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3 years ago

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Pretend you (80 kg) are making repairs on the outside of the International Space Station. You are floating 20 meters away from t

he space station when your safety rope comes loose. You throw your 10 kg tool-bag away from the space station in order to return to the station. If you are able to throw the bag at 5 m/s, how long does it take you to reach the station safely?

1 answer:

djyliett [7]3 years ago

7 0

Answer:

32 seconds

Explanation:

m1 = 80 kg

m2 = 10 kg

v2 = 5m/s

According to the property of conservation of momentum, assuming that both you and the bag are stationary before the safety rope comes lose:

$m_{1} v_{1} =m_{2} v_{2} \\80v_{1} =10*5 \\v_{1} = 0.625\ m/s$

Since the space station is 20 meters away, the time taken to reach it is given by:

$t = \frac{20}{0.625}\\t=32\ s$

It takes you 32 seconds to reach the station.

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Armand is monitoring a large sealed tank by way of a sensor that records the liquid level over time on a graph. He looks at the

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Answer:

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3 years ago

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If a system has 225 kcal of work done to it, and releases 5.00 × 102 kj of heat into its surroundings, what is the change in int

vovikov84 [41]

We can solve the problem by using the first law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change in internal energy of the system

$Q$ is the heat absorbed by the system

$W$ is the work done by the system on the surrounding

In this problem, the work done by the system is

$W=-225 kcal=-941.4 kJ$

with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is

$Q=-5 \cdot 10^2 kJ=-500 kJ$

with a negative sign as well because it is released by the system.

Therefore, by using the initial equation, we find

$\Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ$

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3 years ago

A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of

Vladimir79 [104]

Answer:

$m = 2.23 \times 10^{-32} kg$

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

orbital speeds = 220 km/s

we know that

acceleration due to centripetal force is $a = \frac{F}{m} = \frac{V^2}{r}$

Gravitational force $F= \frac{Gm m}{d^2}$

we know that

$v = \frac{2\pi R}{T}$

solving for

$R = \frac{vT}{2\pi}$

$F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}$

$F = G\times \frac{\pi m}{(vT)^2}$

$a = \frac{v^2}{\frac{vT}{2\pi}}$

$a = \frac{2\pi v}{T}$

we know that

f =ma

$G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}$

solving for m

$m = \frac{2Tv^3}{\pi G}$

$m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}$

$m = 2.23 \times 10^{-32} kg$

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3 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

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This lab is investigating the relationship between mass, ________, and momentum.

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This lab is investigating the relationship between mass, <u>Speed </u>, and momentum.

Momentum is manufactured from the mass and speed of an object. it's miles a vector quantity, owning a significance and a direction. If m is an object's mass and v is its speed, then the object's momentum is p.

Momentum in an easy way is a quantity of movement. right here amount is measurable because if an item is moving and has mass, then it has momentum. If an object no longer flows then it has no momentum. however, in regular existence, it has an important but many people didn't understand it.

Momentum gives the connection between the mass, pace, and direction of an object. Any exchange in momentum results in pressure. So, an exchange in momentum is used to determine the force appearing upon the item.

Learn more about momentum here:-brainly.com/question/1042017

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1 year ago