Velocity is the rate of change of position with respect to time, whereas acceleration is the rate of change of velocity. Both are vector quantities (and so also have a specified direction), but the units of velocity are meters per second while the units of acceleration are meters per second squared.
- Weight (W) = 110 N
- Acceleration due to gravity (g) = 9.8 m/s^2
- Let the mass of the object be m.
- By using the formula, W = mg, we get,
- 110 N = 9.8 m/s^2 × m
- or, m = 110 N ÷ 9.8 m/s^2
- or, m = 11.2 Kg
<u>Answer:</u>
<em><u>The </u></em><em><u>mass </u></em><em><u>of </u></em><em><u>the </u></em><em><u>object </u></em><em><u>is </u></em><em><u>1</u></em><em><u>1</u></em><em><u>.</u></em><em><u>2</u></em><em><u> </u></em><em><u>Kg.</u></em>
Hope you could get an idea from here.
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Answer:
Explanation:
Maximum force of friction possible = μmg
= .65 x 3.8 x 9.8
= 24.2 N
u = 72 x 1000 / 60 x 60
= 20 m /s
v² = u² - 2as
a = 20 x 20 / (2 x 30)
= 6.67 m / s²
force acting on it
= 3.8 x 6.67
= 25.346 N
Friction force possible is less .
So friction will not be able to prevent its slippage
It will slip off .
a = 7.8 m/s^2
Explanation:
Let Fnet = net force = ma
m = mass of the skydiver
a = acceleration caused by Fnet
W = weight = mg
f(air) = frictional force due to air resistance
Fnet = W - f(air)
= (100 kg)(9.8 m/s^2) - (200 N)
= 780 N
Therefore, the acceleration of the skydiver due to Fnet is
a = Fnet/m
= (780 N)/(100 kg)
= 7.8 m/s^2
Yes because you would have at least 3 car spaces
