For an adiabatic process of gas PV(^r)= Constant, where r = Cp/Cv , Cp, and Cv are constants for a particular gas.
What is an ideal gas?It is an imaginary gas for which the volume occupies by it is negligible, this gas does not exist in a practical situation and the concept of an ideal gas is only the theoretical one.
Universal Gas Equation for the ideal gas,
PV=nRT
By differentiating on both sides: P.dV + V.dP = n.R.dT…. (1)
For the adiabatic process, the expansion or compression of gas occurs at a very fast rate which makes no heat transfer. Therefore, dQ= 0
A/c to First Law of TD: dQ=dU+ dW, implies dU= -dW=-P.dV ….. (2)
But dU= n.Cv. dT …. (3) Combining 2 and 3, we get: n.dT= -P.dV/Cv …..(3)
Using R= Cp-Cv, the gas law gives: from 1 and 3.
n.dT= (P.dV + V.dP)/(Cp-Cv) = -PdV/Cv
On rearranging terms we get, (dP/P)+((Cp/Cv)*(dV/V))=0
Cp and Cv are constants for a particular gas, hence integrating,
Ln(P)+ r.Ln(V)=constant (replacing Cp/Cv = r or Gamma)
Hence, P.V(^r)= constant is termed an adiabatic equation.
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Which is true of the force pair of Newton's third law?a.)The two forces are in the same direction.b.)The two forces never produce an acceleration. c.)The two forces act on different objects.
Given:
Force pair of Newton's third law.
Required:
To find the true statement for the force pair of Newton's third law.
Explanation:
According to Newton's third law
This can be understood by taking an example.
Suppose a person hits the wall with his hand, the person hitting the wall is the action.
This action results in pain in the person's hand which is the reaction.
Here, the hand exerted force on the wall and the wall exerted an equal force in opposite direction.
The conclusion can be drawn as
Action and reaction act on two different objects.
The forces act in opposite direction.
Final Answer: The two forces act on different objects.
16. On the moon, Bob weighs 160 N while on earth Fred weighs 882 N Who has thegreater mass?
We are given the weight of two people, one on the moon and the other on earth. To determine the mass we will use the following formula:
[tex]W=mg[/tex]Where:
[tex]\begin{gathered} W=\text{ weight,}\lbrack N\rbrack \\ m=\text{ mass,}\lbrack kg\rbrack \\ g=\text{ acceleration of gravity, }\lbrack\frac{m}{s^2}\rbrack \end{gathered}[/tex]Now, we solve for the mass "m" by dividing both sides by "g":
[tex]\frac{W}{g}=m[/tex]Now, for the case of the moon we have that the acceleration of gravity is:
[tex]g_{moon}=1.62\frac{m}{s^2}[/tex]Plugging in the values:
[tex]\frac{160N}{1.62\frac{m}{s^2}}=m[/tex]Solving the operations:
[tex]98.77kg=m[/tex]Now, for the case of the earth we have:
[tex]g_{\text{earth}}=9.8\frac{m}{s^2}[/tex]Plugging in the values:
[tex]\frac{882N}{9.8\frac{m}{s^2}}=m[/tex]Solving the operations:
[tex]90kg=m[/tex]Therefore, the greater mass is the mass of Bob.
3. A parallel-plate capacitor has a capacitance of 1.35 pF. If a 12.0 V battery is connected to this capacitor, how much electrical potential energy would it store?
Answer:
9.72 x 10^(-11)
Explanation:
The energy stored in a capacitor can be calculated as:
[tex]E=\frac{CV^2}{2}[/tex]Where C is the capacitance and V is the Voltage. So, replacing C by 1.35pF or 1.35 x 10^(-12) F, and V by 12.0 V, we get:
[tex]E=\frac{1.35\times10^{-12}F(12.0V)^2}{2}=9.72\times10^{-11}J[/tex]Therefore, the capacitot would store 9.72 x 10^(-11) J of energy.
The electrical potential energy that would be stored in the capacitor is 9.72 x 10^(-11) J. It is the energy which is present in the object when it is at rest.
