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Alex777 [14]
3 years ago
7

A 0.5 kg ball is dropped vertically onto a floor, hitting with a speed of 20 m/s. It rebounds with a speed of 15 m/s immediately

after the collision. (a) What is the impulse in N-s acting on the ball during the contact
Physics
1 answer:
Nitella [24]3 years ago
5 0

Answer:

2.499Ns

Explanation:

Impulse is defined as change I'm momentum of a body. It is expressed as;

According to Newton's second law;

F = ma

F = m(v-u)/t

Ft = m(v-u) = Impulse

Impulse = Ft

Impulse = m(v-u)

F is the force acting on the ball

t is the time taken

m is the mass of the body

v is the final velocity

u is the initial velocity

To get impulse, we need to get time t first using the equation of motion v = u+gt

20 = 15+(9.8)t

20-15 = 9.8t

5 = 9.8t

t = 5/9.8

t = 0.51secs

Since Impulse = Ft

F = mg = 0.5(9.8)

F = 4.9N

Impulse = 4.9×0.51

Impulse acting on the ball during contact = 2.499Ns

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If the potential due to a point charge is 490 V at a distance of 10 m, what are the sign and magnitude of the charge?
uysha [10]

Answer:

+5.4×10⁻⁷ C

Explanation:

Electric potential: This can be defined as the work done in bringing a unit charge from infinity to that point against the action of the field. The S.I unit of potential is volt (V)

The formula for potential is

V = kq/r............................ Equation 1

Where V = electric potential, k = proportionality constant, q = charge, r = distance.

making q the subject of the equation,

q = Vr/k............................ Equation 2

Given: V = 490 V, r = 10 m,

Constant: k = 9×10⁹ Nm²/C²

Substitute into equation 2

q = 490(10)/(9×10⁹)

q = 5.4×10⁻⁷ C

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Hence the charge is +5.4×10⁻⁷ C

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A coin completes 18 spins in 12 seconds. The centripetal acceleration of the edge of the coin is 2.2 m/s2. The radius of the coi
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Centripetal acceleration, ac = r x w^2

2.2 = r x ( 2 x 3.14 x 1.5) ^2

2.2 = r x 88.7364

r = 0.025 m = 2.5 cm

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