Power transmitted by the steel propeller shaft = 5.5 MW = 5.5 x 10^6 W.
Speed of rotation = 180 rpm.
Shearing stress = 60 MPa.
Maximum angle of twist = 1°
Length of the steel propeller shaft = 25 diameters.
Given that modulus of rigidity of the steel propeller shaft G = 83 GPa.
We know that the power transmitted by the shaft, P = 2πNT/60,where N = speed of rotation in rpm and T = torque in N-m.
Substituting the given values, we get,5.5 x 10^6 = 2π x 180T/60.
T = 2.05 x 10^7 N-m. Now, we know that the maximum shearing stress τmax = 16T/πd^3 and maximum angle of twist θmax = TL/Gd^4.
Now, substituting the given values : we get,τmax = 16T/πd^3 = 60 MPa.θmax = TL/Gd^4 = 1° x π/180 x 25d = 25d.
Solving for diameter d, we get, τmax = 16T/πd^3⇒ 60 x 10^6 = 16 x 2.05 x 10^7/πd^3
⇒ d^3 = 2.69 x 10^-3
⇒ d = 0.144 m or 144 mm.
Answer: Diameter of the steel propeller shaft = 144 mm.
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For f = (2y-z)³ i + x² j - (3x²+1)k, is f conservative
at point (1,4,6)?
is there a curl?
is there a divergence?
For f = (2y-z)³ i + x² j - (3x²+1)k, is f conservative
at point (1,4,6)?
Curl (or rotation) is the curl of a vector field, which describes the magnitude and direction of the rotation of a particle at a point. To find whether f is conservative, we must find the curl of f and check whether it is zero or not.
The curl of the given function is: curl(f) = (∂Q/∂y - ∂P/∂z) i + (∂R/∂z - ∂P/∂x) j + (∂P/∂y - ∂Q/∂x) k
Where, P = (2y - z)³Q = x²R = -(3x² + 1)∂P/∂x = 0∂P/∂y = 6(2y - z)²∂P/∂z = -3(2y - z)²∂Q/∂x = 2x∂Q/∂y = 0∂Q/∂z = 0∂R/∂x = -6x∂R/∂y = 0∂R/∂z = 0
Therefore, curl(f) = (12z - 24y) i + 0 j + 6x k
At point (1, 4, 6),curl(f) = (12(6) - 24(4)) i + 0 j + 6(1) k= -72 i + 6 k
Therefore, the curl of f at point (1, 4, 6) is not zero. Therefore, f is not conservative at point (1, 4, 6).
Divergence is the measure of the magnitude of a vector field's source or sink at a given point in the field. To determine if there is a divergence, we must take the divergence of the function.
The divergence of the given function is:div(f) = ∂P/∂x + ∂Q/∂y + ∂R/∂z= 0 + 0 - 6
Therefore, the divergence of f is -6.
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A block is pushed with a force of 100N along a level surface. The block is 2 kg and the coefficient of friction is 0.3. Find the blocks acceleration.
The block's acceleration is 4.85 m/s².
To find the block's acceleration, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration (F = ma). In this case, the net force is the force applied to the block minus the force of friction.
1. Determine the force of friction. The force of friction can be calculated using the formula Ffriction = μN, where μ is the coefficient of friction and N is the normal force. In this case, the normal force is equal to the weight of the block, which can be calculated as N = mg, where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s²). Therefore, N = 2 kg × 9.8 m/s² = 19.6 N. Plugging in the values, we get Ffriction = 0.3 × 19.6 N = 5.88 N.
2. Calculate the net force. The net force is equal to the applied force minus the force of friction. The applied force is given as 100 N. Therefore, the net force is Fnet = 100 N - 5.88 N = 94.12 N.
3. Determine the acceleration. Now that we know the net force acting on the block, we can use Newton's second law (F = ma) to find the acceleration. Rearranging the formula, we get a = Fnet / m. Plugging in the values, we get a = 94.12 N / 2 kg = 47.06 m/s².
Thus, the block's acceleration is 4.85 m/s² (rounded to two decimal places).
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Score on last try: 0.67 of 2 pts. See Details for more. You can retry this question below A mass is placed on a frictionless, horizontal table. A spring (k=115 N/m), which can be stretched or compressed, is placed on the table. A 3-kg mass is anchored to the wall. The equilibrium position is marked at zero. A student moves the mass out to x=7.0 cm and releases it from rest. The mass oscillates in simple harmonic motion. Find the position, velocity, and acceleration of the mass at time t=3.00 s. x(t=3.00 s)=cm
v(t=3.00 s)=cm/s
a(t=3.00 s)= Enter an integer or decimal number cm/s 2
The position, velocity, and acceleration of a mass on a frictionless, horizontal table with a spring is -1.97 cm, 13.68 cm/s, [tex]50.96 cm/s^2[/tex].
For finding the position of the mass at t=3.00 s, we can use the equation for the simple harmonic motion: [tex]x(t) = A * cos(\omega t + \phi)[/tex], where A is the amplitude, [tex]\omega[/tex]is the angular frequency, t is the time and [tex]\phi[/tex] is the phase constant. In this case, the equilibrium position is marked at zero, so the amplitude A is 7.0 cm.
The angular frequency can be calculated using the formula [tex]\omega = \sqrt(k / m)[/tex], where k is the spring constant (115 N/m) and m is the mass (3 kg). Plugging in the values, we get [tex]\omega = \sqrt(115 / 3) \approx 7.79 rad/s[/tex].
For finding the phase constant [tex]\phi[/tex], consider the initial conditions. The mass is released from rest, so its initial velocity is zero. This means that at t=0, the mass is at its maximum displacement from the equilibrium position (x = A) and is moving in the negative direction. Therefore, the phase constant [tex]\phi[/tex] is [tex]\pi[/tex].
Now calculate the position at t=3.00 s using the equation: [tex]x(t) = A * cos(\omega t + \phi)[/tex].
Plugging in the values,
[tex]x(t=3.00 s) = 7.0 cm * cos(7.79 rad/s * 3.00 s + \pi) \approx -1.97 cm[/tex].
To find the velocity and acceleration at t=3.00 s, differentiate the position equation with respect to time.
The velocity [tex]v(t) = -A\omega * sin(\omega t + \phi)[/tex] and the acceleration [tex]a(t) = -A\omega^2 * cos(\omega t + \phi)[/tex].
Plugging in the values,
[tex]v(t=3.00 s) \approx 13.68 cm/s and a(t=3.00 s) \approx 50.96 cm/s^2[/tex].
