A 2.5 kg stone is released from rest and falls toward Earth. After 4.0 s, magnitude of its momentum in kg.m/s is : A-78 B-98

Question

A 2.5 kg stone is released from rest and falls toward Earth. After 4.0 s, magnitude of its momentum in kg.m/s is :
A-78
B-98
C-122.5
D-24.7

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Khoii Minh 1 day 2021-07-22T06:25:05+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-22T06:26:10+00:00

    Answer:B

    Explanation:

    Initial velocity(u)=0

    Time(t)=4seconds

    Mass(m)=2.5kg

    acceleration due to gravity(g)=9.8m/s^2

    Final velocity(v)=u+gxt

    Final velocity(v)=0+9.8×4

    Final velocity(v)=39.2m/s

    Momentum=mass x velocity

    Momentum=2.5 x 39.2

    Momentum=98kgm/s

    0
    2021-07-22T06:26:28+00:00

    Answer:

    B

    Explanation:

    momentum = mass * velocity

    velocity = initial speed + acceleration * time where initial speed =0, acceleration = 9.81, time = 4

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