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Aleks [24]
3 years ago
8

The local church is hosting a carnival which includes a bumper car ride. Bumper car A and its driver have a mass of 300 kg; bump

er car B and its driver have a mass of 200 kg. Bumper car A has a velocity to the right of 2 m/s and bumper car B is at rest. At t = 0 s, bumper car A and B are separated by 10 m. Bumper car A accelerates at 1 m/s2 to a velocity of 4 m/s and continues at this constant speed until colliding with bumper car B.
Calculate the time required for bumper car A to travel the 10 m to collide with bumper car B.
Calculate the speed of bumper car A following the collision with bumper car B, which now has a velocity to the right of 3 m/s.
Is the direction of motion for bumper car A following the collision with bumper car B to the right, to the left, or is bumper car A at rest?
Is the collision elastic? Justify your answer.
Physics
1 answer:
lukranit [14]3 years ago
8 0

Answer:

a. 20 s

b. 0 m/s  

c. right

d.no its inelastic because when the car b was at rest and a was coming in at it, since b had no force what so ever car a swept it away with it moving to the right

Explanation:

im not sure though

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3 years ago
A very large sheet of insulating material has had an excess of electrons placed on it to a surface charge density of –3.00nC/m2
lys-0071 [83]

Answer: sheet of charge

Explanation:

a )

Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.

b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge.   The equi- potential surface is always perpendicular to electric field.

C ) Electric field which is almost uniform near the sheet of charge is equal t the following

E = σ / ε₀ where  σ is charge density of surface and  ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²

E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²

= .3389 x 10³

= 338.9 V / m

spacing between 1 V

= 1 / 338.9 m

= 2.95 X 10⁻3 m

= 2.95 mm.

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QUESTION 1
Molodets [167]
Question 1: C Question 2: B, Hope this Helps!
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