Position at t=3.00 s: -1.97 cm
Velocity at t=3.00 s: 13.68 cm/s
Acceleration at t=3.00 s: [tex]50.96 cm/s^2[/tex]
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In the following circuit, determine the current flowing through the \( 2 k \Omega \) resistor, \( i \). You can do this via Nodal analysis or the Mesh method.
The current flowing through the 2 kΩ resistor is 1.4 A.
Let's follow these steps to determine the current flowing through the 2 kΩ resistor using the Mesh Method:
Step 1: Define mesh currents, i1 and i2. The mesh current in clockwise direction is assumed to be positive.
Step 2: Apply KVL to each mesh separately. For Mesh 1:i1 * 4 kΩ - i2 * 2 kΩ - 2 V = 0For Mesh 2:i2 * 2 kΩ - i1 * 4 kΩ + 8 V = 0.
Step 3: Write equations for i. The current flowing through the 2 kΩ resistor can be found as: i = -i1 + i2
Step 4: Substitute the mesh equations in step 2 to solve for i1 and i2 in terms of the voltage. To solve the equation, consider the following steps: Subtract (1) from (2) and get:i2 * 4 kΩ - i1 * 2 kΩ + 10 V = 0Add (1) and (2) and get:5 i1 = 8 V or i1 = 1.6 A. Substitute this value in equation 1:i1 * 4 kΩ - i2 * 2 kΩ - 2 V = 0(1.6 A) * 4 kΩ - i2 * 2 kΩ - 2 V = 0i2 = (1.6 A * 4 kΩ - 2 V) / 2 kΩi2 = 3 A
Step 5: Finally, calculate i using the equation :i = -i1 + i2i = -1.6 A + 3 Ai = 1.4 A.
The current flowing through the 2 kΩ resistor is 1.4 A.
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How do you get the mass of a star or planet? Kepler's third law Kepler's second law Kepler's first law
To determine the mass of a star or planet, Kepler's third law is used. Kepler's third law states that the square of the orbital period of a planet or satellite is directly proportional to the cube of the semi-major axis of its orbit.
Kepler's third law provides a relationship between the mass of a star or planet and the orbital parameters of its satellites or planets. The law states that the square of the orbital period (T) is directly proportional to the cube of the semi-major axis (a) of the orbit. Mathematically, it can be expressed as T^2 ∝ a^3.
By measuring the orbital period and the semi-major axis of a planet or satellite, we can determine the mass of the star or planet using Kepler's third law. This is possible because the mass of the star or planet affects the gravitational force acting on the orbiting body, which in turn influences its orbital period and semi-major axis.
By observing the motion of satellites or planets around a star or planet and applying Kepler's third law, astronomers can estimate the mass of celestial objects in the universe, providing valuable information for understanding their properties and dynamics.
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Is the force between parallel conductors with currents in the same direction an attraction or a repulsion? Give a detailed explanation with drawing of why this is expected.
When two long, straight, parallel conductors, carrying currents in the same direction are placed close to each other, the magnetic fields around the conductors interact, creating a force.
The force between parallel conductors with currents in the same direction is a repulsion. Detailed explanation with drawing: When electric current flows through a conductor, it produces a magnetic field that surrounds the conductor.
When two parallel conductors carrying currents in the same direction are brought closer to each other, the magnetic field around the conductors will interact.Inside each conductor, the current flows in a clockwise direction. The arrows in the figure show the direction of the magnetic fields around the conductors. The interaction between the magnetic fields of the conductors produces a force that acts on the conductors and is either attractive or repulsive. In this case, the force is a repulsion. The reason why the force is repulsive is that the magnetic field produced by the current in each conductor is circular and perpendicular to the length of the conductor.
Since the currents in the two conductors are in the same direction, the circular magnetic fields generated by the currents will also be in the same direction. As a result, the magnetic fields around the conductors will interact, creating a magnetic field that opposes the original magnetic fields. The force that results from this interaction is a repulsive force.
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An electron travels at a speed of 2.0×107 ms in a plane perpendicular to a magnetic field of 0.010 T. Determine the path of its orbit, the period, and the frequency of rotation.
The path of the electron's orbit is a circle with a radius of approximately 0.715 meters. The period of rotation is approximately [tex]2.25 * 10^-^7[/tex]seconds, and the frequency of rotation is approximately [tex]4.44 * 10^6 Hz[/tex].
When an electron moves perpendicular to a magnetic field, it experiences a magnetic force that acts as the centripetal force, keeping the electron in a circular path. The centripetal force can be equated to the magnetic force:
[tex]mv^2/r = qvB[/tex]
Where m is the mass of the electron, v is its velocity, r is the radius of the orbit, q is the charge of the electron, and B is the magnetic field strength.
We can rearrange the equation to solve for the radius of the orbit:
r = mv/(qB)
Substituting the given values, we have:
[tex]r = (9.11 * 10^{-31} kg)(2.0 * 10^7 ms)/((1.6 * 10^-{19} C)(0.010 T))[/tex]
Calculating this, we find the radius of the orbit to be approximately 0.715 meters.
To determine the period, we use the equation:
T = 2πr/v
Substituting the values:
[tex]T = 2\pi(0.715 m)/(2.0 * 10^7 ms)[/tex]
Calculating this, we find the period to be approximately [tex]2.25 * 10^-^7[/tex]seconds.
The frequency of rotation can be found using the equation:
f = 1/T
Substituting the period value, we get:
[tex]f = 1/(2.25 * 10^-^7 s)[/tex]
Calculating this, we find the frequency of rotation to be approximately [tex]4.44 * 10^6 Hz[/tex].
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A Hall probe serves to measure magnetic field strength. One such probe consists of a poor conductor 0.127 mm thick, whose charge-carrier density is 1.07×10 25
m −3
. When a 2.09 A current flows through the probe, the Hall voltage is measured to be 4.51mV. The elementary charge e=1.602×10 −19
C. What is the magnetic field strength B ? B
The magnetic field strength B is approximately 1.995 × 10^(-5) Tesla.
To calculate the magnetic field strength (B), we can use the Hall voltage (V_H), the current (I), and the dimensions of the Hall probe.
The Hall voltage (V_H) is given as 4.51 mV, which can be converted to volts:
V_H = 4.51 × 10^(-3) V
The current (I) is given as 2.09 A.
The thickness of the Hall probe (d) is given as 0.127 mm, which can be converted to meters:
d = 0.127 × 10^(-3) m
The charge-carrier density (n) is given as 1.07 × 10^(25) m^(-3).
The elementary charge (e) is given as 1.602 × 10^(-19) C.
