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Aleonysh [2.5K]
4 years ago
12

romeo threw a small bundle toward Juliet on a balcony 10.0 m above him with a velocity of 5.0 m/s. The bundle stopped rising aft

er 1.5 seconds. How high did the bundle travel? Was that high enough for her to catch it?
Physics
1 answer:
choli [55]4 years ago
4 0

I can tell that you've never read much Shakespeare.

If the bundle is pitched with a velocity of 5.0 m/s, it must follow
as the night follows the day that it cannot then continue to rise
for more than barely 1/2 second before it reacheth the zenith
of its arc, reverseth its course, and plungeth Earthward to its
ultimate demise.

But hey !  To thine own self be true, eh ?

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hot-air balloon is ascending at the rate of 14 m/s and is 84 m above the ground when a package is dropped over the side. (a) How
timurjin [86]

Answer:

a) t = 4.14 s

b) Speed with which it hits the ground = 40.58 m/s

Explanation:

Using the equations of motion,

g = 9.8 m/s², y = H = 84 m,

Initial velocity, u = 0 m/s,

final velocity, v = ?

Total Time of fall, t = ?

a) y = ut + gt²/2

84 = 0 + 9.8t²/2

4.9t² = 84

t² = 84/4.9

t = 4.14 s

b) v = u + gt

v = 0 + (9.8 × 4.14)

v = 40.58 m/s

4 0
3 years ago
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A cue ball of mass m1 = 0.325 kg is shot at another billiard ball, with mass m2 = 0.59 kg, which is at rest. The cue ball has an
Roman55 [17]

Answer:

v_{2f} = \frac{2vm_1}{m_2 + m_1}

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

Before the collision

P_i = m_1v

After the collision

P_f = m_1v_{1f} + m_2v_{2f}

So using the law of momentum conservation

P_i = P_f

m_1v = m_1v_{1f} + m_2v_{2f}

We can solve for the speed of ball 1 post collision in terms of others:

v_{1f} = v - v_{2f}\frac{m_2}{m_1}

Their kinetic energy is also conserved before and after collision

m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2

m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2

From here we can plug in v_{1f} = v - v_{2f}\frac{m_2}{m_1}

m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2

m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2

m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2

v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0

v_{2f}(1 + \frac{m_2}{m_1}) = 2v

v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}

8 0
3 years ago
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Answer:

The height of the sound waves increases

Explanation:

3 0
4 years ago
Is a pressure force a body force or surface force?
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It is a surface force
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Calculate the most probable speed of an ozone molecule in the stratosphere
Marysya12 [62]

Answer:

v_{mp}=305.83 m/s

Explanation:

The temperature in stratosphere is generally about 270 K

molecular weight of an ozone molecule = 48 gm/mole

now formula for most probable velocity

v_{mp}= \sqrt{\frac{2RT}{M} }

plugging the values we get

v_{mp}= \sqrt{\frac{2\8.314\times270}{48} }

v_{mp}=305.83 m/s

7 0
4 years ago
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