What is Potential energy?Potential energy is the energy which is stored inside an object which is at rest.
For the parallel plate capacitors, capacitance is dependent upon its geometry, which is given by the formula:
C=ϵ⋅Ad or C = ϵ ⋅ A d ,
where, C is the value of the capacitance,
A is the area of each plate,
d is the distance between the plates, and
ϵ is the permittivity of the material between the plates of the parallel capacitor.
The energy stored in a capacitor can be calculated as:
E = CV²/2
where, C is the capacitance and V is the Voltage.
So, by replacing C with 1.35pF or 1.35 × 10⁻¹²F, and V by 12.0V, we get:
E = 1.35 × 10⁻¹² × (12)²/2
E = 9.72 × 10⁻¹¹ J
Therefore, the capacitor would store 9.72 × 10⁻¹¹ J of energy.
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What would happen if an unbalanced force was acting on an object? Would it move at a constant speed? Explain why or why not.
According to Newton's Second Law of Motion, if a net force acts on an object, the object accelerates acording to the equation:
[tex]\Sigma F=ma[/tex]Where m is the mass of the object.
Then, if an unbalanced force is acting on an object, the net force is different from 0, which means that the object will accelerate in the direction of the net force.
If an object has a uniform acceleration, it won't have a constant velocity, but it may have a constant speed anyway if the force is always perpendicular to the velocity, in that case the object would be moving in a circular trajectory.
In other words, an unbalanced force will cause an acceleration, this can either change the speed of an object or the direction of its velocity (or both).
Then, it might be possible for an object to have a constant speed under the influence of an unbalanced force if that object follows a circular trajectory.
An Earth satellite moves in circular orbit 602 km above the earths surface with a period of 96.53 min. What are the speed and the magnitude of the centripetal acceleration of the satellite.
The speed and the magnitude of the centripetal acceleration of the satellite are 7.568 km/s and 8.205 x 10^-3 m / s^2
Here this problem we are dealing with the speed and centripetal acceleration where the speed of the satellite is the speed required to attain adjustment between gravity's pull on the satellite and the inertia of the satellite's motion whereas centripetal acceleration is referred to as the acceleration of a body navigating a circular way.Since we are given the orbital radius which is 602 km and the period which is 96.53 min.
The radius of the satellite's orbit, r = earth radius + orbit radius
=>r =6378 km + 602 km = 6980 km
Since the formula for the orbital circumference is :
2πR
=>2 • 3.14 •6980 = 43,834.4 km
Now divide the circumference by the time period we get the orbital speed which is,
43,834.4 km / (96.53 min • 60) = 7.568 km/s
the formula we are referring for calculating the centripetal acceleration is:
a = v²/ r ,
=> a = (7.568 ^2) / 6980 = 8.205 x 10^-3 m / s^2
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Which measurements or observations are needed to calculate density
Answer:
mass and volume measurements are required to calculate density.
Provide an example of a situation that can be explained by each of Newton's Three Laws of Motion.
ANSWER and EXPLANATION
First, let us state Newton's three laws of motion.
The first law of motion states that:
The second law of motion states that:
The third law of motion states that:
Let us take an example of a ball that is stationary on a frictionless floor. As long as no force acts on the ball, it will continue to remain at rest but once a person kicks the ball, it begins to move and stay in motion (since the floor is frictionless, hence, no resistive force).
That is an application of the first law of motion.
The acceleration of this ball depends on the force with which it was kicked and the mass of the ball, as shown in the force equation:
[tex]F=ma[/tex]where m = mass, a = acceleration
That is an application of the second law of motion.
Assuming that this ball then comes in contact with a wall and stops moving. The force that the ball exerts on the wall is equal to the force that the wall exerts on the ball, but both forces act opposite one another.
That is an application of the third law of motion.
Please help me
how law of motion applied in basketball explain
Explanation:
According to the first law of motion, the basketball is always moving in one direction, unless acted on by another force.
A force of 540 N keeps a certain ideal spring stretched a distance of 0.300 m.
A) What is the potential energy of the spring when it is stretched 0.300 m?
