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2 years ago

Ball A, moving towards the right, collides with

Physics

1 answer:

Mazyrski [523]2 years ago

8 0

Answer:

ball b has the greater mass.

Explanation:

b only moved because a bumped into it. if a bounced back, it had less mass and b moved slowly.

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A 0.25 kg baseball is in contact with a bat for 0.0075 s and exerts a force or Tour

Tanzania [10]

Answer:.

Explanation:

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3 years ago

True or False:

mart [117]

This is a true statement

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3 years ago

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A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s' for 10 secon

MArishka [77]

Answer:

Distance of 400m.

Explanation:

Use your kinematics equation to solve for distance (we can use kinematics b/c acceleration is constant).

d = (initial velocity x time) + 1/2 at^2

d = (20 x 10) + 1/2 (4) (10)^2

d = 200 + 200

d = 400 m

5 0

2 years ago

Amber asked her roommate to turn down the radio because she was trying to study. Her roommate had increased the volume from a vo

Blizzard [7]

Answer:

Difference threshold or also Just Noticeable Difference

Explanation:

The above mentioned case between room mates, where one room mate was able to detect a minute change in volume shows an instance of the difference threshold.

Difference threshold can be defined as stimulation at its minimum level that can be detected by an individual almost 50 % of the times.

It is the lowest possible level of sound that is detectable by a person.

That is what happened in the mentioned case that when the volume was increased from 14 to 15, Amber was able to detect it.

7 0

3 years ago

A nonconducting container filled with 25 kg of water at 23°C is fitted with a stirrer, which is made to turn by gravity acting o

Paul [167]

Explanation:

Given that,

Weight of water = 25 kg

Temperature = 23°C

Weight of mass = 32 kg

Distance = 5 m

(a). We need to calculate the amount of work done on the water

Using formula of work done

$W=mgh$

$W=32\times9.8\times5$

$W=1568\ J$

The amount of work done on the water is 1568 J.

(b). We need to calculate the internal-energy change of the water

Using formula of internal energy

The change in internal energy of the water equal to the amount of the work done on the water.

$\Delta U=W$

$\Delta U=1568\ J$

The change in internal energy is 1568 J.

(c). We need to calculate the final temperature of the water

Using formula of the change internal energy

$\Delta U=mc_{p}\Delta T$

$\Delta U=mc_{p}(T_{2}-T_{1})$

$T_{2}=T_{1}+\dfrac{\Delta U}{mc_{p}}$

$T_{2}=23+\dfrac{1568}{25\times4.18\times10^{3}}$

$T_{2}=23.01^{\circ}\ C$

The final temperature of the water is 23.01°C.

(d). The amount of heat removed from the water to return it to it initial temperature is the change in internal energy.

The amount of heat is 1568 J.

Hence, This is the required solution.

6 0

2 years ago