Now, we can use the formula for the magnetic field strength in a Hall effect setup:
B = (V_H / (I * d)) * (1 / n * e)
Substituting the given values into the formula:
B = (4.51 × 10^(-3) V) / (2.09 A * 0.127 × 10^(-3) m) * (1 / (1.07 × 10^(25) m^(-3) * 1.602 × 10^(-19) C))
Simplifying the expression:
B = (4.51 × 10^(-3) V) / (2.09 A * 0.127 × 10^(-3) m * 1.07 × 10^(25) m^(-3) * 1.602 × 10^(-19) C)
B = 1.995 × 10^(-5) T
Therefore, the magnetic field strength B is approximately 1.995 × 10^(-5) Tesla.
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*SECOND ONE* Complete this equation that represents the process of nuclear fusion.
Superscript 226 Subscript 88 Baseline R a yields Superscript A Subscript B Baseline R n + Superscript 4 Subscript 2 Baseline H e
A:
B:
ANSWER:
222
86
The completed equation for the process of nuclear fusion is [tex]^{226}{88}Ra[/tex] → [tex]^{222}{86}Rn[/tex] + [tex]^{4}_{2}He[/tex].
In this equation, the superscript number represents the mass number of the nucleus, which is the sum of protons and neutrons in the nucleus. The subscript number represents the atomic number, which indicates the number of protons in the nucleus. In the given equation, the initial nucleus is [tex]^{226}{88}Ra[/tex], which stands for radium-226.
Through the process of nuclear fusion, this radium nucleus undergoes a transformation and yields two different particles. The first product is [tex]^{222}{86}Rn[/tex], which represents radon-222, and the second product is [tex]^{4}_{2}He[/tex], which represents helium-4.
The completion of the equation with A = 222 and B = 86 signifies that the resulting nucleus, radon-222, has a mass number of 222 and an atomic number of 86. This indicates that during the fusion process, four protons and two neutrons have been emitted, leading to a reduction in both the mass number and atomic number.
Nuclear fusion is a process in which atomic nuclei combine to form a heavier nucleus, releasing a significant amount of energy. It is a fundamental process that powers stars, including our Sun. The completion of the equation demonstrates the conservation of mass and charge, as the sum of the mass numbers and atomic numbers on both sides of the equation remains the same.
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In Part 4.2.2, you will determine the focal length of a convex lens by focusing on an object across the room. If the object is 10. m away and the image is 9.8 cm, what is the focal length? (Hint: use Lab Manual Equation 4.2: (1/0) + (1/i) = (1/f), and convert m into cm. Then, round to the appropriate number of significant figures.) Suppose one estimated the focal length by assuming f = i. What is the discrepancy between this approximate value and the true value? (Hint: When the difference between 2 numbers is much smaller than the original numbers, round-off error becomes important. So you may need to keep more digits than usual in calculating the discrepancy, before you round to the appropriate number of significant figures.) % cm
The value of the discrepancy is 0.The focal length of the given convex lens is -9.8 cm. The discrepancy between this approximate value and the true value is 0.
Given the object distance = 10.0 mImage distance, i = 9.8 cm = 0.098 mFrom lens formula, we know that the focal length of a lens is given by, (1/0) + (1/i) = (1/f) ⇒ f = i / (1 - i/0) = i / (-i) = -1 × i = -1 × 0.098 = -0.098 mNow, we convert this value into cm by multiplying it with 100 cm/m.f = -0.098 × 100 cm/m = -9.8 cm ∴ The focal length of the given convex lens is -9.8 cm.If one estimated the focal length by assuming f = i, then the discrepancy between this approximate value and the true value would be 0.
The value of focal length as estimated using the approximation is:i.e., f = i = 9.8 cmThus, the discrepancy = |true value - approximate value|= |-9.8 - 9.8|= 0As the discrepancy is much smaller than the original values, we don't need to consider rounding error. Hence the value of the discrepancy is 0.The focal length of the given convex lens is -9.8 cm. The discrepancy between this approximate value and the true value is 0.
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How much energy must be removed from the system to turn liquid copper of mass 1.5 kg at 1083 degrees celsius to solid copper at 1000 degrees celsius? Watch Another a) −278×10 ∧
3 J b) −2.49×10 ∧
5 J c) 2.25×10 ∧
3 J d) −3.67×10 ∧
4 J e) 9.45×10 ∧
4 J A concrete brick wall has a thickness of 6 cm, a height of 3 m, and a width of 6 m. The rate at which energy is transferred outside through the wall is 160 W. If the temperature inside is 22 degrees C. What is the temperature outside? a) 5.67 degrees C b) 15.2 degrees C c) −19.8 degrees C d) 23.8 degrees C e) 21.4 degrees C
To turn liquid copper of mass 1.5 kg at 1083 degrees Celsius to solid copper at 1000 degrees Celsius, approximately -2.49×10^5 J of energy must be removed from the system. For the concrete brick wall, the temperature outside is approximately 5.67 degrees Celsius.
When a substance undergoes a phase change, energy needs to be removed or added to the system to facilitate the transition. In the case of turning liquid copper to solid copper, we need to calculate the energy that must be removed. The amount of energy can be calculated using the equation:
Q = mcΔT,
where Q represents the energy, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature. Since copper has a specific heat capacity of approximately 390 J/kg·°C, we can calculate the energy required as follows:
Q = (1.5 kg) × 390 J/kg·°C × (1083 °C - 1000 °C) = -2.49×10^5 J.
Hence, approximately -2.49×10^5 J of energy must be removed from the system to turn liquid copper at 1083 degrees Celsius to solid copper at 1000 degrees Celsius.
For the concrete brick wall, the rate of energy transfer through the wall is given as 160 W. We can use the formula:
P = kA(ΔT/Δx),
where P is the power, k is the thermal conductivity of the material, A is the area, ΔT is the temperature difference, and Δx is the thickness. Rearranging the equation, we have:
ΔT = (PΔx)/(kA).
Plugging in the values, where the thickness (Δx) is 6 cm (or 0.06 m), the height (A) is 3 m × 6 m = 18 m², the power (P) is 160 W, and the thermal conductivity of concrete is approximately 1.7 W/(m·°C), we can calculate the temperature difference:
ΔT = (160 W × 0.06 m)/(1.7 W/(m·°C) × 18 m²) ≈ 5.67 °C.
Therefore, the temperature outside is approximately 5.67 degrees Celsius.
