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LiRa [457]
2 years ago
15

Suppose that Salim hits a 100 g fastball moving towards him at a speed of 42 m/s with his bat as shown in the image.the impulse

of the fastball is 8 kg.m/s in the antiparallel direction of the initial velocity. Then the final momentum of the fastball is --------------( Suppose that the direction of the fastball toward the bat is negative and away from the bat is positive)
1.8 kg m/s
0.6 kg m/s
3.8 kg m/s
2.4 kg m/s
Physics
1 answer:
FinnZ [79.3K]2 years ago
3 0

The final momentum of the ball is 3.8 kgm/s.

<h3>Change in momentum of the ball</h3>

The impulse received by the ball is equal to change in momentum of the ball.

J = ΔP

where;

  • J is the impulse
  • ΔP is change in momentum

ΔP = P₂ - P₁

P₂ = ΔP + P₁

<h3>Final momentum of the ball</h3>

The final momentum of the ball is calculated as follows;

P₂ = 8 + (- 0.1 x 42)

P₂ = 8 - 4.2

P₂ = 3.8 kgm/s

Learn more about change in momentum here: brainly.com/question/7538238

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4 0
3 years ago
A mover does 422 J of work pushing a crate 8.39 m. How much force did he exert?
kicyunya [14]

Answer:

50.3N

Explanation:

Work done = force x distance

422J. = force x 8.39m

÷8.39 both side to get force

Force is 50.3N to 1 d.p.

Check:

50.3 x 8.39=422.017J

Same as 422J to 1 d.p

4 0
2 years ago
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Make the following conversion.
anastassius [24]
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4 0
3 years ago
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An elevator filled with passengers has a mass of 1683 kg. (a) The elevator accelerates upward from rest at a rate of 1.20 m/s2 f
artcher [175]

Answer:

the tension is 18513N

Explanation:

Given that

mass = 1683kg

acceleration = 1.2m/s^2

acceleration due to gravity = 9.8m/s^2

T-mg =   ma

T = ma + mg

T = m(a +g)

T = 1683 kg(1.20 m/s2 + 9.8)

T = 1683 (11)

T = 18513N

therefore, the tension is 18513N

4 0
3 years ago
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A particle moving in simple harmonic motion with a period T = 1.5 s passes through the equilibrium point at time t0 = 0 with a v
ch4aika [34]

Answer

given,

Time period= T = 1.5 s

If it's moving through equilibrium point at t₀= 0 with v = 1.0 m/s

v_max=1.00 m/s

we know,

v_ max=A ω  

v = A sin (ωt)

-0.50= -1.00 sin (ωt)

sin (ωt)  =  0.5

\omega t = sin^{-1}(0.5)

\dfrac{2\pi}{T}\times t =0.524

\dfrac{2\pi}{1.5}\times t =0.524

t = 0.125 s

we have time period T=1.5 it is the time to complete one oscillation

means from eq to right,then left,then eq,then left,then from right to eq

time taken for left = t/4 = 0.125/4 = 0.375 s

smallest value of time

=0.375 + 0.125

= 0.50 sec

7 0
3 years ago
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