B) What is its potential energy when it is compressed 6.00 cm?
A force of 540 N keeps a certain ideal spring stretched a distance of 0.300 m.
A) The potential energy of the spring when it is stretched 0.300 m will be 81 J
B) The potential energy when it is compressed 6.00 cm will be 3.24 J
Hooke's law, law of elasticity discovered by the English scientist Robert Hooke in 1660, which states that, for relatively small deformations of an object, the displacement or size of the deformation is directly proportional to the deforming force or load.
To find spring constant , using formula
k= F/x
where
F = Force
k = spring constant
x = displacement
k = 540 / .300 = 1800 N/m
A) Potential energy of the spring when it is stretched 0.300 m
U = 1/2 * k * [tex]x^{2}[/tex] = 1/2 * 1800 * [tex]0.300^{2}[/tex] = 81 J
B ) Potential energy of the spring when it is stretched 0.06 m
U = 1/2 * k * [tex]x^{2}[/tex] = 1/2 * 1800 * [tex]0.06^{2}[/tex] = 3.24 J
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d. Use trigonometry to resolve the following vectors into component
Describe the vectors using vector notation.
v ≈ (3.28512, 20.74146) (3.28512, 20.74146)
In trigonometry, what is a vector?
Vectors can be used to depict motion between points in the coordinate plane. A vector is just a line segment in a specified location with an arrow designating its length and direction. The vectors in the following
Detailed explanationv = 21 (cos (81°), sin (81°))
components for v (3.28512, 20.74146) (x, y)
Authors use different notations. You can write the vector as...
(r, θ) = (21, 81°)
r∠θ = 21∠81°
21 cis 81° = r cis
v = 21·e^(i·9π/20)
(x, y) ≈ (3.28512, 20.74146) (3.28512, 20.74146)
v = 3.28512i +20.74146j Maybe this is the vector notation you're looking for (i and j are unit vectors in the x- and y-directions, respectively)
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Jason is pulling a box across the roompulling with a force of 16 newtons and his arm is making a 68 anglet with the horizontal what is the horizontal component of the he is pulling with?
The horizontal component of a force is defined as follows
[tex]F_x=F\cdot\cos \theta[/tex]Where F = 16 N, and theta = 68. Let's use these values to find the horizontal component of F.
[tex]F_x=16N\cdot\cos 68\approx6N[/tex]Therefore, the horizontal component of the pulling force is 6 Newtons.thunder can make objects inside a room vibrate explain what causes the objects to vibrate
Thunder can make objects inside a room vibrate by which our house columns in underground that causes the objects to vibrate.
What causes the sound of thunder?Thunder is caused by the rapid expansion of air around the path of lightning. It only takes a few millionths of a second for lightning to burst through the air from a cloud to a nearby tree or roof. The loud thunder that follows a lightning strike is generally said to come from the lightning itself. However, the rumbling and roaring heard during thunderstorms is actually due to the rapid expansion of air around the lightning.
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A golf ball is hit horizontally at 40 m/s from the top of a hill that is 2.5 m high. If the terrain around the hill is nearly flat,approximately how far will the golf ball fly? Use - 9.81 m/s2 for the acceleration caused by gravity. Ignore air resistance.Round any intermediate calculations to no less than six decimal places, and round your final answer to two decimal places.
ANSWER
28.56 m
EXPLANATION
Let's make a diagram of this situation to understand it better,
The ball has an initial velocity of 40m/s, which is horizontal. This means that there is no vertical initial velocity.