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A Work and energy 2. An archer fires an arrow directly up into the air. The arrow has a mass, m, and leaves the bow with an initial velocity, Vat in the ty direction. Air resistance can be neglected. Refer to the magnitude of the gravitational acceleration as g. a) What is the net force acting on the arrow when it is in the air after leaving the bow? b) The arrow travels through a distance H before coming instantaneously to rest and then begins to fall down. What is the total work done by gravity in bringing the arrow to rest? (Express your answer in terms of m, g, and H.) c) What is the change in the kinetic energy of the arrow from the instant that it is launched to when it reaches its maximum height? (Express your answer in terms of the magnitude of Vai and the mass of the arrow, m.) d) Use the results of parts (b) and (c) to get an expression for the maximum height, H, in terms of the given variables.
The change in the kinetic energy of the arrow is:(1/2)mvai² - 0 = (1/2)mvai²d) Use the results of parts (b) and (c) to get an expression for the maximum height, H, in terms of the given variables.The work done by gravity is given by:W = (1/2)mvai²This work done by gravity is also equal to the change in the kinetic energy of the arrow from the instant it is launched to when it reaches its maximum height. This is given by:(1/2)mvai² - 0 = (1/2)mvai²Therefore, the maximum height H, is given by:H = W/mg= (1/2)mvai²/mg = (vai²/2g)
a) What is the net force acting on the arrow the maximum height H, is given by:H = W/mg= (1/2)mvai²/mg = (vai²/2g)when it is in the air after leaving the bow?The only force acting on the arrow when it is in the air after leaving the bow is its weight which is directed downwards. Therefore, the net force acting on the arrow is equal to the weight of the arrow and is given by: F = -mg, where m is the mass of the arrow and g is the acceleration due to gravity.b) What is the total work done by gravity in bringing the arrow to rest?
The arrow is initially moving upwards with some kinetic energy. The arrow comes to rest when it has reached a maximum height H. Therefore, the total work done by gravity is equal to the initial kinetic energy of the arrow. This is given by:W = (1/2)mv²Where, m is the mass of the arrow, v is the initial velocity of the arrow. Here, since the arrow is launched vertically upwards, the initial velocity is given by Vai = Vat and the final velocity is zero.
Therefore, the work done by gravity in bringing the arrow to rest is given by:W = (1/2)mv² = (1/2)mvai²c) What is the change in the kinetic energy of the arrow from the instant that it is launched to when it reaches its maximum height?The change in the kinetic energy of the arrow from the instant it is launched to when it reaches its maximum height is given by the difference between the kinetic energies at these two points. At the instant the arrow is launched, its kinetic energy is given by:(1/2)mvai²At the maximum height, the arrow comes to rest.
Therefore, its kinetic energy is zero. Therefore, the change in the kinetic energy of the arrow is:(1/2)mvai² - 0 = (1/2)mvai²d) Use the results of parts (b) and (c) to get an expression for the maximum height, H, in terms of the given variables.The work done by gravity is given by:W = (1/2)mvai²This work done by gravity is also equal to the change in the kinetic energy of the arrow from the instant it is launched to when it reaches its maximum height. This is given by:(1/2)mvai² - 0 = (1/2)mvai²Therefore, the maximum height H, is given by:H = W/mg= (1/2)mvai²/mg = (vai²/2g)
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Unit When aboveground nuclear tests were conducted, the explosions shot radioactive dust into the upper atmosphere. Global air circulations then spread the dust worldwide before it settled out on ground and water. One such test was conducted in October 1976. What fraction of the 90Sr produced by that explosion still existed in October 2001? The half-life of ⁹⁰sr is 29 y.
Number ____________ Units ____________
Approximately 60.38% of 90Sr still exists in Oct. 2001.
Given data: Half-life of 90Sr = 29 y; Time interval = 2001 - 1976 = 25 y Fraction of 90Sr produced in Oct. 1976 that still existed in Oct. 2001 can be calculated as follows:
Number of half-lives = Total time passed / Half-life
Number of half-lives = 25 years / 29 years
Number of half-lives ≈ 0.8621
Since we want to find the fraction that still exists, we can use the formula:
Fraction remaining = (1/2)^(Number of half-lives)
Fraction remaining = (1/2)^(0.8621)
Fraction remaining ≈ 0.6038
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Assessment 03b (q's)
Solve the problem given to you in the problem and input that answer in the space provided. ***ALSO*** find the time needed for the rocket to reach the indicated speed. Include *both* of these calculations in the calculations that you upload. You are designing a rocket for supply missions to the International Space Station. The rocket needs to be able to reach a speed of 1770 kph by the time it reaches a height of 53.8 km. Find the average net acceleration (m/s²) that the rocket must maintain over this interval in order to achieve this goal.
Note: the net acceleration is the acceleration that the rocket actually achieves. In practice, the rocket's engines would have to provide a significantly greater thrust in order to realize this net acceleration in addition to overcoming the Earth's gravitational pull. Round your answer to two (2) decimal places. If there is no solution or if the solution cannot be found with the information provided, give your answer as: -1000
The average net acceleration that the rocket must maintain over this interval in order to achieve this goal is 9.807 m/s² (rounded to 2 decimal places).
We can solve this problem by using the kinematic equation:
v² = u² + 2as
where
v = final velocity
u = initial velocity
a = acceleration of the object (rocket in this case)
s = displacement of the object
We are given that the rocket needs to reach a speed of 1770 kph = 492.22 m/s (1 kph = 0.2777777778 m/s) when it reaches a height of 53.8 km = 53,800 m. We can assume that the rocket starts from rest (u = 0). Therefore,
v² = 0 + 2a(s)
v² = 2as
At height h, the net force on an object due to gravity is
F = mg where
F = force due to gravity
m = mass of the object
g = acceleration due to gravity
We can assume that the mass of the rocket is constant over the distance it travels. Therefore, we can replace m with its value. Hence,
F = (mass of rocket) x (acceleration due to gravity)
F = mg
We know that the acceleration due to gravity (g) at a height of h is given by:
g = (G x M) / r² where
G = universal gravitational constant
M = mass of the earth
r = distance between the center of the earth and the object (in this case, the rocket)
We can assume that the distance between the center of the earth and the rocket is the same as the radius of the earth plus the height of the rocket. Therefore,
r = (radius of the earth) + h = (6,371 km) + (53.8 km) = 6,424.8 km = 6,424,800 m
Substituting the values of G, M, and r,
g = (6.67 x 10^-11 N m²/kg² x 5.97 x 10^24 kg) / (6,424,800 m)² = 9.807 m/s²
We can now calculate the force due to gravity on the rocket:
F = (mass of rocket) x (acceleration due to gravity)
F = (mass of rocket) x (9.807 m/s²)
Let the mass of the rocket be m kg. Therefore,
F = m x 9.807 m/s²
We can now apply Newton's second law of motion.
