The body will stay at rest (Option D). It is because a force of magnitude 50 N is pulled towards left and another force is pulling it towards right with same magnitude 50 N. So, the direction of force is opposite and magnitude is same i.e. 50 N. So, they will cancel each other and net force is 0. Hence, there would be no acceleration.

Option A - Showing acceleration
Option B - Showing acceleration
Option C - Change of direction due to Net force

Hence, these options are incorrect because they are only possible when net external force is non-zero. Staying at rest i.e. Option D means there is no motion and hence no acceleration, this shows that net force is 0 $.$

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3 0

3 years ago

A cargo spacecraft has been launched to rendezvous with the International Space Station. The cargo ship must attain the same spe

Charra [1.4K]

The answer to this question is b

8 0

4 years ago

Identify a necessary part of the process by which legal immigrants become citizens.

beks73 [17]

The Naturalization Test

5 0

3 years ago

Read 2 more answers

You are rushing to the train station to catch your morning commute. The train leaves the train station from rest with an acceler

Vanyuwa [196]

Answer:

a) t= 4.81 s,and t= 23.51.

b)d=6.88 m

c)v=4.8 m/s

Explanation:

Acceleration of train ,a= 0.6 m/s²

u = 0 m/s

Your speed ,V= 8.5 m/s

Lets take after t time he you catch the train

Distance travel by train in t time

$d=ut+\dfrac{1}{2}at^2$

$d=\dfrac{1}{2}\times 0.6\times t^2$ ----------1

d= V ( t- 4)

d= 8.5 ( t- 4) --------2

By equating equation 1 and 2

0.3 t² = 8.5 ( t- 4)

0.3 t² -8.5 t + 34 = 0

t= 4.81 s and t= 23.51

It means that you will catch train after t= 4.81 s,and t= 23.51.

t=23.51 sec means that you will catch the train after 23.51 sec also because acceleration of train is low.

Distance travel

d= 8.5 ( t- 4)

t= 4.81 s

d= 8.5 ( 4.81- 4) m

d=6.88 m

Lets speed = v

0.3 t² = v ( t- 4)

0.3 t² - v t + 4 v = 0

To have one solution

$D=\sqrt{b^2-4ac}$

D= 0 Should be zero.

v²- 4 x 0.3 x 4 v

v = 4 x 0.3 x 4

v=4.8 m/s

6 0

3 years ago