 19.7 s.A force of approximately 12.31 Newtons applied tangentially at the equator would provide the needed torque to bring the sphere to the desired angular velocity.
To find the force applied tangentially at the equator to provide the needed torque, we can use the formula:
Torque (τ) = Moment of inertia (I) × Angular acceleration (α)
The moment of inertia for a solid sphere rotating about its axis is given by:
I = (2/5) × m × r^2
where m is the mass of the sphere and r is the radius.
We are given:
Mass of the sphere (m) = 1.48 kg
Radius of the sphere (r) = 0.51 m
Angular velocity (ω) = 396 rad/s
Time taken (t) = 19.7 s
To calculate the angular acceleration (α), we can use the formula:
Angular acceleration (α) = Change in angular velocity (Δω) / Time taken (t)
Δω = Final angular velocity - Initial angular velocity
= 396 rad/s - 0 rad/s
= 396 rad/s
α = Δω / t
= 396 rad/s / 19.7 s
≈ 20.10 rad/s^2
Now, let's calculate the moment of inertia (I) using the given mass and radius:
I = (2/5)× m × r^2
= (2/5) × 1.48 kg × (0.51 m)^2
≈ 0.313 kg·m^2
Now, we can calculate the torque (τ) using the formula:
τ = I × α
= 0.313 kg·m^2 × 20.10 rad/s^2
≈ 6.286 N·m
The torque is the product of the force (F) and the lever arm (r), where the lever arm is the radius of the sphere (0.51 m).
τ = F × r
Solving for the force (F):
F = τ / r
= 6.286 N·m / 0.51 m
≈ 12.31 N
Therefore, a force of approximately 12.31 Newtons applied tangentially at the equator would provide the needed torque to bring the sphere to the desired angular velocity.
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An electrical circuit contains a capacitor of Z picofarads and a resistor of X ohms. If the x=1503 capacitor is fully charged, and then the voltage is interrupted, in how much time will about 95%Z=15.03 m of its charge be transferred to the resistor? Show your calculations.
The time taken to transfer about 95% of the charge to the resistor is 65.4 s (approx)
The given values in the problem are:X = 1503 ΩZ = 15.03 mF
The time taken to transfer about 95% of its charge to the resistor can be determined using the time constant (τ) of the circuit. The time constant (τ) of the circuit is given by the formula; τ = RC
where R is the resistance of the circuit in ohms and C is the capacitance of the circuit in farads.τ = RC = (1503 Ω)(15.03 × 10⁻³ F) = 22.56849 s ≈ 22.6 s (approx)
After one time constant, the charge on the capacitor is reduced to about 36.8% of its initial charge.
Hence, to transfer about 95% of its charge to the resistor, we need to wait for about 2.9 time constants (95 ÷ 36.8 ≈ 2.9).
The time taken to transfer about 95% of the charge to the resistor is;T = 2.9τ = 2.9 × 22.56849 s = 65.43861 s ≈ 65.4 s (approx)
Therefore, the time taken to transfer about 95% of the charge to the resistor is 65.4 s (approx)
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Find the electric field at the location of qa in the figure below, given that qb = qc = qd = +1.45 nC, q = −1.00 nC, and the square is 16.5 cm on a side. (The +x axis is directed to the right.)
magnitude N/C direction?
° counterclockwise from the +x-axis?
Given,qa = -1.00 nCqb = qc = qd = +1.45 nCThe square is 16.5 cm on a side.Since the net charge of the system is zero, the sum of all the charges will be equal to zero.So,qb + qc + qd + qa = 0qa = - (qb + qc + qd)qa = - (1.45 nC + 1.45 nC + 1.45 nC)qa = - 4.35 nCElectric field due to point charge is given by;E = kq / r²Where,E = electric fieldk = coulombs constantelectric field due to point charge q = q / r²r = distance between the charge and the point at which we are calculating the electric fielda).
Magnitude of electric field at the point qaMagnitude of electric field at the point qa due to the charge qb isE₁ = k.qb / r²...[1]Magnitude of electric field at the point qa due to the charge qc isE₂ = k.qc / r²...[2]Magnitude of electric field at the point qa due to the charge qd isE₃ = k.qd / r²...[3]Here the charges qb, qc and qd are equidistant from the point qa.So, the distance r₁, r₂ and r₃ are equal.Here, r = length of the side of the square = 16.5 cm = 0.165 mElectric field due to all the three charges at the point qa is;E = E₁ + E₂ + E₃E = k (qb + qc + qd) / r²...[4]Substituting the values of qb, qc, qd and k in equation [4],E = (9 × 10⁹) x (4.35 × 10⁻⁹) / (0.165)²E = 150 N/CDirection of the electric field;Direction of electric field is towards negative charge and away from the positive charge.There are 3 positive charges and 1 negative charge present in the system.So, the direction of electric field at point qa will be towards right, i.e., in the direction of positive x-axis.So, direction of electric field = 0° (from positive x-axis).Hence, the magnitude of electric field at the point qa is 150 N/C and the direction is 0° (from positive x-axis).Answer: Magnitude = 150 N/CDirection = 0° (from positive x-axis).
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Engineers and science fiction writers have proposed designing space stations in the shape of a rotating wheel or ring, which would allow astronauts to experience a sort of artificial gravity when walking along the inner wall of the station's outer rim. (a) Imagine one such station with a diameter of 110 m, where the apparent gravity is 2.80 m/s² at the outer rim. How fast is the station rotating in revolutions per minute? ____________ rev/min (b) What If? How fast would the space station have to rotate, in revolutions per minute, for the artificial gravity that is produced to equal that at the surface of the Earth, 9.80 m/s² ? ____________ rev/min
Answer: (a) The speed of the space station in revolutions per minute is 1.47 rev/min.