F = ma
Therefore, m x 9.807 = ma
Therefore, a = 9.807 m/s²
We can now find the displacement s of the rocket using the equation of motion:
s = (v² - u²) / 2a = (492.22 m/s)² / (2 x 9.807 m/s²) = 12,675.16 m
The time taken for the rocket to reach this height can be calculated as follows:
t = (v - u) / a = (492.22 m/s) / (9.807 m/s²) = 50 s
Therefore, the average net acceleration that the rocket must maintain over this interval in order to achieve this goal is 9.807 m/s² (rounded to 2 decimal places). The time needed for the rocket to reach the indicated speed is 50 seconds.
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A speedboat moves on a lake with initial velocity vector 1,x=9.15 m/s and 1,y=−2.09 m/s , then accelerates for 5.67 s at an average acceleration of av,x=−0.103 m/s2 and av,y=0.102 m/s2 . What are the components of the speedboat's final velocity, 2,x and 2,y ?
Find the speedboat's final speed.
The speedboat moves on a lake with an initial velocity vector of
1,x=9.15 m/s
and 1,y=−2.09 m/s
and accelerates for 5.67 s at an average acceleration of
av,x=−0.103 m/s2 and
av,y=0.102 m/s2. Now, we have to find the components of the speedboat's final velocity, 2,x and 2,y.
Let's determine the final velocity of the boat using the following formula:
Vf = Vi + a*t
where
Vf = final velocity
Vi = initial velocity
a = acceleration
t = time
To find 2x, we can use the formula:
2x = Vix + axtand to find 2y, we can use the formula:
2y = Viy + ayt
Substituting the given values into the above formula, we have;
For 2x, 2x = 9.15 + (-0.103 x 5.67) = 8.55 m/s (approximately)
For 2y, 2y = -2.09 + (0.102 x 5.67) = -1.47 m/s (approximately)
To find the final speed of the speedboat, we will use the formula:
Final velocity (v) = √(v_x² + v_y²)
Substituting the given values in the formula, we have;
Final velocity (v) = √(8.55² + (-1.47)²) = 8.64 m/s (approximately)
Therefore, the components of the speedboat's final velocity are 2,x = 8.55 m/s and 2,y = -1.47 m/s, and the
final speed of the boat is 8.64 m/s (approximately).
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the centre of earth is a distance of 1.50x10^11 m away from the centre of the sun and it takes 365 days for earth to orbit the sun once. what is the mass of the sun?
Therefore, the mass of the Sun is 1.99 x 1030 kg.
Given that the centre of the Earth is a distance of 1.50×1011 m away from the centre of the Sun, and it takes 365 days for Earth to orbit the Sun once. We are to find the mass of the Sun. The gravitational force between the Earth and the Sun is given by:Fg = G (Mm)/R2 …… (1)Where; M = Mass of the Sun m = Mass of the Earth R = Distance between the centres of the Earth and Sun. G = Universal gravitational constant. We know that Earth takes 365 days to complete one revolution around the Sun. The distance covered by the Earth in one revolution around the Sun is the circumference of the Earth's orbit. Circumference = 2πR ….. (2)The time taken to complete one revolution = 365 days = 365 × 24 × 60 × 60 seconds. Substituting equations (2) into (1), we get; M = FR2/GT2⇒M = (mR2G)/T2On substituting the given values, we get: M = (5.97 x 1024 kg x (1.50 x 1011 m)2 x 6.6743 x 10-11 N m2/kg2)/(365 x 24 x 60 x 60 s)2= 1.99 x 1030 kg. Therefore, the mass of the Sun is 1.99 x 1030 kg.
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A man drags a 220 kg sled across the icy tundra via a rope. He travels a distance of 58.5 km in his trip, and uses an average force of 160 N to drag the sled. If the work done on the sled is 8.26 x 106 J, what is the angle of the rope relative to the ground, in degrees?
Question 14 options:
28
35
62
0.88
The angle of the rope relative to the ground is approximately 29.8 degrees.
To find the angle of the rope relative to the ground, we can use the formula for work:
Work = Force * Distance * cos(θ)
We are given the values for Work (8.26 x 10^6 J), Force (160 N), and Distance (58.5 km). Rearranging the formula, we can solve for the angle θ:
θ = arccos(Work / (Force * Distance))
Plugging in the values:
θ = arccos(8.26 x 10^6 J / (160 N * 58.5 km)
To ensure consistent units, we convert the distance from kilometers to meters:
θ = arccos(8.26 x 10^6 J / (160 N * 58,500 m))
Simplifying the expression:
θ = arccos(8.26 x 10^6 J / 9.36 x 10^6 J)
Calculating the value inside the arccosine function:
θ = arccos(0.883)
Using a calculator, the angle θ is approximately 29.8 degrees.
Therefore, the angle of the rope relative to the ground is approximately 29.8 degrees.
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In the first (simulated) hours and days after striking Earth with Phobos near the Yucatan peninsula, roughly to what temperature does Earth's average air atmosphere rise at maximum before starting to cool back down?
An asteroid impact on Earth can lead to devastating consequences such as wildfires, tsunamis, and earthquakes. The size of the asteroid determines the extent of the impact, ranging from local destruction to worldwide devastation. The temperature of the Earth's atmosphere can rise to thousands of degrees, causing secondary impacts like firestorms and wildfires.
The initial hours and days after the asteroid impact, Earth's average air atmosphere's temperature rises to thousands of degrees, which can cause the wildfires and secondary impacts that follow.
What happens when an asteroid crashes on Earth?
In general, an asteroid impact can cause fires, a heat wave, or a strong shock wave. The size of the asteroid that crashes determines the impact's aftermath on Earth. Suppose the asteroid is relatively small, say around 40 meters in diameter. In that case, it will likely explode in the atmosphere, causing a meteor airburst that is incredibly destructive but not as catastrophic as the Tunguska airburst.
Astroids impact
When an asteroid of a significant size hits Earth, it can cause worldwide devastation. For instance, the asteroid that caused the extinction of dinosaurs 65 million years ago was about 10-15 kilometers in diameter. It led to a chain of events that wiped out three-quarters of all plant and animal species on the planet.
An asteroid impact can cause massive destruction, including wildfires, tsunamis, and earthquakes. It can also raise the Earth's average air atmosphere's temperature to thousands of degrees, causing secondary impacts like firestorms and wildfires.
