Answer:
156.26N
Explanation:
The data needed are incomplete. Let the acceleration of the body be 3.5m/s²
Other given parameters
Mass = 1.35×10^1 = 13.5kg
coefficient of friction between the tires and the road = 0.850
Acceleration due to gravity = 9.8m/s²
According to Newton's second law:
Fnet = ma
Fnet = Fapp - Ff
Fapp is the applied force
Ff is the frictional force = umg
The equation becomes:
Fapp - Ff = ma
Fapp-umg = ma
Fapp - 0.85(13.5)(9.8) = 13.5(3.5)
Fapp - 109.0125 = 47.25
Fapp = 47.25+109.0125
Fapp = 156.2625N
Hence the applied force that caused the acceleration is 156.26N
Note that the acceleration of the car was assumed. Any value of acceleration can be used for the calculation.
Mike enters a revolving door that is not moving. Mike should
push at the edge of the door where it is largest distance from the pivot point
in order to produce a torque with the least amount of force. Torque is equal to
t = force x distance.
Answer:
Force F = 69.35 N
Explanation:
given data
Ball Initial speed u = 0
Ball Final speed v = 42.1 m/s
average power generate = 2920 W
solution
Power generate is express as
P= 
..............1
here W is work done and t is time
and work w = F × d
so
P= 
and we know speed v = 
so here
Power P = F × v
put here value and we get force
Force F = 
Force F = 69.35 N
Answer:
10 km/hr/s
Explanation:
The acceleration of an object is given by
where
v is the final velocity
u is the initial velocity
t is the time
For the car in this problem:
u = 0
t = 6 s
Substituting in the equation,