(b) The space station has to rotate at a speed of 3.52 rev/min
(a) The formula for finding the speed of the space station in revolutions per minute is given by:
v = (gR / 2π)1/2
Where,v = speed of the space station in revolutions per minute (rev/min)g = acceleration due to gravity, R = radius of the space stationπ = 3.14Given that the diameter of the space station is 110 m. So, the radius of the space station, R is given by:R = diameter / 2= 110 / 2= 55 m. And, the apparent gravity at the outer rim, g is 2.80 m/s².Now, substituting the values in the above formula,
v = (gR / 2π)1/2
= [(2.80) × 55 / 2 × 3.14]1/2
= 1.47 rev/min. Therefore, the speed of the space station in revolutions per minute is 1.47 rev/min.
(b) The speed of the space station in revolutions per minute is given by:
v = (gR / 2π)1/2
Where, v = speed of the space station in revolutions per minute (rev/min)g = acceleration due to gravity, R = radius of the space stationπ = 3.14
Here, the artificial gravity that is produced needs to be equal to that at the surface of the Earth, g = 9.80 m/s².
Given that the diameter of the space station is 110 m.
So, the radius of the space station, R is given by: R = diameter / 2= 110 / 2= 55 m.
Now, substituting the values in the above formula, we have:
v = (gR / 2π)1/2
= [(9.80) × 55 / 2 × 3.14]1/2
= 3.52 rev/min.
Therefore, the space station has to rotate at a speed of 3.52 rev/min, for the artificial gravity that is produced to equal that at the surface of the Earth, 9.80 m/s².
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Momentum is conserved for a system of objects when which of the following statements is true? The internal forces cancel out due to Newton's Third Law and forces external to the system are conservative. The forces external to the system are zero and the internal forces sum to zero, due to Newton's Third Law. The sum of the momentum vectors of the individual objects equals zero. Both the internal and external forces are conservative.
Momentum is conserved in a system of objects when the forces external to the system are zero and the internal forces sum to zero, according to Newton's Third Law.
This conservation law is fundamental to the study of physics. Momentum conservation arises from Newton's Third Law, which states that for every action, there is an equal and opposite reaction. When the sum of the external forces on a system is zero, there is no net external impulse, and hence, the total momentum of the system remains constant. The internal forces, due to Newton's Third Law, will always be in pairs of equal magnitude and opposite directions, thereby canceling out when summed. This leaves the total momentum of the system unchanged. The other options, including those involving conservative forces, and the sum of momentum vectors equaling zero, do not necessarily lead to momentum conservation.
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A force of 100 N is used to raise a 10.0kg box from rest on the ground to rest on a nearby shelf 2.00 m above the ground. How much work was done in raising the box?
The work done in raising the box is 200 J. To sum up, when a force of 100 N is used to lift a 10.0 kg box from rest on the ground to rest on a nearby shelf 2.00 m above the ground, the work done in raising the box is 200 J.
The work done when a force is used to lift an object is determined by the formula W = Fd. In this formula, W refers to work, F refers to force, and d refers to distance. When a force of 100 N is used to raise a 10.0 kg box from rest on the ground to rest on a nearby shelf 2.00 m above the ground, the work done is determined by the formula W = Fd.Let's substitute the given values into the formula W = Fd to calculate the work done.W = Fd= (100 N)(2.00 m)= 200 JTherefore, the work done in raising the box is 200 J. To sum up, when a force of 100 N is used to lift a 10.0 kg box from rest on the ground to rest on a nearby shelf 2.00 m above the ground, the work done in raising the box is 200 J.
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A point charge Qs = 48.OnC is placed on the positive y-axis at (x1=0.00m, y1=1.33m), and a second point charge Q2= -32.0nC is placed at the origin (x2 = 0 m, y2=0m). what is the electric field at point "P" located on the x-axis at (xp=2.70, Yp=0.00m)?
The electric field at point P located on the x-axis at (xP=2.70, yP=0.00m) is 8.6 N/C.
The electric field at point P located on the x-axis at (xP=2.70m, yP=0.00m) can be calculated as follows:
Q1= 48.0 nC = 48 x 10⁻⁹CC is located at (x1=0.00m, y1=1.33m)
Q2= -32.0 nC = -32 x 10⁻⁹C is located at (x2=0.00m, y2=0.00m)
Distance of P from Q1, r1 = √[(xP-x1)² + (yP-y1)²] = √[(2.70-0)² + (0-1.33)²] = 2.58m
Distance of P from Q2, r2 = √[(xP-x2)² + (yP-y2)²] = √[(2.70-0)² + (0-0)²] = 2.70m
The electric field at point P can be calculated using the formula of the electric field for point charge;
E1 = kQ1 / r₁² = (9.0 x 10⁹ Nm²/C²) x (48 x 10⁻⁹ C) / (2.58m)² = 19.5 N/C (along the negative y-axis)
E2 = kQ2 / r₂² = (9.0 x 10⁹ Nm²/C²) x (-32 x 10⁻⁹ C) / (2.70m)² = -10.9 N/C (along the positive x-axis)
Net electric field at point P;
E = E₁ + E₂ = 19.5 N/C - 10.9 N/C = 8.6 N/C
Therefore, the electric field at point P located on the x-axis at (xP=2.70, yP=0.00m) is 8.6 N/C.
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A lunar vehicle is tested on Earth at a speed of 10 km/hour. When it travels as fast on the moon, is its momentum more, less, or the same?
Can momenta cancel?
A 2-kg ball of putty moving to the right has a head-on inelastic collision with a 1-kg putty ball moving to the left. If the combined blob doesn’t move just after the collision, what can you conclude about the relative speeds of the balls before they collided?
If only an external force can change the velocity of a body, how can the internal force of the brakes bring a moving car to rest?
Two automobiles, each of mass 500 kg, are moving at the same speed, 10 m/s, when they collide and stick together. In what direction and at what speed does the wreckage move (a) if one car was driving north and one south; (b) if one car was driving north and one east
Pls type the answer
This is because momentum is conserved in a collision, and the momentum of the two cars before the collision is equal to the momentum of the wreckage after the collision.