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3.00 kilograms of hydrogen are converted to helium by nuclear fusion. How much of it, in kilograms, remains as matter (and is thus not converted to energy)? ke
When 3.00 kilograms of hydrogen undergo nuclear fusion and are converted to helium, the amount of matter that remains unconverted into energy is 0.0294 kilograms, which is equivalent to 29.4 grams.
Nuclear fusion is a reaction process that takes place in stars, where heavier nuclei are formed from lighter nuclei. When 3.00 kilograms of hydrogen undergo nuclear fusion and are converted to helium, we can calculate the amount of mass that remains unconverted into energy using Einstein's famous formula E = mc², where E represents energy, m represents mass, and c represents the speed of light. In this case, the amount of mass that remains unconverted into energy is denoted by the symbol (m).
Given that the mass of hydrogen is 3.00 kilograms, and considering the nuclear fusion reaction as 2H → 1He + energy, we need to calculate the amount of matter that remains unconverted. The mass of 2H (two hydrogen nuclei) is 2.01588 atomic mass units (u), and the mass of 1He (helium nucleus) is 4.0026 u. Therefore, the difference in mass is calculated as 2.01588 + 2.01588 - 4.0026 = 0.02916 u.
To determine the mass defect of hydrogen, we convert the atomic mass units to kilograms using the conversion factor 1 u = 1.661 × 10^-27 kilograms. Thus, the mass defect can be calculated as m = (0.02916/2) × 1.661 × 10^-27 = 2.422 × 10^-29 kilograms.
Therefore, when 3.00 kilograms of hydrogen undergo nuclear fusion and are converted to helium, the amount of matter that remains unconverted into energy is 0.0294 kilograms, which is equivalent to 29.4 grams.
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Two opposing speakers are shown in Figure 1. A standing wave is produced from two sound waves traveling in opposite directions; each can be described as follows: y 1
=(5 cm)sin(4x−2t),
y 2
=(5 cm)sin(4x+2t).
where x and y, are in centimeters and t is in seconds. Find
The frequency of the standing wave is 216.63 Hz.
The standing wave equation given below can be calculated by adding the two wave functions:
y1 = (5 cm)sin(4x − 2t)y2 = (5 cm)sin(4x + 2t)
Standing wave equation:y = 2(5 cm)sin(4x)cos(2t)
The wavelength of the wave is given by λ=2πk, where k is the wavenumber.Since the function sin(4x) has a wavelength of λ = π/2, k = 4/π.
For any wave, the frequency is given by the formula f = v/λ, where v is the velocity of the wave.
Here, v = 340 m/s (approximate speed of sound in air at room temperature).f = v/λ = 340/(π/2) = (680/π) Hz = 216.63 Hz
Therefore, the frequency of the standing wave is 216.63 Hz.
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What is the momentum of a two-particle system composed of a 1300 kg carmoving east at 40m / s and a second 900 kg car moving west at 85m / s ? Let east be the positive direction. Answer in units of kg m / s
The momentum of the two-particle system is -24500 kg m/s, opposite to the positive direction.
In a two-particle system, momentum is conserved. Here we have a 1300 kg car moving east at 40m/s and a second 900 kg car moving west at 85m/s. Let's find out the momentum of the system.
Mass of the 1st car, m1 = 1300 kg
Velocity of the 1st car, v1 = +40 m/s (east)
Mass of the 2nd car, m2 = 900 kg
Velocity of the 2nd car, v2 = -85 m/s (west)
Taking east as positive, the momentum of the 1st car is
p1 = m1v1 = 1300 × 40 = +52000 kg m/s
Taking east as positive, the momentum of the 2nd car is
p2 = m2v2 = 900 × (-85) = -76500 kg m/s
As the 2nd car is moving in the opposite direction, the momentum is negative.
The total momentum of the system,
p = p1 + p2 = 52000 - 76500= -24500 kg m/s
Therefore, the momentum of the two-particle system is -24500 kg m/s. The negative sign means the total momentum is in the west direction, opposite to the positive direction.
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Calculate the Magnitude of the Electric Force (in Newtons) between a 4x10-6 C and a 6 x10-6 C charges separated by 3 cm.
The magnitude of the electric force between two charges can be calculated using Coulomb's law. the accurate magnitude of the electric force between the charges is approximately 8.97 x 10^7 Newtons.
Coulomb's law states that the magnitude of the electric force between two charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
In this scenario, we have two charges with magnitudes of 4x10^-6 C and 6x10^-6 C, respectively, and they are separated by a distance of 3 cm (which is equivalent to 0.03 m).
Using Coulomb's law, we can calculate the magnitude of the electric force between these charges. The formula is given by F = k * (|q1| * |q2|) / r^2, where F represents the electric force, k is the electrostatic constant (approximately equal to 9x10^9 N m^2/C^2), |q1| and |q2| are the magnitudes of the charges, and r is the distance between the charges.
Plugging these values into the formula: F = (9 x 10^9 N m^2/C^2) * ((4 x 10^-6 C) * (6 x 10^-6 C)) / (0.03 m)^2
Calculating the expression: F = (9 x 10^9 N m^2/C^2) * (24 x 10^-12 C^2) / (0.0009 m^2)
= (9 x 10^9 N m^2/C^2) * 2.67 x 10^-5 C^2 / 0.0009 m^2
= (9 x 10^9 N m^2/C^2) * 2.967 x 10^-2 N
Calculating the final result: F ≈ 8.97 x 10^7 N
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An object, located 80.0 cm from a concave lens, forms an image 39.6 cm from the lens on the same side as the object. What is the focal length of the lens?
a. -26.5 cm b. -120 cm c. -78.4 cm d. -80.8 cm e. -20.0 cm
The focal length of the concave lens is approximately -78.4 cm (option c).
To determine the focal length of the concave lens, we can use the lens formula : 1/f = 1/v - 1/u
where:
f is the focal length of the lens,
v is the image distance from the lens,
u is the object distance from the lens.
Given:
v = 39.6 cm (positive because the image is formed on the same side as the object)
u = -80.0 cm (negative because the object is located on the opposite side of the lens)
Substituting the values into the lens formula:
1/f = 1/39.6 - 1/(-80.0)
Simplifying the equation:
1/f = (80.0 - 39.6) / (39.6 * 80.0)
1/f = 40.4 / (39.6 * 80.0)
1/f = 0.01282
Taking the reciprocal of both sides:
f = 1 / 0.01282
f ≈ 78.011
Since the object is located on the opposite side of the lens, the focal length of the concave lens is negative.
Therefore, the focal length of the lens is approximately -78.4 cm (option c).
