Answer:0.27
Explanation:
Given
One worker Pushes with force 
other Pulls it with a rope of rope 
mass of crate 
both forces are horizontal and crate slides with a constant speed
Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it
where 
is the friction force
where 
is the coefficient of static friction
Answer:
3054.4 km/h
Explanation:
Using the conservation of momentum
momentum before separation = 5M × 2980 Km/h where M represent the mass of the module while 4 M represent the mass of the motor
initial momentum = 14900 M km/h
let v be the new speed of the motor so that the
new momentum = 4Mv and the new momentum of the module = M ( v + 94 km/h )
total momentum = 4Mv + Mv + 93 M = 5 Mv + 93M
initial momentum = final momentum
14900 M km/h = 5 Mv + 93M
14900 km/h = 5v + 93
14900 - 93 = 5v
v = 2961.4 km/h
the speed of the module = 2961.4 + 93 = 3054.4 km/h
Answer:
1.0s
Explanation:
distance = 1/2 × acceleration × time2 + intial speed × time
