Answer: The motion of the object will remain the same
Explanation:
Answer:
b. 20 sec
Explanation:
y = y₀ + v₀ t + ½ g t²
0 = 0 + (100) t + ½ (-10) t²
0 = 100t − 5t²
0 = t (100 − 5t)
t = 0, t = 20
The body lands after 20 seconds.
Answer:
<em>600N.</em>
Explanation:
From the question, we are to calculate the net force acting on the car.
According to Newton's second law of motion:
F = ma
m is the mass of the car
a is the acceleration = change in velocity/Time
a = v-u/t
F = m(v-u)/t
v is the final velocity = 30m/s
u is the initial velocity = 20m/s
t is the time = 5secs
m = 300kg
Get the net force:
Recall that: F = m(v-u)/t
F = 300(30-20)/5
F = 60(30-20)
F = 60(10)
<em>F = 600N</em>
<em>Hence the net force acting on the car is 600N.</em>
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Answer:
Explanation:
charge, q = 1.6 x 10^-19 C
distance, r = 911 nm = 911 x 10^-9 m
The Coulomb's force is given by
F = 2.78 x 10^-16 N
The force between the electron and the proton is 2.78 x 10^-16 N.
Answer:
The coefficient of kinetic friction between the crate and the floor can be calculated using the formula μ = Ff / N, where Ff is the frictional force, N is the normal force, and μ is the coefficient of kinetic friction.
In this case, the normal force is equal to the weight of the crate, which is 24 kg * 9.8 m/s2 = 235.2 N. The frictional force can be calculated using the formula Ff = μ * N, where μ is the coefficient of kinetic friction and N is the normal force.
If we substitute the values for N and Ff into the formula for the coefficient of kinetic friction, we get:μ = 53 N / 235.2 N = 0.225
Therefore, the coefficient of kinetic friction between the crate and the floor is 0.225.
