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alisha [4.7K]
4 years ago
8

By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. If the mass of th

e vehicle is 155.1× greater than the astronaut by what factor would the astronaut's acceleration be affected? Think of a method in which recoil of the vehicle is avoided. original text = Discuss how this would affect the measurement of the astronaut’s acceleration.
Physics
1 answer:
yaroslaw [1]4 years ago
4 0

Answer:

The astronaut's acceleration is 155.1 times the vehicle's acceleration

Explanation:

These effects due to Newton's third law of action and reaction. Since the forces are equal but in the opposite direction and each acting on a different body. We distance that the Force is F let's calculate the acceleration of the vehicle and the astronaut  

Astronaut

        F = m_{ast}  a₁

Vehicle

        F = m_{veh} a₁

F = 555.1 m_{ast} a₂

Let's match the equation

m_{ast} a₁ = 155.1 m_{ast} a₂

a₁ = 155.1 a₂

a₁ / a₂ = 155.1

The astronaut's acceleration is 155.1 times the vehicle's acceleration

We see that even when the acceleration of the vehicle is small, there is a very high multiplicative factor.

One method to improve this situation is that the vehicles fear some small retro-rocket vehicles to reduce their acceleration. This would have a very favorable impact on the astronaut's mission.

Another method would be for the astronaut himself to have the retro-rocket and control his acceleration.

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Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is this case is 0 as the road is level

Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}

Here

  • SSD is the stopping sight distance which is to be calculated.
  • u is the speed which is given as 60 mi/hr
  • t is the perception-reaction time given as 2.5 sec.
  • a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
  • G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

                              SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

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Answer:

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Explanation:

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emf_{max}=NBA\omega

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

A=\pi r^2 =\pi(0.23m)^2=0.16m^2

emf_{max}=(100000)(0.3*10^{-4}T)(0.166m^2)(140\frac{rev}{s})=69.72V

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I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA

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we have that

I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A

P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W

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