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Elenna [48]
3 years ago
8

A ball is dropped from a height of 1.20 m and hits the floor. The ball compresses and then reforms to spring upwards from the fl

oor to a height of 0.86 m. The time from the ball's first contact to when it left contact with the floor was 0.091 s. Calculate the ball's acceleration while it is in contact with the floor.
Physics
1 answer:
Zina [86]3 years ago
4 0

The ball's acceleration while it is contact with the floor is 98.4m/s²

<u>Explanation:</u>

V1 = -√(2gh1) = -4.85 m/s

V2 = √(2gh2) = 4.1056 m/s

dV = V2 - V1 = +8.9556 m/s

a = dV/dt = 8.9556/0.091 = 98.4 m/s²

Therefore, the ball's acceleration while it is contact with the floor is 98.4m/s²

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A source charge generates an electric field of 4286 N/C at a distance of 2. 5 m. What is the magnitude of the source charge? (Us
svp [43]

The magnitude of the source charge is 3 μC which generates 4286 N/C of the electric field. Option B is correct.

What does Gauss Law state?

It states that the electric flux across any closed surface is directly proportional to the net electric charge enclosed by the surface.

Q = \dfrac {ER^2}k

Where,

E = electric force = 4286 N/C

k = Coulomb constant = 8.99 \times  10^9 \rm\ N m ^2 /C ^2

Q\\&#10;     = charges = ?

r = distance of separation = 2.5 m

Put the values in the formula,

Q  = \dfrac {4286\times  2.5 ^2}{8.99 \times  10^9 }\\\\&#10;Q  = 3\rm \  \mu C

Therefore, the magnitude of the source charge is 3 μC.

Learn more about Gauss's law:

brainly.com/question/1249602

8 0
2 years ago
A quarterback throws a football toward a receiver with an initial speed of 20 m/s at an angle of 30∘ above the horizontal. At th
lana66690 [7]

Answer:

a) In order to catch the ball at the level at which it is thrown in the direction of motion.

b)Speed of the receiver will be 7.52m/s

Explanation:

Calculating range,R= Vo^2Sin2theta/g

R= (20^2×Sin(2×30)/9.8 = 35.35m

Let receiver be(R-20) = 35.35-20= 15.35m

The horizontal component of the ball is:

Vox= Vocostheta= 20× cos30°

Vox= 17.32m/s

Time taken to coverR=35.35m with 17.32m/s will be:

t=R/Vox= 35.35/17.32

t= 2.04seconds

b)Speed required to cover 15.35m at 2.04seconds

Vxreciever= d/t = 15.35/2.04 = 7.52m/s

7 0
3 years ago
Read 2 more answers
5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on
kari74 [83]

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)

(0.1)(v1) + 0 = 30.6

(0.1)(v1) = 30.6

v1 = 306 m/s

8 0
3 years ago
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