We have to find the horizontal distance the golf ball flies. Since the horizontal velocity is constant - this is because there is no acceleration in that direction, the distance the ball travels is,
[tex]\Delta x=v_{ox}\cdot t[/tex]The horizontal initial velocity is given, but we have to find the time the ball was in the air. To find it, we use the vertical distance the ball travels - which we know is the height of the hill. In this case, we do have vertical acceleration - the acceleration of gravity, so the vertical distance the ball travels, as shown in the diagram, has a parabolic form and it is given by the equation,
[tex]\Delta y=v_{oy}\cdot t+\frac{1}{2}gt^2[/tex]The initial vertical velocity is zero because the ball is hit horizontally,
[tex]\Delta y=\frac{1}{2}gt^2[/tex]Solve for t. Multiply both sides by 2/g,
[tex]\begin{gathered} \Delta y\frac{2}{g}=\frac{1}{2}g\cdot\frac{2}{g}\cdot t^2 \\ \frac{2\Delta y}{g}=t^2 \end{gathered}[/tex]And take the square root to both sides of the equation,
[tex]t=\sqrt[]{\frac{2\Delta y}{g}}[/tex]Δy is the height of the hill, 2.5m, and g = 9.81m/s²,
[tex]t=\sqrt[]{\frac{2\cdot2.5m}{9.81m/s^2}}=\sqrt[]{\frac{5m}{9.81m/s^2}}=\sqrt[]{0.509684s^2}=0.713922s[/tex]This is the time the ball was in the air for. Now we can find the distance it traveled,
[tex]\Delta x=40m/s\cdot0.713922s=28.56m[/tex]The golf ball flew 28.56 m horizontally.
what does it mean to have a velocity of 10 m/s?
3. Find for the total resistance of the series-parallel circuit below
In order to find the total resistance, first let's calculate the total resistance between R2 and R3, which are in series, so the total resistance is the sum:
[tex]R23=R2+R3=3+2=5\text{ ohms}[/tex]Now, let's calculate the total resistance between R23 and R1, which are in parallel, so we use the equation below:
[tex]R123=\frac{R23\cdot R1}{R23+R1}=\frac{5\cdot6}{5+6}=\frac{30}{11}=2.727\text{ ohms}[/tex]Finally, we calculate the total resistance between R123 and R4, which are in series:
[tex]RT=R4+R123=3+2.727=5.727\text{ ohms}[/tex]Find the y-component of this vector.
80.0°
B-18.6 m
0 = 170°
By = [?] m
The y-component of vector B is 3.23 m.
What is y - component of a vector?The y-component of a vector is the vertical component of the vector. It the vector that points in the y-direction.
The y-component of a vector is measured along the y-axis of the Cartesian coordinate.
Mathematically, the y-component of a vector is given as;
By = B sinθ
where;
By is the vertical or y-component of vector BB is the magnitude of vector Bθ is the angle of inclination of vector BThe y-component of vector B is calculated as follows;
By = 18.6 m x sin(170)
By = 3.23 m
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if the mass of the paper is 0.003 kg, what force does the boxer except on it?
Answer:
1.15 N
Explanation:
You want to know the force exerted on a mass of 0.003 kg to accelerate it from 0 to 23 m/s in a period of 0.06 s.
AccelerationThe acceleration of the mass is the change in velocity divided by the change in time:
a = ∆v/∆t
a = ((23 -0) m/s)/(0.06 s) = 1150/3 m/s²
ForceThe force applied is the product of mass and acceleration:
F = ma
F = (0.003 kg)(1150/3 m/s²) = 1.15 kg·m/s² = 1.15 N
The applied force is 1.15 newtons.
A 2 kg mass hangs motionless from a spring with spring constant 8 N/m. Howfar will the spring stretch if the mass is motionless?
a 10 kg block of ice slides a ramp 20 m long. inclined at 10 degrees to the horizontal, if the ramp frictionless what is the acceleration of the ice
The acceleration of the ice is obtained as 0.97 m/s².
What is the acceleration?We have to recall that the acceleration has to do with the rate at which the velocity is change per unit time. Now, we know that from the parameters that have been given in the question that is before us here;
μmgcos(θ) = ma
μgcos(θ) = a
Hence;
μ = coefficient of friction
g = acceleration due to gravity
θ = the angle of inclination of the plane
We can now find the acceleration by the use of the formula as shown above hence we have;
a = (0.1)(9.8) cos(10)
a = 0.97 m/s²
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Calculate the wavelength of a light that has an energy of 9.55x10^-19J.
