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miv72 [106K]
1 year ago
11

Two astronauts, A and B, both with mass of 60Kg, are moving along a straight line in the same direction in a weightless spaceshi

p. Relative to the spaceship the speed of A is 2 m/s and that of B is 1 m/s. A is carrying a bag of mass 5 Kg with him. To avoid collision with B A throws the bag with a speed v relative to the spaceship towards B catches it. Find the minimum value of v.
Physics
1 answer:
sveta [45]1 year ago
7 0

Answer:

V=14m/s

Explanation:

From the Question we are told that

Mass of A and B is 60kg

Speed of A=2m/s

Speed of B=1m/s

Mass of bag =5kg

Generally the momentum of the astronaut  A and bag is mathematically given as

  M_A=(60+5)*2

   M_A=130kgm/s

Generally to avoid collision the speed of astronaut be should be less than or equal to that of astronaut A with the bag

Therefore for minimum requirement speed of astronaut A should be given by astronaut B's speed which is equal to 1

Therefore

  130=(60*1)=(5*v)

   V=14m/s

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2 years ago
A proton of mass mp is released from rest just above the lower plate and reaches the top plate with speed vp. An electron of mas
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Answer:

v_e=\sqrt{\frac{m_pv_p^2}{m_e}}

Explanation:

You can consider that the force that acts over the proton is the same to the force over the electron. This is because the electric force is given by:

F=qE

F_p=F_e

where E is the constant electric field between the parallel plates, and is the same for both electron and proton. Also, the charge is the same.

by using the Newton second law for the proton, and by using kinematic equation for the calculation of the acceleration you can obtain:

m_pa_p=qE\\\\a_p=\frac{v_p^2}{2d}\\\\\frac{m_pv_p^2}{2d}=qE

(it has been used that vp^2 = v_o^2+2ad) where d is the separation of the plates, ap the acceleration of the proton, vp its velocity and mp its mass.

By doing the same for the electron you obtain:

\frac{m_ev_e^2}{2d}=qE

we can equals these expressions for both proton and electron, because the forces qE are the same:

\frac{m_pv_p^2}{2d}=\frac{m_ev_e^2}{2d}\\\\v_e=\sqrt{\frac{m_pv_p^2}{m_e}}

4 0
2 years ago
Which of the following diagrams involves a virtual image ?
sergiy2304 [10]

Answer:

The third diagram

Explanation:

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  • <u>A convex lens</u> can form both virtual and real image depending on the position of the object from the lens.
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  • In the third diagram, a virtual image is formed because the position of the object is between the focus and the optical center of the convex lens.
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2 years ago
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