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shutvik [7]
3 years ago
11

A basketball has a mass of 1 kg and is traveling 15 m/s. How fast would a 5 kg bowling ball have to travel to have the same mome

ntum?
A) 3 m/s

B) 15 m/s

C) 1 m/s

D) 5 m/s
Physics
2 answers:
DiKsa [7]3 years ago
8 0
A) 3 m/s. Hope this helps.
Andrei [34K]3 years ago
5 0

Answer:

A 3m/s

Explanation:

Took the test (k12)

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zavuch27 [327]
<span>A) x = 41t
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3 years ago
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A 0.350kg bead slides on a curved fritionless wire,
LuckyWell [14K]

Answer:

h2 = 0.092m

Explanation:

From a balance of energy from point A to point B, we get speed before the collision:

m1*g*h-\frac{m1*V_B^2}{2}=0  Solving for Vb:

V_B=\sqrt{2gh}=6.56658m/s

Since the collision is elastic, we now that velocity of bead 1 after the collision is given by:

V_{B'}=V_B*\frac{m1-m2}{m1+m2} = \sqrt{2gh}* \frac{m1-m2}{m1+m2}=-1.34316m/s

Now, by doing another balance of energy from the instant after the collision, to the point where bead 1 stops, we get the distance it rises:

m1*g*h2-\frac{m1*V_{B'}^2}{2}=0 Solving for h2:

h2 = 0.092m

6 0
3 years ago
A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. T
Mamont248 [21]

Let y_0 be the height of the building and thus the initial height of the ball. The ball's altitude at time t is given by

y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2

where g=9.80\frac{\rm m}{\mathrm s^2} is the acceleration due to gravity.

The ball reaches the ground when y=0 after t=2.9\,\mathrm s. Solve for y_0:

0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2

\implies y_0\approx16.258\,\mathrm m

so the building is about 16 m tall (keeping track of significant digits).

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The reasoning which is in use when large, angular rocks are interpreted to have originated from the outcrop at the top of the hill is; Fossil succession

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Read more on fossil succession;

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Answer:

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Explanation:

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