T=2s
g=10m/s2
h=?
free fall: h=gt2/2
h= 10*4/2
h=40/2
h=20m
Answer:
4500 N
Explanation:
When a body is moving in a circular motion it will feel an acceleration directed towards the center of the circle, this acceleration is:
a = v^2/r
where v is the velocity of the body and r is the radius of the circumference:
Therefore, a body with mass m, will feel a force f:
f = m v^2/r
Therefore we need another force to keep the body(car) from sliding, this will be given by friction, remember that friction force is given a the normal times a constant of friction mu, that is:
fs = μN = μmg
The car will not slide if f = fs, i.e.
fs = μmg = m v^2/r
That is, the magnitude of the friction force must be (at least) equal to the force due to the centripetal acceleration
fs = (1000 kg) * (30m/s)^2 / (200 m) = 4500 N
Answer:
a) 19440 km/h²
b) 10 sec
Explanation:
v₀ = initial velocity of the car = 45 km/h
v = final velocity achieved by the car = 99 km/h
d = distance traveled by the car while accelerating = 0.2 km
a = acceleration of the car
Using the kinematics equation
v² = v₀² + 2 a d
99² = 45² + 2 a (0.2)
a = 19440 km/h²
b)
t = time required to reach the final velocity
Using the kinematics equation
v = v₀ + a t
99 = 45 + (19440) t
t = 0.00278 h
t = 0.00278 x 3600 sec
t = 10 sec
Answer:
As we keep on increasing the radius the value of the gravitation force of attraction decreases and as we decrease the radius the gravitation force increases.
Explanation:
Like the coulombs law of electrostatics, the law of gravitation also depends inversely on the square of the value of r. Therefore, as we keep on increasing the value of r the value of the gravitation force decreases and as we decrease the value of the r the value of gravitation force increases.
Gravitation Force=
Coulombs's Law= 
