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Art [367]
3 years ago
8

An astronaut in a spacecraft looks out her window and observes a comet travel in the opposite direction at a relative speed of 2

37 m/s. The velocity of the spacecraft is 114 m/s directly away from the sun. What is the comet's velocity relative to the sun? Assume that motion away from the sun is in the positive direction.
Physics
2 answers:
Sergeu [11.5K]3 years ago
7 0
The velocity of the spacecraft with respect to the Sun is
v_s = +114 m/s
the velocity of the comet in the reference system of the spacecraft is
v_c' = 237 m/s

So, the velocity of the comet in the reference system of the Sun will be
v_c = v_c'-v_s = -237 m/s - 114 m/s=-351 m/s
And the negative sign tells us that the comet is directed toward the Sun.
Umnica [9.8K]3 years ago
4 0

Answer: The relative velocity of the comet with respect to sun is '-351 m/s'.

Explanation:

Let the magnitude of the velocity of the spaceship moving away from the sun be |v_s|=114 m/s

Let the magnitude of the velocity of the comet moving towards the sun be |v_c| m/s

Assuming the direction away from the sun to be positive then :

The velocity of the space ship :v_s=+114 m/s

The  velocity of the comet :-v_c

v_r = relative velocity of spaceship with respect to comet

v_r=-237 m/s=v_s+(-v_c)=114 m/s-v_c

v_c=351 m/s

Velocity of the comet.v_c=351 m/s

v_R = Relative velocity of comet with respect to sun:

v_R=v_s+v_c=0 m/s+(-351 m/s)=-351 m/s

v_s= 0 m/s(Sun is fixed at one point)

The negative sign means that comet is moving towards the sun.The relative velocity of the comet with respect to sun is '-351 m/s'.

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The following picture displays a map of potential difference (vertical axis) for an unknown configuration of charges as a functi
Slav-nsk [51]

Answer:

Direction 1: Force is Non-zero and Not- constant

Direction 2: Force is Non-zero but constant  

Explanation:

Given:

The picture of the map is attached. ( Missing from the question ).

Find:

The effect of force as it travels along each direction.

Solution:

- We know the relationship between change in potential and the force acting on the charge particle is given by:

                               F = - q*dV/ dr

Where,

q : Charge of the particle

V : Volt potential

dV/dr : Potential difference along a direction.

Direction 1:

- The color code of the map changes as the particle moves along this direction. Each color code represents a potential difference. So as the particle moves between different potential difference then according to the relationship given above The force varies along varies as particle moves from one color to another. Hence, a non zero force but not constant.

Direction 2:

- In the direction 2, the charged particle moves along the same color. The potential difference for each color is constant. Hence, according to the relationship of potential difference and force. If potential difference is constant then the Electrostatic Force on the charge is also constant. Hence, Force is non-zero and constant.

8 0
3 years ago
The tibia is a lower leg bone (shin bone) in a human. The maximum strain that the tibia can experience before fracturing corresp
IRISSAK [1]

Answer:

a

   k    =  11600000 N/m

b

   \Delta  L  =  3.2323 *10^{-5} \ m

c

  F =  3750.28 \  N  

Explanation:

From the question we are told that

    The Young modulus is  E =  1.4 *10^{10} \  N/m^2

     The length is  L  =  0.35 \ m

      The  area is  2.9 \ cm^2  =  2.9 *10^{-4} \ m ^2

   

Generally the force acting on the tibia is mathematically represented as

       F =  \frac{E *  A  *  \Delta  L }{L}    derived from young modulus equation

Now this force can also be mathematically represented as

      F =  k *  \Delta  L    

So

     k    =  \frac{E *  A  }{L}

substituting values

     k    =  \frac{1.4 *10^{10} *  2.9 *10^{-4}  }{ 0.35}

     k    =  11600000 N/m

    Since the tibia support half the weight then the force experienced by the tibia is  

        F_k  =  \frac{750 }{2}  =  375 \  N

 From the above equation the extension (compression) is mathematically represented as

          \Delta  L  =  \frac{ F_k  *  L  }{ A *  E }        

substituting values

           \Delta  L  =  \frac{  375   *  0.35  }{ (2.9 *10^{-4}) *   1.4*10^{10} }

           \Delta  L  =  3.2323 *10^{-5} \ m

From the above equation the maximum force is  

        F =  \frac{1.4*10^{10} *  (2.9*10^{-4})  *  3.233*10^{-5} }{ 0.35}  

         F =  3750.28 \  N  

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where Mp and Me are the masses, and ap and ae are the accelerations of the proton and the electron, respectively. Since the mass of the electron is much smaller than the mass of the proton, in order for the equation above to hold, the acceleration of the electron at that moment must be considerably larger than the acceleration of the proton at that moment. Since electrons have much greater acceleration than protons, they achieve a faster rate than protons and therefore first reach the origin.
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