Density =mass/volume
density= (0.044kg/(11 x10^-6)
4000kg/m^3
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Given:
V1 = initial velocity = 24 m/s
a = acceleration = 2.0 m/s^2
s = time = 8 s
V2 = final velocity = ?
For linear-motion problems with those given terms, the following formula is used:
V2 = V1 + as
Substituting the given values:
V2 = 24 + 2(8)
V2 = 24 + 16
V2 = 40 m/s
Therefore the car will have a speed of 40 m/s as it leaves the tunnel.
<span>motion of truck constitutes of 3 travels.
1. accelerating uniformly with acceleration a1 = 2 m/s^2 until its velocity reached 20 m/s travelling a
distance of 's1' meters.
2. uniform motion with 20 m/s for a time duration t1 = 20s travelling a distance of 's2' meters.
3. uniform deceleration for t2 = 5 sec which stops the truck after travelling a distance of 's3' meters.. </span>
No. No because if you use force on something that doesn't move it doesn't change. therefore you are not doing work.