ANSWER
[tex]2.08\cdot10^{-7}m[/tex]EXPLANATION
To calculate the wavelength of light, we have to apply the formula for the energy of light:
[tex]E=\frac{hc}{\lambda}[/tex]h = Planck's constant
c = speed of light
λ = wavelength
Therefore, we have that:
[tex]\lambda=\frac{hc}{E}[/tex]The wavelength of the light is:
[tex]\begin{gathered} \lambda=\frac{6.626\cdot10^{-34}\cdot3\cdot10^8}{9.55\cdot10^{-19}} \\ \lambda=2.08\cdot10^{-7}m \end{gathered}[/tex]What in the final position of a robot that starts from (0, 0) m and makes displacements of (4.4) * m (3.3) * m and (- 4, - 2) m?
The final position of the robot is (3, 5) m
Explanation:The initial position of the robot = (0, 0)
The position of the robot after making a displacement of (4, 4)m = (0+4, 0+4)
The position of the robot after making a displacement of (4, 4)m = (4, 4) m
The new position of the robot after making a displacement of (3, 3)m
= (4+3, 4+3)m
The new position of the robot after making a displacement of (3, 3)m =(7, 7)m
The final position of the robot after making a displacement of (-4, -2)m
= (7-4, 7-2)
The final position of the robot after making a displacement of (-4, -2)m
= (3, 5)m
Therefore, the final position of the robot is (3, 5) m
Shiyon just received his first credit card in the mail. His mom reminded him that he needs to order a new pair of baseball cleats. In order to make sure his browser is secure for the transaction, he should
(A) use the same password for all accounts
(B) install Microsoft Excel
(C) enable pop-ups
(D) deactivate background scripts
Answer:
the answer is D
Explanation:
i got it right on the test
In order to make sure his browser is secure for the transaction, Shiyon should deactivate background scripts. Hence, option (D) is correct.
What is background scripts?Without the requirement to create a trigger or script like a business row, a background script provides a free-form method of instantly running server-side code. When you want to clean a modest amount of data that doesn't call for a lot of intricate adjustments, we recommend using a background script.
You could get around that if you need to update records that may be subject to several scripts running on them—as long as they are small or decent in size, not excessive. You'll discover a few techniques to work around that.
Your transaction may run for hours when you run a background script, but this one will end it after four hours.
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if the camera left his hands at 21.9 m/s and took 2.009s to hit the ground, how far down the cliff did it fall?
The camera falls down 25.835 meters under gravity.
While a frame falls, it happens because of gravity, and the force of air will increase with speed. This continues till the force of air equals the weight. Now the accelerates but falls at a constant speed known as the terminal speed.
A motion under gravity refers back to the motion of an item whose vertical movement is laid low with the presence of gravity. The pressure that attracts gadgets downwards is gravity. In fact, gravity works towards the center of the Earth.
S = ut - 1/2at²
= 21.9 × 2.009s - 1/2 × 9.8 × (2.009)²
= 43.997 - 18.162
= 25.835 meter.
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Which of the following statements is always true?
a.
Pressure always increases when force increases or the area acted on increases.
b.
Pressure always increases when force increases or the area acted on decreases.
c.
Pressure always increases when force decreases or the area acted on increases.
d.
Pressure always increases when force decreases or the area acted on decreases
The statement that is true is that pressure always increases when force increases or the area acted on decreases (option B).
What is pressure?Pressure is the amount of force that is applied over a given area divided by the size of this area.
The pressure applied to a surface can be calculated by dividing the force required by the area of the surface as follows:
P = F/A
The above shows that the pressure of a body is inversely proportional to the area but directly proportional to the force.
This means that as the pressure of an object increases, the force increases while the area acted on decreases.
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A truck with mass m has a brake failure while going down an icy mountain road of constant downward slope angle α (Figure 1). Initially the truck is moving downhill at speed v0. After careening downhill a distance L with negligible friction, the truck driver steers the runaway vehicle onto a runaway truck ramp of constant upward slope angle β. The truck ramp has a soft sand surface for which the coefficient of rolling friction is μr.
What is the distance that the truck moves up the ramp before coming to a halt? Solve using energy methods.