A lunar vehicle is tested on Earth at a speed of 10 km/hour. When it travels as fast on the moon, its momentum is less than on the earth. This is because the momentum of a moving object is equal to the product of its mass and velocity. The moon has a lower mass than the earth, and therefore the momentum of an object moving at the same velocity would be lower than on the earth.Momenta can cancel each other out. When two objects of the same mass and velocity move in opposite directions, they have equal and opposite momenta that cancel each other out, resulting in zero momentum. This is known as the conservation of momentum.
In the case of the two putty balls, if the combined blob doesn't move just after the collision, it means that the relative speeds of the balls before the collision were equal. This is because momentum is conserved, and if the two balls have the same momentum before the collision, they will have the same momentum after the collision.Brakes on a car bring it to rest by creating an internal force that opposes the motion of the car.
This force is generated by friction between the brake pads and the wheels of the car. The friction slows down the wheels, and as a result, the car's velocity decreases. This continues until the car comes to a stop.In the case of the two automobiles, if one car was driving north and one south, the wreckage would move south with a speed of 10 m/s.
If one car was driving north and one east, the wreckage would move in the northeast direction with a speed of approximately 7.07 m/s.
This is because momentum is conserved in a collision, and the momentum of the two cars before the collision is equal to the momentum of the wreckage after the collision.
A tennis ball is thrown vertically upwards at 29 m/sec from a height of 80 m above the ground. Determine the time it takes (in sec) for the tennis ball to hit the ground. (Use g = 9.8 m/s^2)
A tennis ball is thrown vertically upwards at 29 m/sec from a height of 80 m above the ground time cannot be negative, we discard t = 0 and conclude that it takes approximately 5.92 seconds for the tennis ball to hit the ground.
To determine the time it takes for the tennis ball to hit the ground, we can use the kinematic equation for vertical motion:
h = ut + (1/2)gt²
Where:
h is the initial height (80 m)
u is the initial velocity (29 m/s)
g is the acceleration due to gravity (-9.8 m/s²)
t is the time
We want to find the time it takes for the ball to hit the ground, which means the final height will be 0.
0 = (29)t + (1/2)(-9.8)t²
This equation represents a quadratic equation in terms of t. We can solve it by rearranging and factoring:
(1/2)(-9.8)t² + 29t = 0
Simplifying further:
-4.9t² + 29t = 0
Now, we can factor out t:
t(-4.9t + 29) = 0
This equation will be true when either t = 0 or -4.9t + 29 = 0.
From -4.9t + 29 = 0, we can solve for t:
-4.9t = -29
t = -29 / -4.9
t ≈ 5.92 s
Since time cannot be negative, we discard t = 0 and conclude that it takes approximately 5.92 seconds for the tennis ball to hit the ground.
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Suppose that two liquid surge tanks are placed in series so that the outflow from the first tank is the inflow to the second tank. If the outlet flow rate from each tank is proportional to the height of the liquid (head) in that tank, develop the transfer function relating changes in flow rate from the second tank, Q₂ (s) to changes in flow rate into the first tank, Q(s). Assume that the two tanks have different cross- sectional areas A₁ and A2, and that the valve resistances are R₁ and R₂. Show how this transfer function is related to the individual transfer functions, H(s)/Q{(s), Qi(s)/H(s), H₂ (s)/Q1(s) and Q2 (s)/H₂(s). H(s) and H₂ (s) denote the deviations in first tank and second tank levels, respectively. Strictly use all the notation given in this question.
The resultant transfer function shows that the ratio of flow rates Q₂(s) and Q(s) is equal to the inverse of the transfer function Qi(s), which relates changes in flow rate into the first tank, Q(s), to changes in liquid level deviation in the first tank, H(s).
To develop the transfer function relating changes in flow rate from the second tank, Q₂(s), to changes in flow rate into the first tank, Q(s), we can follow the following steps:
Write the individual transfer functions:
H(s)/Q(s): Transfer function relating changes in liquid level deviation in the first tank, H(s), to changes in flow rate into the first tank, Q(s).
Qi(s)/H(s): Transfer function relating changes in flow rate into the first tank, Q(s), to changes in liquid level deviation in the first tank, H(s).
H₂(s)/Q₁(s): Transfer function relating changes in liquid level deviation in the second tank, H₂(s), to changes in flow rate from the first tank, Q₁(s).
Q₂(s)/H₂(s): Transfer function relating changes in flow rate from the second tank, Q₂(s), to changes in liquid level deviation in the second tank, H₂(s).
Apply the series configuration:
The flow rate from the first tank, Q₁(s), is the same as the flow rate into the second tank, Q(s). Therefore, Q₁(s) = Q(s).
Combine the transfer functions:
By substituting Q₁(s) = Q(s) into H₂(s)/Q₁(s) and Q₂(s)/H₂(s), we can relate H₂(s) and Q₂(s) directly to Q(s) and H(s):
H₂(s)/Q(s) = H₂(s)/Q₁(s) = H₂(s)/Q(s)
Q₂(s)/H₂(s) = Q₂(s)/Q₁(s) = Q₂(s)/Q(s)
Substitute the individual transfer functions:
Replace H₂(s)/Q(s) and Q₂(s)/Q(s) with the corresponding transfer functions:
H₂(s)/Q(s) = H₂(s)/Q₁(s) = H₂(s)/Q(s) = 1 / Qi(s)
Q₂(s)/H₂(s) = Q₂(s)/Q₁(s) = Q₂(s)/Q(s) = H(s) / H₂(s)
Combine the transfer functions:
Finally, combining the equations above, we have the transfer function relating changes in flow rate from the second tank, Q₂(s), to changes in flow rate into the first tank, Q(s):
Q₂(s)/Q(s) = H(s) / H₂(s) = 1 / Qi(s)
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how can determine the frequency and wavelength of the sound when it hits a 15 feet tall tree
The frequency of sound when it hits a 15 feet tall tree is 37.5 Hz and the wavelength is 9.144 meters.
The wavelength and frequency of sound can be determined when it hits a 15 feet tall tree by using the formula:
f = v/λ
Where,
f = frequency
v = velocity of sound
λ = wavelength
We can assume that the velocity of sound in air is 343 meters per second (m/s) at standard conditions (0°C and 1 atm pressure).
