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maksim [4K]
2 years ago
11

A baseball has a mass of 0. 45 kg and is thrown with a speed of 25 m/s. What is the momentum of the baseball? 0. 018 kg • mete

rs squared per second 11 kg • meters per second 56 kg • meters per second 140 kg • meters per second.
Physics
1 answer:
aleksandr82 [10.1K]2 years ago
3 0

Answer:

11 Kg.m/s

Explanation:

P=mv

P=0.45*25

P=11.25 Kg.m/s

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Question #15).
A). is true, but has nothing to do with particle behavior
B). Diffraction is a purely wave thing, so it doesn't help.
C). is true, but doesn't say anything about particle behavior
D). is true, but isn't an indication of particle behavior.

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3 0
4 years ago
A calorimeter uses the temperature change of water to determine the _____ of another substance.
Evgesh-ka [11]

Answer:

A calorimeter uses the temperature change of water to determine the <u>specific heat </u> of another substance.

Explanation:

8 0
3 years ago
Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in
charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

  • t = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
  • s = distance to be moved by the rock long the horizontal = 98 yards
  • y = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

  • u = magnitude of initial velocity of the rock
  • \theta = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\

On dividing equation (1) by (2), we have

\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ

Now, putting this value in equation (2), we have

u\cos 51.34^\circ\times  5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

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3 years ago
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First thing to do is to draw the system described above. Then, write an equation for the forces present.
<span>
</span>Σ<span>F = Fg - Ff
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u = tan</span><span>∅
</span>∅(max) = tan^-1 (u)<span>
</span>
7 0
3 years ago
Read 2 more answers
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