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2 years ago

A person is standing 110m below the top of cliff B, and thetwo

Physics

1 answer:

kondor19780726 [428]2 years ago

3 0

Answer:

8.86m/s

Explanation:

The attachment below contain the explanation and answer

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for every 120 joules of energy input a car waste 85 joules, find the useful energy output of the car?

harina [27]

Answer:

Useful Output = 35 J

Explanation:

The useful energy output of the car must be equal to the difference between the total input energy supplied to the car and the energy wasted by the car:

Useful Output = Total Input - Waste

where,

Total Input = 120 J

Waste = 85 J

Therefore,

Useful Output = 120 J - 85 J

<u>Useful Output = 35 J</u>

7 0

2 years ago

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

2 years ago

Read 2 more answers

WILL MARK BRAINLIEST IF RIGHT & HAS EXPLINATION

Colt1911 [192]

Answer:

s= 64m I'm not 100% sure

Explanation: like 97.9%

7 0

2 years ago

A 72.8-kg swimmer is standing on a stationary 265-kg floating raft. The swimmer then runs off the raft horizontally with a veloc

nalin [4]

Answer:

-1.43 m/s relative to the shore

Explanation:

Total momentum must be conserved before and after the run. Since they were both stationary before, their total speed, and momentum, is 0, so is the total momentum after the run off:

$m_sv_s + m_rv_r = 0$

where $m_s = 72.8, m_r = 265$ are the mass of the swimmer and raft, respectively. $v_s = 5.21 m/s, v_r$ are the velocities of the swimmer and the raft after the run, respectively. We can solve for $v_r$