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Answer:
The normal force the seat exerted on the driver is 125 N.
Explanation:
Given;
mass of the car, m = 2000 kg
speed of the car, u = 100 km/h = 27.78 m/s
radius of curvature of the hill, r = 100 m
mass of the driver, = 60 kg
The centripetal force of the driver at top of the hill is given as;
where;
Fc is the centripetal force
is downward force due to weight of the driver
is upward or normal force on the drive
Therefore, the normal force the seat exerted on the driver is 125 N.
Answer:
v =7.1 m/s
Explanation:
Given that
u = 3.35 m/s
t= 5 s
a= 0.75 m/s²
The final velocity = v
We know v = u +at
v=final velocity
u=initial velocity
Now by putting the values in the above equation
v = 3.35 + 0.75 x 5 m/s
v =7.1 m/s
Therefore the final velocity will be 7.1 m/s
An atom. Hope this helped
Mohs hardness scale. hope this helps
