Ask question

Ask question

All categories

3 years ago

12

A 87.0 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. The astron

aut turns away to look at Earth and several seconds later is 38.1 m behind the ship, at rest relative to the spaceship. The only way to return to the ship without a thruster is to throw a wrench directly away from the ship. The wrench has a mass of 0.570 kg, and the astronaut throws the wrench with a speed of 22.4 m/s. How long does it take the astronaut to reach the ship?

1 answer:

Ket [755]3 years ago

4 0

Answer:

259.62521 seconds

Explanation:

$m_1$ = Mass of astronaut = 87 kg

$m_2$ = Mass of wrench = 0.57 kg

$v_1$ = Velocity of astronaut

$v_2$ = Velocity of wrench = 22.4 m/s

Here, the linear momentum is conserved

$m_1v_1=m_2v_2\\\Rightarrow v_1=\frac{m_2v_2}{m_1}\\\Rightarrow v_1=\frac{0.57\times 22.4}{87}\\\Rightarrow v_1=0.14675\ m/s$

Time = Distance / Speed

$Time=\frac{38.1}{0.14675}=259.62521\ s$

The time taken to reach the ship is 259.62521 seconds

You might be interested in

How do you calculate acceleration

AfilCa [17]

A=f/m
Example a=10/2
A=5

7 0

4 years ago

Read 2 more answers

A team of engineering students is testing their newly designed 200 kg raft in the pool where the diving team practices. The raft

elena-s [515]

Answer:

The water level rises more when the cube is located above the raft before submerging.

Explanation:

These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.

Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.

When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.

Vc = 0.45*0.45*0.45 = 0.0911 [m^3]

Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.

$Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]$

Density is given by:

Ro = m/V

where:

m= mass [kg]

V = volume [m^3]

The buoyancy force can be calculated using the following equation:

$F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]$

Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.

7 0

4 years ago

A dart gun consists of a horizontal spring with k = 52 Newtons/m that is compressed 43

gulaghasi [49]

Answer:

$299 m/s^2$

Explanation:

When a spring is compressed, the force exerted by the spring is given by:

$F=kx$

where

k is the spring constant

x is the compression of the spring

In this problem we have:

k = 52 N/m is the spring constant

x = 43 cm = 0.43 m is the compression

Therefore, the force exerted by the spring on the dart is

$F=(52)(0.43)=22.4 N$

Now we can apply Newton' second law of motion to calculate the acceleration of the dart:

$F=ma$

where

F = 22.4 N is the force exerted on the dart by the spring

m = 75 g = 0.075 kg is the mass of the dart

a is its acceleration

Solving for a,

$a=\frac{F}{m}=\frac{22.4}{0.075}=299 m/s^2$

7 0

3 years ago

A 50-kg copper block initially at 140Â°c is dropped into an insulated tank that contains 90 l of water at 10Â°c. Determine the f

xxMikexx [17]

Answer:

$T_f=24.71$

Explanation:

From the question we are told that:

Mass of block $m=50$

Temperature of block $T_b =140 \textdegree C$

Volume of water $V= 90L$

Temperature of water $T_w=10 \textdegree C$

Density of water $\rho=1000kg/m^3$

Specific heat of water $C_w=4.18KJ/kg-k$

Specific heat of copper $C_p=0.96KJ/kg-k$

Generally the equation for equilibrium stage is mathematically given by

$mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)$

$50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)$

$48(140-T_f)=376.2(T_f-10)$

$140-T_f=7.8375(T_f-10)$

$140-T_f=7.8375T_f-78.375$

$-8.8375T_f=-218.375$

$T_f=\frac{-218.375}{-8.8375}$

$T_f=\frac{-218.375}{-8.8375}$

$T_f=24.71$

6 0

3 years ago

why does diving 30m below sea level affect our bodies more than being in a building 30m above sea level

Genrish500 [490]

Imagine you are in a swimming pool 30m deep. Assuming you know that water is denser than air, you would know that the 30m of water above you will carry more weight, and press down on your body. Say you were in a swimming pool 60m deep, you would be sandwiched between 30m of water pressing down on you, and the upthrust created by the 30m of water below you.

In a building 30m up, the pressure will be regulated, as you are in a building. The floor will be strong enough to support the weight of the body, and the body will not recoil into itself.

5 0

3 years ago