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Juli2301 [7.4K]
1 year ago
13

A pipe of length 10.0 m increases in length by 1.5 cm when its temperature is increased by 90°F. What is its coefficient of line

ar expansion?
Physics
1 answer:
azamat1 year ago
3 0

The coefficient of linear expansion, given that the length of the pipe increased by 1.5 cm is 1.67×10¯⁵ /°F

<h3>How to determine the coefficient of linear expansion</h3>

From the question given above, the following data were obtained

  • Original diameter (L₁) = 10 m
  • Change in length (∆L) = 1.5 cm = 1.5 / 100 = 0.015 m
  • Change in temperature (∆T) = 90 °F
  • Coefficient of linear expansion (α) =?

The coefficient of linear expansion can be obtained as illustrated below:

α = ∆L / L₁∆T

α = 0.015 / (10 × 90)

α = 0.015 / 900

α = 1.67×10¯⁵ /°F

Thus, we can conclude that the coefficient of linear expansion is 1.67×10¯⁵ /°F

Learn more about coefficient of linear expansion:

brainly.com/question/28293570

#SPJ1

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Suppose a baseball pitcher throws the ball to his catcher.
amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

\Delta v = |v-u|

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

\Delta v=|v-0|=|v|

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

\Delta v =|0-v|=|-v|

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

\Delta p = m \Delta v

where

m is the mass of the object

\Delta v is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

|\Delta p|=m|\Delta v|

- In situation 1 (pitcher throwing the ball), the change in momentum is

\Delta p = m|\Delta v|=m|v|=mv

- In situation 2 (catcher receiving the ball), the change in momentum is

\Delta p = m\Delta v = m|-v|=mv

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

I=\Delta p

where

I is the impulse

\Delta p is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it  doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

I=F\Delta t

where

I is the impulse

F is the force applied

\Delta t is the time of impact

This can be rewritten as

F=\frac{I}{\Delta t}

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- \Delta t when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to \Delta t, this means that the force is larger when the ball is catched.

6 0
3 years ago
An electrically neutral model airplane is flying in a horizontal circle on a 2.0-m guideline, which is nearly parallel to the gr
amm1812

Answer:

q=3.5*10^-4

Explanation:

<u>concept:</u>

The force acting on both charges is given by the coulomb law:

F=kq1q2/r^2

the centripetal force is given by:

Fc=mv^2/r

The kinetic energy is given by:

KE=1/2mv^2

<u>The tension force:</u>

<u><em>when the plane is uncharged </em></u>

T=mv^2/r

T=2(K.E)/r

T=2(50 J)/r

T=100/r

<u><em>when the plane is charged </em></u>

T+k*|q|^2/r^2=2(K.E)charged/r

100/r+k*|q|^2/r^2=2(53.5 J)/r

q=√(2r[53.5 J-50 J]/k)                                          √= square root on whole

q=√2(2)(53.5 J-50 J)/8.99*10^9

q=3.5*10^-4

5 0
2 years ago
Imagine Two Artificial Satellites Orbiting Earth At The Same Distance. One Satellite Has A Greater Mass Than The Other One? Whic
Bad White [126]

After reading this whole question, I feel like I've already
earned 5 points !

-- Two satellites at the same distance, different masses:

The forces of gravity between two objects are directly
proportional to the product of the objects' masses.  In
other words, the gravitational forces between the Earth
and an object on its surface are proportional to the mass of
the object.  In other words, people with more mass weigh more
on the Earth, and the Earth weighs more on them. 

If the satellites are both at the same distance from Earth,
then the Earth pulls on the one with more mass with greater
force, and also the one with more mass pulls on the Earth
with greater force.

-- Two satellites with the same mass, at different distances:

The forces of gravity between two objects are inversely
proportional to the square of the distance between them. 
In other words, the gravitational forces between the Earth
and an object are inversely proportional to the square of
the distance between the object and the center of the Earth. 

If the satellites both have the same mass, then the Earth
pulls on the nearer one with greater force, and also the
nearer one pulls on the Earth with greater force.

-- Resistor in a circuit when the voltage changes:

The resistance depends on how the resistor was manufactured. 
Its resistance is marked on it, and doesn't change.  It remains
the same whether the voltage changes, the current changes,
the time of day changes, the cost of oil changes, etc.

If you increase the voltage in the circuit where that resistor is
installed, the current through the resistor increases.  If the current
remains constant, then you can be sure that somebody snuck over
to your circuit when you weren't looking, and they either installed
another resistor in series with the original one to make the total
resistance bigger, or else they snipped the original one out of the
circuit and quickly connected one with more resistance in its place.

6 0
3 years ago
Help again please !!!
rusak2 [61]
D I think .... don’t be mad if I’m wrong
5 0
3 years ago
A skydiver is falling at constant velocity. If a force of 600 N is pulling down on the skydiver, how much force must be acting u
skelet666 [1.2K]

Well, if the skydiver is at constant velocity, than there’s no acceleration, as stated by Newton’s first law. Thus the total net force would equate to 0. In order to make this statement true, the answer would have to be exactly 600 N.

7 0
3 years ago
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