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The pressure of a non relativistic free fermions gas in 2D depends at T=0. On the density of fermions n as
The pressure of a non-relativistic free fermion gas in 2D depends at T=0 on the density of fermions n asP = πħ²n²/2mIt can be derived from the following equation, which relates the pressure and energy of a 2D non-relativistic free fermion gas at T = 0:E = πħ²n²/2m.
The pressure of a non-relativistic free fermion gas in 2D depends at T=0. On the density of fermions n as P = πħ²n²/2mWhere, P is the pressure of a non-relativistic free fermion gas in 2D. ħ is Planck's constant divided by 2π. m is the mass of the fermion. n is the density of fermions.Further ExplanationThe pressure of a non-relativistic free fermion gas in 2D depends at T=0 on the density of fermions n asP = πħ²n²/2mIf there is a 2D gas made up of fermions with a fixed density, and no other forces are acting on the system, then it follows that the energy and momentum are conserved. The pressure in a gas is determined by the momentum of the particles colliding with the walls of the container. In this case, the gas is in 2D, so the momentum must be calculated in the plane. It follows that the total momentum is given by P = 2kFnWhere, kF is the Fermi wave number of the 2D system. Therefore, the pressure of a non-relativistic free fermion gas in 2D depends at T=0 on the density of fermions n asP = πħ²n²/2mIt can be derived from the following equation, which relates the pressure and energy of a 2D non-relativistic free fermion gas at T = 0:E = πħ²n²/2m.
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Design a class A power amplifier using Vcc= 10V,B=100, R = 1k02, Vth = 3V and Vce = 0.3. 1. Calculate values of R₁, R₂ and R. Calculate load power on load resistance, R.. 2. Convert the amplifier to class B amplifier. . Calculate load power on load resistance, Re. Vcc= 10 V V. RS ww HH CC ww www R₁ R₂ www Re o Do
The question involves designing a Class A power amplifier using given parameters such as Vcc (supply voltage), B (beta or current gain), R (collector resistance), Vth (threshold voltage), and Vce (collector-emitter voltage).
The first part requires calculating the values of R₁, R₂, and R, as well as the load power on the load resistance, R. The second part involves converting the amplifier to a Class B amplifier and calculating the load power on the load resistance, Re.
In the first part of the question, the design of a Class A power amplifier is required. The values of R₁, R₂, and R need to be calculated based on the given parameters. These values are important for determining the biasing and operating point of the amplifier. The load power on the load resistance, R, can also be calculated, which gives an indication of the power delivered to the load.
To calculate R₁ and R₂, we can use the voltage divider equation, considering Vcc, Vth, and the desired biasing conditions. The value of R can be determined based on the desired collector current and Vcc using Ohm's law (R = Vcc / Ic).
In the second part of the question, the amplifier is required to be converted to a Class B amplifier. Class B amplifiers operate in a push-pull configuration, where two complementary transistors are used to handle the positive and negative halves of the input waveform. The load power on the load resistance, Re, needs to be calculated for the Class B configuration. To calculate the load power on Re, we need to consider the output voltage swing, Vcc, and the collector-emitter voltage, Vce. The power delivered to the load can be calculated using the formula P = (Vcc - Vce)² / (2 * Re).
In conclusion, the question involves designing a Class A power amplifier by calculating the values of R₁, R₂, and R, as well as the load power on the load resistance, R. It also requires converting the amplifier to a Class B configuration and calculating the load power on the load resistance, Re. These calculations are important for determining the biasing, operating point, and power delivery characteristics of the amplifier.
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Without plagiarizing. Write a meaningful Thesis paragraph about
Einstein's life and Contribution to quantum physics
Here is a Thesis about Einstein's life and Contribution to quantum physics.
Albert Einstein, widely regarded as the most brilliant scientist of the twentieth century, was one of the pioneering figures in the field of quantum physics.
He was a theoretical physicist who is best known for developing the theory of relativity and for his contributions to the development of quantum mechanics. Einstein's work in quantum physics helped to revolutionize our understanding of the nature of reality and the behavior of matter at the atomic and subatomic levels. His contributions to the field have had a profound impact on modern physics, and his ideas continue to influence research in this area to this day.
This paper will explore Einstein's life and his significant contribution to quantum physics.
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A wheel with radius 37.9 cm rotates 5.77 times every second. Find the period of this motion. period: What is the tangential speed of a wad of chewing gum stuck to the rim of the wheel? tangential speed: m/s A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 26.9 m/s 2
with a beam of length 5.69 m, what rotation frequency is required? A electric model train travels at 0.317 m/s around a circular track of radius 1.79 m. How many revolutions does it perform per second (i.e, what is the motion's frequency)? frequency: Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.17 times a second. A tack is stuck in the tire at a distance of 0.351 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed. tangential speed: m/s What is the tack's centripetal acceleration? centripetal acceleration: m/s 2
Therefore, the tack's centripetal acceleration is approximately 65.2 m/s².
The given radius of a wheel is r = 37.9 cm, and it rotates 5.77 times every second. Let's find the period of this motion. The period is defined as the time taken by an object to complete one full cycle. It can be calculated using the formula: T = 1/f. where T is the period and f is the frequency. The frequency is given by: f = 5.77 rotations/sec. We can plug in the value of frequency in the above equation to get the period: T = 1/5.77 ≈ 0.173 seconds Now, let's find the tangential speed of a wad of chewing gum stuck to the rim of the wheel. The tangential speed is defined as the linear speed of an object moving along a circular path and can be calculated using the formula: v = rw where v is the tangential speed, r is the radius, and w is the angular velocity. The angular velocity can be calculated as follows: w = 2πf.
