Answer to Question #284477 in Physics for Fruit

Question #284477

1.) How much impulse does a 10 N force produce if it is applied in 0.05s?


2.) A 0.80kg object at rest is given a force of 4.0N during a time interval of 2.0s What is the velocity of the object at the end of this time interval ?


3.) A 5.0kg ball rolls into a pillow at 2.0 m/s and stops in 5.0s, what is the average force exerted on the pillow?


4.) A 1500kg car moving east speeds up uniformly from 5.0 m/s to 30m/s in 8.0s. What constant force acts on the car


5.) How long must a 1500N force act on a 50kg cart to increase its speed from 50 m/s to 150 m/s



6.) A 300g ball moving at 20 m/s is hit by a bat. After the impact, the ball rebounds at the same speed along the same line. What was the impulse exerted by the bat on the ball?



1
Expert's answer
2022-01-03T09:01:19-0500

1)

J=FΔt=10 N×0.05 s=0.5 N×s.J=F\Delta t=10\ N\times0.05\ s=0.5\ N\times s.


2)

FΔt=Δp=m(vfvi)=mvf,F\Delta t=\Delta p = m(v_f-v_i)=mv_f,vf=FΔtm=4 N×2 s0.80 kg=10 ms.v_f=\dfrac{F\Delta t}{m}=\dfrac{4\ N\times2\ s}{0.80\ kg}=10\ \dfrac{m}{s}.


3)

FΔt=Δp=m(vfvi),F\Delta t=\Delta p = m(v_f-v_i),F=m(vfvi)Δt=5 kg×(02 ms)5 s=2 N.F=\dfrac{m(v_f-v_i)}{\Delta t}=\dfrac{5\ kg\times(0-2\ \dfrac{m}{s})}{5\ s}=-2\ N.


4)

F=m(vfvi)Δt=1500 kg×(30 ms5 ms)8 s=4688 N.F=\dfrac{m(v_f-v_i)}{\Delta t}=\dfrac{1500\ kg\times(30\ \dfrac{m}{s}-5\ \dfrac{m}{s})}{8\ s}=4688\ N.


5)

Δt=m(vfvi)F=50 kg×(150 ms50 ms)1500 N=3.33 s.\Delta t=\dfrac{m(v_f-v_i)}{F}=\dfrac{50\ kg\times(150\ \dfrac{m}{s}-50\ \dfrac{m}{s})}{1500\ N}=3.33\ s.


6)

J=m(vfvi)=0.3 kg×(20 ms(20 ms))=12 N×s.J=m(v_f-v_i)=0.3\ kg\times(20\ \dfrac{m}{s}-(-20\ \dfrac{m}{s}))=12\ N\times s.

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