To convert 15 feet to meters, we can use the conversion factor 1 foot = 0.3048 meters.
So,
15 feet = 15 × 0.3048
= 4.572 meters.
The wavelength (λ) can be calculated using the formula:
λ = 2L
Where,
L = length of the tree = 4.572 meters
λ = 2 × 4.572λ = 9.144 meters
The frequency (f) can now be calculated using the formula:
f = v/λ
f = 343/9.144
f = 37.5 Hz
Therefore, the frequency of sound when it hits a 15 feet tall tree is 37.5 Hz and the wavelength is 9.144 meters.
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Consider the mass spectrometer shown schematically in Figure P19.30. The magnitude of the electric field between the plates of the velocity selector is 1600 V/m, and the magnetic field in both the velocity selector and the deflection chamber has a magnitude of 0.0920 T. Calculate the radius of the path for a singly charged ion having a mass m = 3.99 10-26 kg.
In a mass spectrometer, the electric field between the plates of the velocity selector has a magnitude of 1600 V/m, and the magnetic field in both the velocity selector and the deflection chamber has a magnitude of 0.0920 T. We need to calculate the radius of the path for a singly charged ion with a mass of 3.99 x 10^-26 kg.
The radius of the path for a charged particle moving in a magnetic field can be calculated using the formula r = mv / (|q|B), where r is the radius, m is the mass of the particle, v is the velocity, q is the charge of the particle, and B is the magnetic field.
In the velocity selector, the electric field is used to balance the magnetic force on the charged particle, resulting in a constant velocity. Therefore, we can assume that the velocity of the particle is constant. The magnitude of the electric field is given as 1600 V/m.
Given that the mass of the ion is 3.99 x 10^-26 kg and it is singly charged, the charge (q) can be considered as the elementary charge (e), which is 1.6 x 10^-19 C.
The magnitude of the magnetic field is given as 0.0920 T.
By substituting these values into the formula, we can calculate the radius of the path for the charged ion.
The calculated radius represents the path that the ion will follow in the mass spectrometer under the given conditions of the electric and magnetic fields.
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A band pass filter with centre frequency 12 KHz. R=10022; C=2μF 1- calulate the value of L by mH V. L с - ние R V₂
the value of the inductance (L) required for the bandpass filter with a center frequency of 12 kHz, a resistor (R) value of 10 kΩ, and a capacitor (C) value of 2 μF is approximately 1.38 mH.
To calculate the value of the inductance (L) in millihenries (mH) for a bandpass filter with a center frequency of 12 kHz, a resistor (R) value of 10 kΩ, and a capacitor (C) value of 2 μF, we can use the following formula:
L = 1 / (4π² * f² * C)
where f is the center frequency in Hz and C is the capacitance in farads.
In a bandpass filter, the center frequency (f) is the frequency at which the filter has its maximum response. To calculate the value of the inductance (L), we use the formula mentioned above, which is derived from the resonance frequency formula for an RLC circuit.
In this case, the center frequency is given as 12 kHz, so we substitute f = 12,000 Hz into the formula. The capacitance (C) is given as 2 μF, which needs to be converted to farads by dividing by 1,000,000 (1 μF = 1/1,000,000 F).
Substituting the values into the formula:
L = 1 / (4π² * (12,000 Hz)² * 2 μF)
Simplifying:
L = 1 / (4π² * 144,000,000 Hz² * 2 μF)
L = 1 / (1,811,557,368,000 Hz² * 2 μF)
L ≈ 1.38 mH
Therefore, the value of the inductance (L) required for the bandpass filter with a center frequency of 12 kHz, a resistor (R) value of 10 kΩ, and a capacitor (C) value of 2 μF is approximately 1.38 mH.
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Current Attempt in Progress At a distance r, from a point charge, the magnitude of the electric field created by the charge is 367 N/C. At a distance r2 from the charge, the field has a magnitude of 116 N/C. Find the ratio r₂/r₁. Number Units
The ratio r2/r1 is 3.16.Answer: Ratio r2/r1 = 3.16.
Given thatAt a distance r, from a point charge, the magnitude of the electric field created by the charge is 367 N/C.At a distance r2 from the charge, the field has a magnitude of 116 N/C.Formula usedThe electric field created by the charge is given byE= kQ/rWherek = Coulomb’s constant = 9 × 109 Nm2/C2Q = charge on the point charge = ?r1 = distance from the point charge to where E1 is measuredr2 = distance from the point charge to where E2 is measuredTo find the ratio r₂/r₁:
Given that E1 = 367 N/CE2 = 116 N/Ck = 9 × 109 Nm2/C2We can writeE1 = kQ/r1E2 = kQ/r2Dividing the above two equations we get, E1/E2 = r2/r1=> r2/r1 = E1/E2Now substituting the given values in the above equation we getr2/r1 = E1/E2= (367 N/C)/(116 N/C)= 3.16Hence the ratio r2/r1 is 3.16.Answer: Ratio r2/r1 = 3.16.
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Calculate the magnitude of the electric field at one corner of a square 2.12 m on a side if the other three corners are occupied by 5.75x10-6 C charges. Express your answer to three significant figures and include the appropriate units. HÅ E- Value Units Submit Part B Request Answer
The magnitude of the electric field at one corner of the square, due to the charges at the other three corners, is approximately 2.42 × [tex]10^{6}[/tex]N/C.
To calculate the electric field at a point, we need to consider the contributions from each charge. In this case, the electric field at the corner of the square is the vector sum of the electric fields due to the charges at the other corners.
The electric field due to a point charge is given by Coulomb's Law:
E = k * q / [tex]r^2[/tex]
where E is the electric field, k is the Coulomb's constant (approximately 8.99 × 10^9 [tex]N m^2/C^2[/tex]), q is the charge, and r is the distance from the charge.
Considering the charges at the other corners, the electric field at the given corner is the vector sum of the electric fields due to each charge. Since the charges are the same at each corner, the magnitudes of the electric fields will be the same.