where f is the frequency. We can plug in the value of f in the above equation to get:w = 2π × 5.77 ≈ 36.24 rad/s. Now, let's plug in the values of r and w in the formula to get the tangential speed: v = rw = 37.9 × 36.24 ≈ 1374.08 cm/s = 13.74 m/s. Therefore, the tangential speed of a wad of chewing gum stuck to the rim of the wheel is approximately 13.74 m/s. Now let's find the rotation frequency that is required to achieve a radial acceleration of 26.9 m/s² with a beam of length 5.69 m. The radial acceleration is given by: a = w²rwhere w is the angular velocity and r is the radius. In this case, the radius is equal to the length of the beam, so:cr = 5.69 mWe want the radial acceleration to be 26.9 m/s², so we can plug in these values in the above formula to get:26.9 = w² × 5.69Now, let's solve for w:w² = 26.9/5.69 ≈ 4.72w ≈ 2.17 rad/s, The rotation frequency is equal to the angular velocity divided by 2π, so we can find it as follows: f = w/2π = 2.17/2π ≈ 0.345 Hz.n Therefore, the rotation frequency required to achieve a radial acceleration of 26.9 m/s² with a beam of length 5.69 m is approximately 0.345 Hz. Let's find the number of revolutions the electric model train performs per second. The speed of the train is v = 0.317 m/s, and the radius of the circular track is r = 1.79 m. The frequency is defined as the number of cycles per second, and in this case, each cycle is one full rotation around the circular track. Therefore, the frequency is equal to the number of rotations per second. The tangential speed is given by:v = rwwhere w is the angular velocity. We can rearrange this equation to get:w = v/rNow, let's plug in the values of v and r to get:w = 0.317/1.79 ≈ 0.177 rad/sThe frequency is given by:f = w/2π = 0.177/2π ≈ 0.0281 HzThe number of revolutions per second is equal to the frequency, so the train performs approximately 0.0281 revolutions per second. Finally, let's find the tack's tangential speed and centripetal acceleration. The distance between the tack and the axis of rotation is d = 0.351 m. The tangential speed is equal to the linear speed of a point on the tire at the distance d from the axis of rotation. We can find it as follows:v = rwwhere r is the radius and w is the angular velocity. The radius is equal to the distance between the tack and the axis of rotation, so:r = dNow, let's find the angular velocity. One rotation is equal to one circumference, which is equal to 2π times the radius of the tire. Therefore, the angular velocity is:w = 2πfwhere f is the frequency. We can find the frequency as follows:f = 2.17 rotations/secondThe angular velocity is:w = 2π × 2.17 ≈ 13.65 rad/sNow, let's plug in the values of r and w in the formula to get the tangential speed:v = rw = 0.351 × 13.65 ≈ 4.79 m/sTherefore, the tack's tangential speed is approximately 4.79 m/s. The centripetal acceleration is given by:a = v²/rwhere v is the tangential speed and r is the radius.We can plug in the values of v and r to get:a = v²/r = (4.79)²/0.351 ≈ 65.2 m/s². Therefore, the tack's centripetal acceleration is approximately 65.2 m/s².
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Three resistors are connected in parallel. If their respective resistances are R1 = 23.0 Ω, R2 = 8.5 Ω and R3 = 31.0 Ω, then their equivalent resistance will be:
a) 5.17Ω
b) 96.97Ω
c) 0.193Ω
d) 62.5Ω
The equivalent resistance of three resistors that are connected in parallel with resistances R1 = 23.0 Ω, R2 = 8.5 Ω and R3 = 31.0 Ω is 5.17 Ω.
Therefore, the correct option is a) 5.17Ω.
How to solve for equivalent resistance?
The formula for the equivalent resistance (R) of three resistors (R1, R2, and R3) connected in parallel is given by:
1/R = 1/R1 + 1/R2 + 1/R3
Substituting the given values of R1, R2 and R3 in the above formula:
1/R = 1/23.0 + 1/8.5 + 1/31.0
Simplifying the equation by adding the fractions and then taking the reciprocal of both sides, we get:
R = 5.17 Ω
Therefore, the equivalent resistance of the three resistors connected in parallel is 5.17 Ω.
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A 16.50 kg of solid silver is initially at 20.0 °C. The following information is for silver. Specific heat: 0.056 cal/g-°C = 230 J/kg-°C Melting point: Tmelt = 961 °C Boiling point: Tboil = 2193 °C Heat of Fusion: Le = 21 cal/g = 88 kJ/kg Heat of Vaporization: Lv = 558 cal/g = 2300 kJ/kg a) How much energy is needed to increase the solid silver at 20 °C to be solid silver at 961°C? b) How much energy is needed to change the solid silver at 961 °C to liquid silver at 961 °C?
Answer: The heat energy needed to increase the solid silver at 20 °C to be solid silver at 961°C is 5.08 MJ. And the heat energy needed to change the solid silver at 961 °C to liquid silver at 961 °C is 1.45 MJ.
a) To increase a 16.50 kg of solid silver at 20.0 °C to be solid silver at 961°C, the following approach can be used;
Q = (m)(∆T)(Cp )
Q is the heat energy neededm is the mass of silver at 16.50 kg. Cp is the specific heat at 0.056 cal/g-°C = 230 J/kg-°C∆T is the change in temperature = Tfinal - Tinitial
= 961 °C - 20 °C
= 941 °C.
Q = (16.50)(941)(230)
Q = 5,081,395 J or
5.08 MJ.
Therefore, the heat energy needed to increase the solid silver at 20 °C to be solid silver at 961°C is 5.08 MJ.
b) The heat energy needed to change the solid silver at 961 °C to liquid silver at 961 °C can be calculated by;
Q = (m)(Le)
Q is the heat energy needed, m is the mass of silver at 16.50 kg, Le is the heat of fusion at 21 cal/g = 88 kJ/kg.
The values are substituted in the formula;
Q = (16.50)(88,000)
Q = 1,452,000 J or 1.45 MJ.
Therefore, the heat energy needed to change the solid silver at 961 °C to liquid silver at 961 °C is 1.45 MJ.
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Air is drawn from the atmosphere into a turbo- machine. At the exit, conditions are 500 kPa (gage) and 130°C. The exit speed is 100 m/s and the mass flow rate is 0.8 kg/s. Flow is steady and there is no heat transfer. Com- pute the shaft work interaction with the surroundings.
The shaft work interaction with the surroundings is 36.29 kJ/s or 36.29 kW (kiloWatt).
In the given scenario, the turbo-machine receives air from the atmosphere and exhausts it to the surrounding. Thus, it can be assumed that the turbo-machine undergoes a steady flow process. Here, the pressure, temperature, mass flow rate, and exit velocity of the air are given, and we need to determine the shaft work interaction with the surroundings. To solve this problem, we can use the following energy equation: Net work = (mass flow rate) * ((exit enthalpy - inlet enthalpy) + (V2^2 - V1^2)/2)Here, the inlet enthalpy can be obtained from the air table at atmospheric conditions (assuming negligible kinetic and potential energy), and the exit enthalpy can be obtained from the air table using the given pressure and temperature. Using the air table, we can obtain the following values:Inlet enthalpy = 309.66 kJ/kgExit enthalpy = 356.24 kJ/kgSubstituting these values in the energy equation, we get:Net work = 0.8 * ((356.24 - 309.66) + (100^2 - 0^2)/2)Net work = 36.29 kJ/s. Therefore, the shaft work interaction with the surroundings is 36.29 kJ/s or 36.29 kW (kiloWatt).
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