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alisha [4.7K]
3 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]3 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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The superheroine Xanaxa, who has a mass of 67.5 kg , is pursuing the 70.3 kg archvillain Lexlax. She leaps from the ground to th
Paladinen [302]

Answer:

there is a net loss of potential energy  is Ut = -0.11 10⁵ J

Explanation:

The gravitational potential energy is

        ΔU = mg (y2- y1)

Where y1 is the initial position where the energy is defined as zero, in general in problems on the Earth's surface the reference system is taken at the lowest point, whereby y1 = 0 and consequently U1 = 0, this It is a matter of comfort since what matters are the changes in energy and not its value at one point explained.

Let's apply these concepts to our problem.

Jump from the ground to 179 m, let's calculate the energy change

     ΔU1 = mg (y2-y1)

     ΔU1 = 67.5 9.8 (179 -0)

     ΔU1 = 1.18 10⁵ J           gains energy

Dives up to Y3 = -15.9m below ground, therefore, this height is negative

The total height is y = 179 - (-15.9) = 194.9 m

       ΔU2 = mg (y3-y2)

These are the points where the dive begins and where it arrives

       ΔDU2 = 67.5 9.8 (- 194.9)

       ΔU2 = -1.29 105 J        loses energy

The total change in potential energy is

       Ut = ΔU1 + ΔU2

       Ut = 1.18 10⁵ - 1.29 10⁵

       Ut = -0.11 10⁵ J

Therefore, there is a net loss of potential energy

You can also calculate ΔU = mg (Δy)

Where ΔY is the change in body position, that is -15.9 m

         ΔU = 67.5 9.8 (-15.9)

         ΔU = -1.1 10⁴ J = -0.11 10⁵ J

You can see that it is the same value.

4 0
3 years ago
A heavy mirror that has a width of 2 m is to be hung on a wall as shown in the figure below. The mirror weighs 700 N and the wir
loris [4]

The translational equilibrium condition allows finding that the response for cable length with a maximum tension is

      L = 2.56 m

Newton's second law says that the force is proportional to the mass and the acceleration of the body, in the special case that the acceleration is zero, the relationship is called the translational equilibrium condition.

              ∑ F = 0

 

Where the bold indicates vectors, F is the force and the sum is for all external forces.

The reference systems are coordinate systems with respect to which the decomposition of the vectors is carried out and the measurements are made, in this case we will use a system with the horizontal x axis and the vertical y axis.

In the attachment we can see a free body diagram of the system, let's write the equilibrium condition for each axis.

x-axis  

         Tₓ -Tₓ = 0

y-axis

         T_y + T_y - W =0

         2T_y - W = 0

Let's use trigonometry to decompose the tension, we can see from the graph and the adjoint that each string is half the length, let's call the angle θ

          cos θ = \frac{T_x}{T}

          sin θ = \frac{T_y} { T}

          Tₓ = T cos θ

          T_y = T sin θ

We substitute

          2 T sin θ = W          (1)

The text indicates that the length of the block is 2 m, so the distance to the midpoint is

        x = 1 m

Let's use the Pythagoras' Theorem            

            H² = CA² + CO²

           CO = \sqrt{H^2 - CA^2}

           CA = x

           CO = \sqrt{(\frac{L}{2} )^2  - 1 }

Where CO is the opposite leg,  CA is the adjacent leg and H is the hypotenuse indicating H = L / 2,

Let's write the trigonometry functions

           sin θ = \frac{CO}{H}

Let's substitute        

            sin θ = \frac{2 \sqrt{\frac{L^2}{4} -1 } }{L}

Let's subtitute in the equation  1

          2 T  ( \frac{2 \sqrt{\frac{L^2}{4} -1 } }{L}   ) = W

          \sqrt{\frac{L^2}{4}-1  }  = \frac{1}{4}  \ \frac{W}{T} \ L

Let's solve by squaring

         \frac{L^2}{4}  -1 = \frac{1}{4} \ (\frac{W}{T} )^2 \ \frac{L^2}{4}  

         \frac{L^2}{4} ( 1 - \frac{1}{4} (\frac{W}{T})^2  ) -1 =0

They indicate that the maximum tension of the cable is T = 700N and the weight is worth W = 700N, we substitute the values

        \frac{L^2}{4} ( 1- \frac{1}{4}) - 1 =0 \\\frac{3 L^2}{16} = 1 \\ L^2 = \frac{16}{3} \\L = \sqrt{\frac{16}{3} }

        L =   2.31 m    

In conclusion using the translational equilibrium condition we can find that the response for cable length with maximum tension is

      L = 2.31 m

Learn more about translational equilibrium here:

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3 years ago
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How can two machines appear identical and yet not have the same actual mechanical advantage
STatiana [176]
There are many factors which contributes as to how a machine will be processing the input energy and convert it to output energy. Even with identical mechanism, these factors will have major effect on the output. Some factors are deflection, friction and wear. Some system maybe exposed to poor lubrication than the other which'll produce more friction and wear thus lower mechanical advantage.
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The critical angle between a medium and air is 430. What is the speed of light in that medium?
vagabundo [1.1K]

Answer:

The speed of light is that medium is 281907786.2 m/s.

Explanation:

since the critical angle is Фc = 430, we know that the refractive index is given by:

n = 1/sin(Фc)

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then if n is the refractive index of the medium and c is the speed of light, then the speed of light in the medium is given by:

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