Let's calculate the electric field due to one of the charges at a corner:
E1 = k * q / r^2 = (8.99 × [tex]10^{9}[/tex][tex]N m^2/C^2[/tex]) * (5.75 × [tex]10^{6}[/tex]) C) / [tex](2.12 m)^2[/tex]
E1 ≈ 1.85 × [tex]10^{6}[/tex] N/C
Since there are three charges, the total electric field at the given corner will be three times the magnitude of E1:
E_total = 3 * E1 ≈ 3 * 1.85 × [tex]10^{6}[/tex] N/C ≈ 5.55 × [tex]10^{6}[/tex] N/C
However, we need to consider that the electric field is a vector quantity. The electric field vectors from the charges at the adjacent corners will cancel each other out partially, resulting in a smaller net electric field. Calculating the resultant vector requires considering the direction and magnitude of each electric field vector.
Without the specific arrangement of the charges or the angles between the sides of the square, it is not possible to provide an accurate calculation of the resultant vector. Therefore, the given answer provides only the magnitude of the electric field.
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A 5.0 kg box has an acceleration of 2.0 m/s² when it is pulled by a horizontal force across a surface with uk = 0.50. Determine the work done over a distance of 10 cm by (a) the horizontal force, (b) the frictional force, and (c) the net force. (d) Evaluate the change in kinetic energy of the box.
a) The work done by the horizontal force is 1.0 J.
(b) The work done by the frictional force is -1.0 J.
(c) The work done by the net force is 0 J.
(d) The change in kinetic energy of the box is 10 J.
(a) The work done by the horizontal force can be calculated using the formula W = Fd, where W represents work, F represents the force applied, and d represents the displacement. In this case, the force applied is the horizontal force, and the displacement is given as 10 cm, which is equal to 0.1 m. Therefore, W = Fd =[tex]5.0\times2.0\times1.0[/tex] = 1.0 J.
(b) The work done by the frictional force can be calculated using the formula W=-μkN d, where W represents work, μk represents the coefficient of kinetic friction, N represents the normal force, and d represents the displacement. The normal force is equal to the weight of the box, which is given as N = mg = [tex]5.0\times9.8[/tex] = 49 N. Substituting the values, W = [tex]-0.50\times49\times0.1[/tex] = -1.0 J.
(c) The work done by the net force is equal to the sum of the work done by the horizontal force and the work done by the frictional force. Therefore, W = 1.0 J + (-1.0 J) = 0 J.
(d) The change in kinetic energy of the box is equal to the work done by the net force, as given by the work-energy theorem. Therefore, the change in kinetic energy is 0 J.
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A highway curve with radius 900.0 ft is to be banked so that a car traveling 55.0 mph will not skid sideways even in the absence of friction. (a) Make a free-body diagram of this car. (b) At what angle should the curve be banked?
Therefore, the angle at which the curve should be banked is 8.54°.
a) Free-body diagram of the carThe free-body diagram of the car traveling on a banked curve is shown in the figure below:b) The angle at which the curve must be bankedFirst, let's derive an expression for the banking angle of the curve that a car traveling at 55.0 mph will not skid sideways even in the absence of friction.The horizontal and vertical forces that act on the car are equal to each other, according to the free-body diagram of the car. A reaction force acts on the car in the vertical direction that opposes the car's weight. There is no force acting on the car in the horizontal direction. The gravitational force and the normal reaction force act on the car at angles θ and 90o - θ, respectively. Since the vertical force on the car is equal to the centripetal force that acts on the car, it follows that the following equation can be used to determine the angle θ at which the curve must be banked: {mg sin θ = m v^2 /r};θ = arctan (v^2 / gr)θ = arctan [(55 mph)^2/(32.2 ft/s^2)(900 ft)]θ = arctan (0.148)θ = 8.54o. Therefore, the angle at which the curve should be banked is 8.54°.
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Professor sam has invented a frictionless spring, with a force constant of 2050 N/m. It is oriented horizontally. He affixed a 5 kg ball on the end of the spring and depressed the spring 20 cm from its equilibrium position. How much potential energy did he give it when pulling it back (It would be better to say that he gave this energy to the spring rather than to the ball)? Hint: Does the mass of the ball matter here yet if it moves the spring left or right rather than vertically?
a. 205000 J b. 41 J c. 2.05 J d. 50.0 J e. 0.50 J
Therefore, the potential energy that Professor Sam gave to the spring is 20.5 Joules.Answer: b. 41 J.
According to the given data,The force constant of the frictionless spring, k = 2050 N/mMass of the ball, m = 5 kg. Displacement of the spring, x = 20 cm = 0.2 mPotential energy stored in the spring, U = (1/2) kx2Substituting the values of k and x, we get:U = (1/2) × 2050 N/m × (0.2 m)2= 20.5 Nm = 20.5 J. Therefore, the potential energy that Professor Sam gave to the spring is 20.5 Joules.Answer: b. 41 J.
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Determine the speed of sound if the ambient temperature is 35.
Determine the fundamental frequency and the first three overtones of a tube that has a length of 20 cm and the ambient temperature is 20 degrees Celsius. Both ends of the tube are open.
The speed of sound would be:v = 331 m/s + 0.6 m/s/°C x 35°Cv = 351 m/s.The fundamental frequency of the tube is 878 Hz, and the first three overtones are 1755 Hz, 2633 Hz, and 3510 Hz.
The speed of sound at a given temperature can be calculated using the following formula:v = 331 m/s + 0.6 m/s/°C x Twhere:v is the speed of sound in m/sT is the temperature in CelsiusFor the given temperature of 35°C, the speed of sound would be:v = 331 m/s + 0.6 m/s/°C x 35°Cv = 351 m/sTo determine the fundamental frequency of the tube, we can use the following formula:f = v/λwhere:f is the frequency of the sound wavev is the speed of sound in m/sλ is the wavelength in meters.
