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Amanda
1 month ago
8

In space, astronauts don’t have gravity to keep them in place. That makes doing even simple tasks difficult. Gene Cernan was the

first astronaut who worked on a task outside a spaceship. He said of the experience, “Every time I’d push or turn a valve, it would turn my entire body at zero gravity. I had nothing to hold on to.” As he worked, Gene Cernan’s heart rate and temperature went so high that his fellow astronauts worried that he wouldn’t survive.
Think about routine tasks that astronauts might need to do inside and outside a spaceship. Choose several tasks, and describe the features the ship and spacesuits should have to account for zero gravity as the astronaut completes the task. Use Newton’s laws of motion in your analysis.
Physics
2 answers:
ValentinkaMS [3.4K]1 month ago
8 0

Response:

Newton's laws include:

-Newton's First Law: A body will stay at rest or in motion along a straight path unless acted upon by a force.

-Newton's Second Law: The effect of force on motion is proportional and parallel to it.

-Newton's Third Law: For every action, there is a reaction that is equal and opposite.

Some tasks that could be problematic include:

Exercising is difficult since, without gravity, weights cannot be lifted, making body-weight exercises ineffective; thus, they rely on treadmills secured with ropes.

Conducting work outside the spaceship is quite hazardous because any initiated movement continues in that direction due to the first law; hence, a force is necessary to alter movement in space. Certain space suits are equipped with small propulsors in their hands for better navigation in the void.

Manipulating valves or screws can be complicated, as, according to the third law, turning a valve exerts an equal force back on the person; without the weight that anchors them as on Earth, astronauts will start to spin when attempting to turn it. Spacecraft are often equipped with handles to assist astronauts in maintaining their position.

Keith_Richards [3.2K]1 month ago
4 0

Newton's First Law: A body remains in its current state of motion or at rest unless a force acts upon it.

Newton's Second Law: Motion changes are proportional to the applied force and oriented in the same direction.

Newton's Third Law: Every action has a corresponding and opposite reaction.

Tasks that would be challenging to perform in orbit include:

-operating a valve

-navigating on foot

-attempting to take a shower

-remaining still


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Yuliya22 [3333]

Answer:

The voltage across the bulb measures 3.0 V,

Explanation:

The bulb's voltage aligns with the voltage of the batteries, as they are the only power source for the bulb. Therefore, the voltage across the batteries is 3.0 V.

3 0
2 months ago
A Honda Civic travels in a straight line along a road. The car’s distance x from a stop sign is given as a function of time t by
serg [3582]

a) Average velocity: 2.8 m/s

b) Average velocity: 5.2 m/s

c) Average velocity: 7.6 m/s

Explanation:

a)

The car's position over time t can be described by

x(t)=\alpha t^2 - \beta t^3

where

\alpha = 1.50 m/s^2

\beta = 0.05 m/s^3

To find the average velocity, we divide the displacement by the elapsed time:

v=\frac{\Delta x}{\Delta t}

At time t = 0, the position is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

At time t = 2.00 s, the position is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

This leads us to the displacement of

\Delta x = x(2)-x(0)=5.6-0=5.6 m

The duration for this interval is

\Delta t = 2.0 s - 0 s = 2.0 s

Therefore, the average velocity during this period is

v=\frac{5.6 m}{2.0 s}=2.8 m/s

b)

At time t = 0, the position is:

x(0)=\alpha \cdot 0^2 - \beta \cdot 0^3 = 0

At time t = 4.00 s, the position is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

Thus, the displacement is

\Delta x = x(4)-x(0)=20.8-0=20.8 m

The time interval is

\Delta t = 4.0 - 0 = 4.0 s

This yields an average velocity of

v=\frac{20.8}{4.0}=5.2 m/s

c)

The position at t = 2 s is:

x(2)=\alpha \cdot 2^2 - \beta \cdot 2^3=5.6 m

And at t = 4 s it is:

x(4)=\alpha \cdot 4^2 - \beta \cdot 4^3=20.8 m

This gives us a displacement of

\Delta x = 20.8 - 5.6 = 15.2 m

While the time interval is

\Delta t = 4.0 - 2.0 = 2.0 s

So the resulting average velocity is

v=\frac{15.2}{2.0}=7.6 m/s

Find out more about average velocity:

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2 months ago
On its own, a certain tow-truck has a maximum acceleration of 3.0 m/s2. what would be the maximum acceleration when this truck w
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F= ma_1
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Upon attaching a bus that has double the mass of the truck, the total mass of the entire system (truck+bus) becomes (m+2m)=3m. In this situation, the force generated by the engine is
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m a_1 = 3 m a_2
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Read 2 more answers
A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at
Ostrovityanka [3204]

Answer

Given:

Wavelength = λ = 18.7 cm

                  = 0.187 m

Amplitude, A = 2.34 cm

Velocity, v = 0.38 m/s

A)  Calculate the angular frequency.

     f = \dfrac{v}{\lambda}

     f = \dfrac{0.38}{0.187}

     f =2.03\ Hz

Angular frequency,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) Calculate the wave number:

      K = \dfrac{2\pi}{\lambda}

     K= \dfrac{2\pi}{0.187}

    K =33.59\ m^{-1}

C)

Since the wave is traveling in the -x direction, the sign is positive between x and t

y (x, t) = A sin(k x - ω t)

y (x, t) = 2.34 sin(33.59 x - 12.75 t)

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2 months ago
In an amusement park rocket ride, cars are suspended from 3.40-m cables attached to rotating arms at a distance of 5.90 m from t
ValentinkaMS [3465]

Answer:

The rotational angular speed is measured at 1.34 rad/s.

Explanation:

Considering the following parameters,

Length = 3.40 m

Distance = 5.90 m

Angle = 45.0°

We are tasked with finding the angular speed of rotation

Using the balance equation

Horizontal component

T\cos\theta=mg

T=\dfrac{mg}{\cos\theta}

Vertical component

T\sin\theta=m\omega^2 r

Substituting the tension value

mg\tan\theta=m\omega^2(d+L\sin\theta)

\omega=\sqrt{\dfrac{g\tan\theta}{(d+L\sin\theta)}}

Substituting the value into the equation

\omega=\sqrt{\dfrac{9.8\tan45.0}{5.90+3.40\sin45.0}}

\omega=1.34\ rad/s

Thus, the angular speed of rotation computes to 1.34 rad/s.

7 0
2 months ago
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