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Varvara68 [4.7K]

3 years ago

13

Suppose all the people of the Earth go to the North Pole and, on a signal, all jump straight up. Estimate the recoil speed of th

e Earth. The mass of the Earth is 6 1024 kg, and there are about 6 billion people (6 109). Take the average mass of a person to be 74 kg and the distance the average person's center of mass rises after leaving the ground to be 0.2 m. m/s Additional Materials

1 answer:

Makovka662 [10]3 years ago

4 0

Explanation:

It is known that relation between velocity and height is as follows.

v = $\sqrt{2gh}$

where, g = acceleration due to gravity = 9.8 $m/s^{2}$

h = height = 0.2 m

Therefore, velocity is calculated as follows.

v = $\sqrt{2gh}$

= $\sqrt{2 \times 9.8 \times 0.2}$

= 3.92 m/s

Also,

$m_{people}v_{people} = m_{earth}v_{earth}$

$v_{earth} = \frac{m_{people}v_{people}}{m_{earth}}$

Putting the given values into the above formula as follows.

$v_{earth} = \frac{m_{people}v_{people}}{m_{earth}}$

= $\frac{6 \times 10^{9} \times 74 kg/person}{6 \times 10^{24} kg}$

= $\frac{444 \times 10^{9}}{6 \times 10^{24}}$

= $74 \times 10^{-15}$ m/s

or, = $7.4 \times 10^{-14}$ m/s

Thus, we can conclude that recoil speed of the Earth is $7.4 \times 10^{-14}$ m/s.

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Leni [432]

Answer: 12.0 m/s^2

Explanation:

Let $\alpha$ be the angular acceleration of the end of the rod

Taking torque about the link, we have:

$\tau = W \times OM\\ or\\ \tau = mg\times \left(\dfrac{L}{2}\right)\sin 55^\circ ....(i)$

Torque is also given in terms of moment of inertia of the rod and its angular acceleration i.e.

$\tau = I_{rod}\ \alpha......(ii)$

From equations (i) and (ii) we have:

$mg\times \left(\frac{L}{2}\right)\sin 55^\circ = \left(\frac{mL^2}{3}\right)\alpha\ \ \ \ \ \ \ (\because I_{rod} = \dfrac{mL^2}{3}) \\ \\ \alpha = \left(\frac{3g}{2L}\right)\sin 55^\circ\\ \\ \alpha = \left(\frac{3\times 9.8}{2\times 2.4}\right)\sin 55^\circ\\ \\ \boxed{\alpha = 5.02\ rad/s^2} }$

The acceleration of the end of the rod farthest from the link is given by:

$a = L\alpha\\ \\ a = (2.4\ m)(5.02\ rad/s^2)\\ \\ \boxed{a = 12.0\ m/s^2}$

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2 years ago

Olenka [21]

Answer:−4.05

Explanation:

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3 years ago

Sachi wants to throw a water balloon to knock over a target and win a prize. The target will only fall over if it is hit with a

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3 years ago

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3 years ago

Two cars, one of mass 1100 kg, and the second of mass 2500 kg, are moving at right angles to each other when they collide and st

Pie

Answer:

$v_{f}=17.47 m/s$

Explanation:

Let's use the conservation of momentum to solve it.

$p_{initial}= p_{final}$ (1)

The total initial momentum will be: $m_{1}v_{1i}+m_{2}v_{2i}$
The total final momentum will be: $m_{1}v_{1f}+m_{2}v_{2f}$ , but as they stick together after the collision, v1f = v2f = vf.

So we can rewrite (1), using the above information:

$m_{1}v_{1i}+m_{2}v_{2i}=m_{1}v_{f}+m_{2}v_{f}$

$m_{1}v_{1i}+m_{2}v_{2i}=v_{f}(m_{1}+m_{2})$

$v_{f}=\frac{m_{1}v_{1i}+m_{2}v_{2i}}{m_{1}+m_{2}}$

$v_{f}=\frac{1100\cdot 14+2500\cdot 19}{1100+2500}$

Finally, the magnitude of the velocity of the wreckage of the two cars immediately after the collision is:

$v_{f}=17.47 m/s$

I hope it helps you!

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3 years ago