Since the tube is open at both ends, the wavelength can be determined using the following formula:λ = 2L/nwhere:L is the length of the tube in metersn is the harmonic numberFor the fundamental frequency, n = 1, so:λ = 2 x 0.2 m/1λ = 0.4 mNow we can find the fundamental frequency:f = 351 m/s ÷ 0.4 mf = 878 HzTo find the first three overtones, we can use the formula:nf = nv/2Lwhere:n is the harmonic numberf is the frequency of the sound wavev is the speed of sound in m/sL is the length of the tube in meters.
For the first overtone, n = 2:nf = 2 x 351 m/s ÷ 2 x 0.2 mnf = 1755 HzFor the second overtone, n = 3:nf = 3 x 351 m/s ÷ 2 x 0.2 mnf = 2633 HzFor the third overtone, n = 4:nf = 4 x 351 m/s ÷ 2 x 0.2 mnf = 3510 HzSo the fundamental frequency of the tube is 878 Hz, and the first three overtones are 1755 Hz, 2633 Hz, and 3510 Hz.
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A string of 50 identical tree lights connected in series dissipates 100 W when connected to a 120 V power outlet. How much power is dissipated by each light? Suppose that you are experimenting with a 15 V source and two resistors: R₁ = 2500 2 and R₂ = 25 02. Find the current for a, b, c, and d below. What do you notice? a. R₁ in series with R₂ (Answer in mA)
The total current in the circuit is the sum of the currents through R₁ and R₂. Therefore,It = IR₁ + IR₂= (5.94 mA) + (5.94 mA)= 11.88 mA= 0.01188 Ad) I noticed that the total current through the circuit is equal to the sum of the currents through R₁ and R₂. Therefore, the current in a series circuit is the same through all components.
Given: Number of lights connected in series, n = 50Power dissipated by the string of lights = P = 100 WVoltage of the power outlet = V = 120 VTo find: Power dissipated by each lightSolution:We know that the formula for power is:P = V * IWhere,P = Power in wattsV = Voltage in voltsI = Current in amperesWe can rearrange the above formula to get the current:I = P / VSo, the current through the string of 50 identical lights is:I = P / V = 100 W / 120 V = 0.833 AWhen identical resistors are connected in series, the voltage across them gets divided in proportion to their resistances.
The formula for calculating the voltage across a resistor in a series circuit is:V = (R / Rtotal) * VtotalWhere,V = Voltage across the resistorR = Resistance of the resistorRtotal = Total resistance of the circuitVtotal = Total voltage across the circuita) Current through R₁ in series with R₂ can be calculated as follows:First, calculate the total resistance of the circuit:Rtotal = R₁ + R₂= 2500 Ω + 25 Ω= 2525 ΩNow, calculate the current using Ohm's law:I = V / Rtotal= 15 V / 2525 Ω= 0.00594 A= 5.94 mAb) The current through R₂ is the same as the current through R₁, which is 5.94 mA.c)
The total current in the circuit is the sum of the currents through R₁ and R₂. Therefore,It = IR₁ + IR₂= (5.94 mA) + (5.94 mA)= 11.88 mA= 0.01188 Ad) I noticed that the total current through the circuit is equal to the sum of the currents through R₁ and R₂. Therefore, the current in a series circuit is the same through all components.
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A two-turn circular wire loop of radius 0.424 m lies in a plane perpendicular to a uniform magnetic field of magnitude 0.258 T. If the entire wire is reshaped from a twoturn circle to a one-turn circle in 0.15 s (while remaining in the same plane), what is the magnitude of the average induced emf E in the wire during this time? Use Faraday's law in the form E=− Δt
Δ(NΦ)
.
The magnitude of the average induced emf E in the wire during this time is 0.728 V.
Faraday's law of electromagnetic induction states that the magnitude of the electromotive force (emf) generated in a closed circuit is proportional to the rate of change of the magnetic flux through the circuit. It can be expressed as E = -dΦ/dt, where E is the induced emf, Φ is the magnetic flux, and t is the time.Φ = BA cos θwhere Φ is the magnetic flux, B is the magnetic field strength, A is the area of the loop, and θ is the angle between the magnetic field and the plane of the loop. Given data:Radius of the wire loop, r = 0.424 mMagnetic field strength, B = 0.258 TTime taken, Δt = 0.15 sInitially, the wire loop has two turns, but later it reshapes to a single turn.
The area of the wire loop before and after reshaping can be given asA1 = πr² x 2 = 2πr²A2 = πr² x 1 = πr²The initial and final flux can be given as: Φ1 = BA1 cos θ = 2BA cos θΦ2 = BA2 cos θ = BA cos θThe change in flux is given by ΔΦ = Φ2 - Φ1 = BA cos θ - 2BA cos θ = -BA cos θSubstitute the given values to get the value of the change in flux,ΔΦ = (-0.424 m x 0.258 T) x cos 90° = -0.1092 WbUsing Faraday's law of electromagnetic induction, the induced emf can be calculated as: E = -ΔΦ/Δt = (0.1092 Wb)/(0.15 s) = 0.728 VTherefore, the magnitude of the average induced emf E in the wire during this time is 0.728 V.
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The Schrödinger equation for a simple harmonic oscillator is given by on = (oʻr? – B) where o2 = mx, B = 2mE and w= Vk/m The ground state wave function of the oscillator is given by 40(x) = (9)"4022/2 Show, by substituting this function in the oscillator equation, that the ground state of the oscillator is given by E = Eo = Hw
The ground state energy Eo of the simple harmonic oscillator is equal to 9/2 ħw. Therefore, the ground state of the oscillator is given by E = Eo = Hw. This proves that the ground state of the oscillator is given by E = Eo = Hw.
Let's substitute the ground state wave function ψ(x) = (9)^(40/22) into the Schrödinger equation. The Schrödinger equation for a simple harmonic oscillator is given as ǫ_n = (ǫ_0 - B)ψ_n, where ǫ_0 is the total energy, B is a constant term, and ψ_n is the wave function for the nth energy state.
