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CaHeK987 [17]
2 years ago
11

An astronaut working with many tools some distance away from a spacecraft is stranded when the "maneuvering unit" malfunctions.

How can the astronaut return to the spacecraft by sacrificing some of the tools? (Note: the maneuvering unit is connected to the astronaut's spacesuit and is not removable)
Physics
1 answer:
N76 [4]2 years ago
6 0

Answer:

He can return to the spacecraft by sacrificing some of the tools employing the principle of conservation of momentum.

Explanation:

By carefully evaluating his direction back to the ship, the astronaut can throw some of his tools in the opposite direction to that. On throwing those tools of a certain mass, they travel at a certain velocity giving him velocity in the form of recoil in the opposite direction of the velocity of the tools. This is same as a gun and bullet recoil momentum conservation. It is also the principle on which the operational principles of their maneuvering unit is designed.

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A horizontal spring with spring constant 750 N/m is attached to a wall. An athlete presses against the free end of the spring, c
ANTONII [103]

Answer:

37.5 N Hard

Explanation:

Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.

Using the expression for hook's law,

F = ke.............. Equation 1

F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.

Given: k = 750 N/m, e = 5.0 cm = 0.05 m

Substitute into equation 1

F = 750(0.05)

F = 37.5 N

Hence the athlete is pushing 37.5 N hard

4 0
2 years ago
Read 2 more answers
Why is it important for a muscle to be attached to a fixed origin at one end and a moving insertion at the other? Discuss how th
vlabodo [156]
Muscles function only by contracting. This makes it necessary for one end of the muscle to be fixed and the other mobile.
Take the bicep for example.
Its origin is at the shoulder and its two heads connect to the bones of the forearm, the radius and ulna.
Now, had the muscle not been fixed at one end, and contracted, it would pull both our shoulder and forearm together resulting in an ineffective movement. The desired motion is to lift the forearm (proximal and distal movement) which can only be achieved if the bicep is fixed at the shoulder and allowed to move at the forearm.
6 0
2 years ago
¿Alguien me puede ayudar? Problema: Un niño le pide gastada a su papá y éste le contesta que le dará los $120 que tiene en su bo
VLD [36.1K]

Answer: there are 15 coins of $2 and 18 coins of $5

Explanation:

I will answer in English.

X is the number of $5 coins.

Y is the number of $2 coins.

We have the system of equations:

Y + X = 33

Y*2 + X*5 = 120

first, we must isolate one of the variables in one of the equations and then replace it in the other equation, let's isolate Y in the first equation:

Y = 33 - X.

Then we can replace it in the other equation:

(33 - X)*2 + X*5 = 120

66 - X*2 + X*5 = 120

X*3 = 54

X = 54/3 = 18

and using the equation for Y.

Y = 33 - X = 33 - 18 = 15

So there are 15 coins of $2 and 18 coins of $5

3 0
2 years ago
Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the crit
Dennis_Churaev [7]

Complete Question:

A beam of white light is incident on the surface of a diamond at an angle \theta_{a},  since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. For example, the indices of refraction in diamond are n_{red} = 2.410 for red light and n_{blue} = 2.450 for blue light. Thus, blue light and red light are refracted at different angles inside the diamond. The surrounding air has n_{air} = 1.000.

Now consider θc, the angle at which the blue refracted ray hits the bottom surface of the diamond. If θc is larger than the critical angle θcrit, the light will not be refracted out into the air, but instead it will be totally internally reflected back into the diamond. Find θcrit. Express your answer in degrees to four significant figures.

Answer:

\theta_{crit} = 24.09^{0}

Explanation:

Only the blue refracted ray is related to the critical angle in this question

n_{air} = 1.000

n_{blue} = 2.450

The relationship between the critical angle(\theta_{crit}), n_{air} and n_{blue} can be given as sin \theta_{crit} = \frac{n_{air} }{n_{blue} }

sin \theta_{crit} = \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} \frac{1 }{2.450 }\\\theta_{crit} = sin^{-1} 0.4082^{0}\\  \theta_{crit} = 24.09^{0}

6 0
2 years ago
A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac
GREYUIT [131]

Answer:

The flux through the surface of the cube is 2.314\ Nm^{2}/C

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = 8.0\times 10^{- 2}\ m

Volume Charge density, \rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}

Now,

To calculate the electric flux:

\phi = \frac{q}{\epsilon_{o}}                                                      (1)

where

\phi = electric flux

\epsilon_{o} = 8.85\times 10^{- 12}\ F/m = permittivity of free space  

Volume Charge density for the given case is given by the formula:

\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}                  (2)

Volume of cube, V = a^{3}

Thus

V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}

Thus from eqn (2), the total charge is given by:

q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}

q = 2.048\times 10^{-11}\ F = 20.48\ pF

Now, substitute the value of 'q' in eqn (1):

\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C

5 0
2 years ago
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