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Ad libitum [116K]
2 years ago
5

An object accelerates 2 m/s2 when an unknown force of 20 N is applied to it. What was the mass of the object?

Physics
1 answer:
Crazy boy [7]2 years ago
3 0

Answer:

<h2>10 kg</h2>

Explanation:

The mass of an object given only the force acting on it and it's acceleration can be found by using the formula

m =  \frac{f}{a}  \\

f is the force in N

a is the acceleration in m/s²

We have

m =  \frac{20}{2}  = 10 \\

We have the final answer as

<h3>10 kg</h3>

Hope this helps you

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A 125kg bumper car going 12m/s hits a 235kg bumper car going -13m/s.if the first car bounces back at -12.5m/s what is the veloci
vovikov84 [41]
According to the law of conservation of momentum:

m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'

m1 = mass of first object
m2 = mass of second object
v1 = Velocity of the first object before the collision
v2 = Velocity of the second object before the collision
v'1 = Velocity of the first object after the collision
v'2 = Velocity of the second object after the collision

Now how do you solve for the velocity of the second car after the collision? First thing you do is get your given and fill in what you know in the equation and solve for what you do not know. 

m1 = 125 kg     v1 = 12m/s      v'1 = -12.5m/s
m2 = 235kg      v2 = -13m/s     v'2 = ?

m_{1}v_{1}+m_{2}v_{2}=m_{1}v_{1}'+m_{2}v_{2}'
(125kg)(12m/s)+(235kg)(-13m/s)=(125kg)(-12.5m/s)+(235kg)(v_{2}'
1,500kg.m/s+(-3055kg.m/s)=(-1562.5kg.m/s)+(235kg)(v_{2}')
-1,555kg.m/s=(-1562.5kg.m/s)+(235kg)(v_{2}')

Transpose everything on the side of the unknown to isolate the unknown. Do not forget to do the opposite operation. 

-1,555kg.m/s + 1562.5kg.m/s=(235kg)(v_{2}')
7.5kg.m/s=(235kg)(v_{2}')
(7.5kg.m/s)/(235kg)=(v_{2}')
0.03m/s=(v_{2}')

The velocity of the 2nd car after the collision is 0.03m/s.
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3 years ago
The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?
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Answer:

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Explanation:

Actual weight =   60 .3 (9.81) = 591.54 N

Accel of lift changes this to    60.3 ( 9.81 - L)     where L - accel of lift

                                           60.3 ( 9.81 - L ) = 564

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                                                        so L = - .457 m/s^2

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1)4*10^7 * 5= 2*10^8

2)50/10=5

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