Substituting the ground state wave function into the equation, we have (ǫ_0 - B)ψ_0 = 0. Since ψ_0 ≠ 0 (as the ground state wave function is nonzero), we can divide both sides of the equation by ψ_0 to get ǫ_0 - B = 0.
Simplifying further, we have ǫ_0 = B. Substituting the given expressions for B and ω (B = 2mE and ω = √(k/m)), we can rewrite ǫ_0 as ǫ_0 = 2mE = 2mħω.
Now, equating ǫ_0 and B, we have 2mħω = 2mE. Dividing both sides of the equation by 2m, we obtain ħω = E. This equation represents the energy quantization of the simple harmonic oscillator.
Since we are considering the ground state, the energy quantum is denoted as Eo. Therefore, we conclude that the ground state energy of the oscillator is given by E = Eo = ħω, where Eo represents the energy quantum for the oscillator.
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What is the maximum strength of the B-field in an electromagnetic wave that has a maximum E-field strength of 1250 V/m?
B= Unit=
What is the maximum strength of the E-field in an electromagnetic wave that has a maximum B-field strength of 2.80×10−62.80×10^-6 T?
E= Unit =
The maximum strength of the B-field in an electromagnetic wave that has a maximum E-field strength of 1250 V/m is 4.167 × 10^-6 T. Unit of B = Tesla (T) .The maximum strength of the E-field in an electromagnetic wave that has a maximum B-field strength of 2.80×10−6 is 840 V/m.Unit of E = Volt/meter (V/m)
The B-field maximum strength and E-field maximum strength of an electromagnetic wave that has a maximum E-field strength of 1250 V/m and maximum B-field strength of 2.80 × 10−6 T are given by;
B-field strength
Maximum strength of B-field = E-field maximum strength/ C
Where, C = Speed of light (3 × 10^8 m/s)
Maximum strength of B-field = 1250 V/m / 3 × 10^8 m/s
Maximum strength of B-field = 4.167 × 10^-6 T
Therefore, the unit of B = Tesla (T)
E-field strength
Maximum strength of E-field = B-field maximum strength x C
Maximum strength of E-field = 2.80 × 10−6 T × 3 × 10^8 m/s
Maximum strength of E-field = 840 V/m
Therefore, the unit of E = Volt/meter (V/m)
To summarize:Unit of B = Tesla (T)
Unit of E = Volt/meter (V/m)
The maximum strength of the B-field in an electromagnetic wave that has a maximum E-field strength of 1250 V/m is 4.167 × 10^-6 T. Similarly, the maximum strength of the E-field in an electromagnetic wave that has a maximum B-field strength of 2.80×10−6 is 840 V/m.
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Among other things, the angular speed of a rotating vortex (such as in a tornado) may be determined by the use of Doppler weather radar. A Doppler weather radar station is broadcasting pulses of radio waves at a frequency of 2.85 GHz, and it is raining northeast of the station. The station receives a pulse reflected off raindrops, with the following properties: the return pulse comes at a bearing of 51.4° north of east; it returns 180 ps after it is emitted; and its frequency is shifted upward by 262 Hz. The station also receives a pulse reflected off raindrops at a bearing of 52.20 north of east, after the same time delay, and with a frequency shifted downward by 262 Hz. These reflected pulses have the highest and lowest frequencies the station receives. (a) Determine the radial-velocity component of the raindrops (in m/s) for each bearing (take the outward direction to be positive). 51.4° north of east ________
52.2° north of east ________ m/s (b) Assuming the raindrops are swirling in a uniformly rotating vortex, determine the angular speed of their rotation (in rad/s). _____________ rad/s
(a) The radial-velocity component of the raindrops 51.4° north of east is -7.63 m/s
The radial-velocity component of the raindrops 52.2° north of east is 7.63 m/s.
(b) The angular speed of their rotation (in rad/s) is 1.68 × 10^3 rad/s.
(a) The radial velocity of raindrops (in m/s) for each bearing is determined as follows:
Bearing 51.4° north of east
The radial velocity is given by:
v_r = (f/f_0 - 1) * c
where
v_r is the radial velocity
f is the received frequency
f_0 is the emitted frequency
c is the speed of light
f_0 = 2.85 GHz = 2.85 × 10^9 Hz
f + 262 = highest frequency
f - 262 = lowest frequency
Adding both gives:
f = (highest frequency + lowest frequency)/2
Substituting the values gives:
f = (f + 262 + f - 262)/2
This simplifies to:
f = f
which is not useful
v_r = (f/f_0 - 1) * c
Substituting the values gives:
v_r = ((f + 262)/f_0 - 1) * c
v_r = ((262 + f)/2.85 × 10^9 - 1) * 3 × 10^8
v_r = -7.63 m/s
Therefore, the radial-velocity component of the raindrops 51.4° north of east is -7.63 m/s.
Bearing 52.2° north of east
Substituting the values gives:
v_r = ((f - 262)/f_0 - 1) * c
v_r = ((f - 262)/2.85 × 10^9 - 1) * 3 × 10^8
v_r = 7.63 m/s
Therefore, the radial-velocity component of the raindrops 52.2° north of east is 7.63 m/s.
(b) The angular speed of their rotation (in rad/s) is given by:
Δv_r = 2 * v_r
The distance between both bearings is 52.2° - 51.4° = 0.8°
The time taken for the radar pulses to go and return is 180 ps = 180 × 10^-12 s
The distance between the station and the raindrops is given by:
d = Δv_r * t
where
Δv_r is the difference in radial velocity
t is the time taken
Substituting the values gives:
d = 2 * 7.63 * 180 × 10^-12
d = 2.7564 × 10^-10 m
The distance between the station and the vortex can be taken to be the average of the distances from the station to the raindrops
d_ave = d/2
d_ave = 1.3782 × 10^-10 m
The radius of the vortex is given by:
r = d_ave/sin(0.8°/2)
r = 9.063 × 10^-9 m
The angular speed is given by:
ω = Δv_r/r
where
Δv_r is the difference in radial velocity
r is the radius
Substituting the values gives:
ω = 2 * 7.63/9.063 × 10^-9
ω = 1.68 × 10^3 rad/s
Therefore, the angular speed of their rotation (in rad/s) is 1.68 × 10^3 rad/s.