Express your answer in terms of m , α , v0 , L , g , β and μr .
((v² ÷ 2g) + LSinα) ÷ (Sinβ + µcosβ) is the distance that the truck moves up the ramp before coming to a halt.
Let the distance the truck moves up the ramp be by x.
The kinetic energy of the truck on an icy road is given by,
K1 = (1÷2)mv²
The potential energy of the truck on an icy road is given by,
U1 = mgLSinα
The kinetic energy of the truck on the tuck ramp is given by,
K2 = 0
The potential Energy of the truck-on-truck ramp is given by,
U2 = mgxSinβ
Work done is given by,
W(others) = -µ×mg×cosФ
Hence, by using the work-energy theorem,
W(others) = (K2 + U2)(K1 + U1)
Therefore, by putting the values we get,
((1÷2)mv² + mgLSinα)(0 + mgxSinβ) = -µ×mg×cosФ
x = (K1 + mgLSinα) ÷ (mg(Sinβ + µcosβ))
x = ((v² ÷ 2g) + LSinα) ÷ (Sinβ + µcosβ)
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A railroad locomotive is at rest with its whistle shrieking and then it starts moving toward you. Describe what happens to the frequency and wavelength of the sound that hits your ear when:A) It is approaching you.B) It pulls away from you.What happens to the amplitude in both of these scenarios?
a) Frequency increases , the time between waves decreases , wavelenghts decrease
b) Frequency decreases , wavelenght increases.
The doppler effect is a change in the frequency of sound waves that occurs when the source of the sound waves is moving relative to a stationary listener. AS the source approaches the listener , the sound waves get closer together, increasing their frequency and decreasing their wavelenghts.
Amplitude remains the same.
The rocket’s velocity just before it hits the ground is the same magnitude as the initial velocity. Use the appropriate kinematics equation to show that this is true.
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The rocket’s velocity just before it hits the ground is the same magnitude as the initial velocity because change in length in y direction will zero.
if u and v are the initial and final velocities of the rocket then from kinematic equations it may be written as
[tex]s=ut+\frac{1}{2} at^{2}\\0=ut+\frac{1}{2} at^{2}\\\frac{1}{2} at^{2} =-u t\\at=-2u[/tex]
since
v=u + at
v=u - 2u
v = - u
Hence the rocket hit the ground with same velocity as it projected.
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Set the cannon to have an initial speed of 15 m/s. For which situation do you think the cannon ball will be in the air for the longest time: if it is set at a 60-degree angle, or if it is set at a 70-degree angle?
Question 6 options:
60 degree
70 degree
The cannon ball will be in the air for the longest time if the the angle of projection is set at 70 degrees
How to determine which angle will result in longest time
Case 1:
Initial velocity (u) = 15 m/sAngle of projection (θ) = 60 ° Acceleration due to gravity (g) = 9.8 m/s²Time of flight (T) =?T = 2uSineθ / g
T = (2 × 15 × Sine 60) / 9.8
T = 25.98 / 9.8
T = 2.65 s
Case 2:
Initial velocity (u) = 15 m/sAngle of projection (θ) = 70 ° Acceleration due to gravity (g) = 9.8 m/s²Time of flight (T) =?T = 2uSineθ / g
T = (2 × 15 × Sine 70) / 9.8
T = 28.19 / 9.8
T = 2.88 s
From the above calculations, we can conclude that the ball will stay longer in air if the angle is 70 °
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A boat is moving across a 100 meter wide stream at 5 m/s. If the stream is flowing south at 10 m/s, how long will it take the boat to cross the river?
answer:
20 minutes for 5 m/s 10 minutes for 10 m/s
Wei drags a heavy piece of driftwood for 940 m along an irregular path. If Wei ends 710 m from where he starts, exerting a constant force of 625 N parallel to his path the entire time, how much work does he do?
Answer:
587500
Explanation:
Since Wei will still hit every point of the path even though it is irregular, you do not have to worry about the displacement in this problem, just the original path (940m).
W= F x D
587500 = 625N x 940M