All categories

3 years ago

Please help with a step by step method

Physics

1 answer:

Elan Coil [88]3 years ago

3 0

Answer:

Explanation:

This is a First Law of thermodynamics problem. We have to remember that the total energy available to a system is constant throughout the whole problem and that energy cannot be created or destroyed. So we need to find the total energy available right at the start. Well it just so happens that we are told that the total energy is 1000J and that it is all potential energy when the sphere is at rest and is 25 m off the ground. If the object isn't moving, all the energy is potential until it starts moving and the energy begins to convert from potential to kinetic a little bit at a time. The thing that we don't know is the mass of the shpere. Begin with the fact that the PE = 1000 (I'm going to se 2 sig fig's since there's only 1 in 1000). If

PE = 1000 and PE = mgh, then

1000 = m(9.8)(25) so

m = 4.1 kg

We also need the height at which this sphere has a PE of 600. Again, if

PE = 600 and PE = mgh, then

600 = (4.1)(9.8)h so

h = 15 Filling in the total energy equation now, using the fact that the total energy available to the system is 1000J:

TE = PE + KE and

1000 = (4.1)(9.8)(15) + $\frac{1}{2}(4.1)v^2$ and we are looking for v.

1000 = 6.0 × 10² + 2.1v² so

$v=\sqrt{\frac{1000-6.0*10^2}{2.1} }$ and

$v=\sqrt{\frac{4.0*10^2}{2.1} }$ gives us

v = 14 m/s

You might be interested in

A graduated cylinder contains 63.0 mL of water. A piece of gold, which has a density of 19.3 g/ cm3, is added to the water and t

icang [17]

Answer:

29.0 g

Explanation:

The mass of the piece of gold is given by:

m = dV

where

m is the mass

d is the density

V is the volume of the piece of gold

The density of gold is

d = 19.3 g/cm^3

while the volume of the sample is equal to the volume of displaced water, so

V = 64.5 mL - 63.0 mL = 1.5 mL

And since

1 mL = 1 cm^3

the volume is

V = 1.5 cm^3

So the mass of the piece of gold is:

m = (19.3 g/cm^3)(1.5 cm^3)=29.0 g

7 0

3 years ago

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:

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.

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)

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.

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.

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:

$I=F\Delta t$