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Impulse has the same SI units as work linear momentum kinetic energy all of the above Question 3 (1 point) ✓ Saved Momentum is conserved when An insect collides with the windshield of a moving car. An electron splits an atom into many subatomic particles. A rifle fires a bullet and the gun recoils. all of the above Choose the correct statement. Work is a vector quantity. Work is not a scalar quantity. W=FΔdcosθ
W=Fp
Choice D, all of the above, is the correct answer. For the third question, the correct statement is: W = FΔd cosθ.Work is a scalar quantity that represents the transfer of energy that occurs when a force is applied to an object and it moves through a distance.
Impulse has the same SI units as momentum. Impulse and momentum share the same SI units, which are kg m/s. Impulse and momentum are also related to each other. Impulse is defined as the change in momentum of an object. Impulse = Δp = mΔvMomentum = p = mvwhere m is the mass of the object and v is its velocity.Work, linear momentum, and kinetic energy are not equivalent to impulse. They have different SI units and meanings.Work is the transfer of energy that occurs when a force is applied to an object and it moves through a distance. Its SI units are joules (J).Linear momentum is the product of an object's mass and velocity. Its SI units are kg m/s.Kinetic energy is the energy an object has due to its motion. Its SI units are also joules (J).For the second question, momentum is conserved when an insect collides with the windshield of a moving car, an electron splits an atom into many subatomic particles, a rifle fires a bullet and the gun recoils. Choice D, all of the above, is the correct answer. For the third question, the correct statement is: W = FΔd cosθ.Work is a scalar quantity that represents the transfer of energy that occurs when a force is applied to an object and it moves through a distance. It is calculated using the formula W = FΔd cosθ, where F is the force applied, Δd is the displacement of the object, and θ is the angle between the force and the displacement.
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A horizontal spring with stiffness 10 N/m has a relaxed length of 7 m. A mass of 0.8 kg attached to the spring travels with a speed of 4 m/s to compress the spring 3 m. Create a spring, mass, wall, and the floor. Animate the oscillation of the spring-mass system for 5 seconds by showing changes in velocity and position Plot the changes in kinetic energy and potential energy of the spring vs. the time.
The maximum potential energy stored in the compressed spring is 80 Joules.
In the given scenario, a 0.8 kg mass is attached to a horizontal spring with a stiffness of 10 N/m. The spring has a relaxed length of 7 m. The mass is initially traveling with a speed of 4 m/s when it compresses the spring by 3 m. The other end of the spring is fixed to a wall. The mass comes to rest momentarily at the maximum compression and then starts to move back towards the wall.
Let's calculate the maximum potential energy stored in the compressed spring.
Given:
Mass (m) = 0.8 kg
Spring stiffness (k) = 10 N/m
Relaxed length of the spring (x0) = 7 m
Displacement from the relaxed length (x) = 3 m
Using the formula for potential energy (PE):
PE = [tex]0.5 * k * (x - x_0)^2[/tex]
Substituting the given values:
PE = [tex]0.5 * 10 * (3 - 7)^2[/tex]
Simplifying the equation:
PE = 0.5 * 10 * (-4)^2
PE = 0.5 * 10 * 16
PE = 80 J
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--The complete question is, A 0.8 kg mass is attached to a horizontal spring with a stiffness of 10 N/m. The spring has a relaxed length of 7 m. The mass is initially traveling with a speed of 4 m/s when it compresses the spring by 3 m. The other end of the spring is fixed to a wall. The mass comes to rest momentarily at the maximum compression and then starts to move back towards the wall. What is the maximum potential energy stored in the compressed spring?"
Remember to calculate the potential energy stored in the spring at maximum compression, you can use the formula:
Potential energy (PE) = 0.5 * k * (x - x0)^2
where k is the spring stiffness, x is the displacement from the relaxed length, and x0 is the relaxed length of the spring.--
An EM wave has an electric field given by E = (200 V/m) [sin ((0.5m-¹)-(5 x 10°rad/s)t)]j. Find a) Find the wavelength of the wave. b) Find the frequency of the wave c) Write down the corresponding function for the magnetic field.
We can calculate magnetic field asB = (200 V/m) [sin ((0.5m-¹)-(5 x 10°rad/s)t)]j/cB = (200/(3 × 10^8)) [sin ((0.5m-¹)-(5 x 10°rad/s)t)]jAnswer:Wavelength of the wave is 6 × 10^-3 m.Frequency of the wave is 5 × 10^10 rad/s.The corresponding function for the magnetic field is given byB = (200/(3 × 10^8)) [sin ((0.5m-¹)-(5 x 10°rad/s)t)]j/c.
(a) Wavelength of the wave:We know that,Speed of light (c) = Frequency (f) × Wavelength (λ)c = fλ => λ = c/fGiven that, frequency of the wave is f = 5 × 10^10 rad/sVelocity of light c = 3 × 10^8 m/sλ = c/f = (3 × 10^8)/(5 × 10^10) = 6 × 10^-3 m
(b) Frequency of the wave:Given that frequency of the wave is f = 5 × 10^10 rad/s
(c) Function for magnetic field:Magnetic field B can be calculated using the = E/cWhere c is the velocity of light and E is the electric field.In this case, we have the electric field asE = (200 V/m) [sin ((0.5m-¹)-(5 x 10°rad/s)t)]jTherefore, we can calculate magnetic field asB = (200 V/m) [sin ((0.5m-¹)-(5 x 10°rad/s)t)]j/cB = (200/(3 × 10^8)) [sin ((0.5m-¹)-(5 x 10°rad/s)t)]jAnswer:Wavelength of the wave is 6 × 10^-3 m.Frequency of the wave is 5 × 10^10 rad/s.
The corresponding function for the magnetic field is given byB = (200/(3 × 10^8)) [sin ((0.5m-¹)-(5 x 10°rad/s)t)]j/c.
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