Answer:
time will elapse before it return to its staring point is 23.6 ns
Explanation:
given data
speed u = 2.45 × m/s
uniform electric field E = 1.18 × N/C
to find out
How much time will elapse before it returns to its starting point
solution
we find acceleration first by electrostatic force that is
F = Eq
here
F = ma by newton law
so
ma = Eq
here m is mass , a is acceleration and E is uniform electric field and q is charge of electron
so
put here all value
9.11 × kg ×a = 1.18 × × 1.602 ×
a = 20.75 × m/s²
so acceleration is 20.75 × m/s²
and
time required by electron before come rest is
use equation of motion
v = u + at
here v is zero and u is speed given and t is time so put all value
2.45 × = 0 + 20.75 × (t)
t = 11.80 × s
so time will elapse before it return to its staring point is
time = 2t
time = 2 ×11.80 ×
time is 23.6 × s
time will elapse before it return to its staring point is 23.6 ns
Answer:
a) 378Ns
b) 477.27N
Explanation:
Impulse is the defined as the product of the applied force and time taken. This is expressed according to the formula
I = Ft = m(v-u)
m is the mass = 70kg
v is the final velocity = 5.4m/s
u is the initial velocity = 0m/s
Get the impulse
I = m(v-u)
I = 70(5.4-0)
I = 70(5.4)
I = 378Ns
b) Average total force is expressed as
F = ma (Newton's second law)
F = m(v-u)/t
F = 378/0.792
F = 477.27N
Hence the average total force experienced by a 70.0-kg passenger in the car during the time the car accelerates is 477.27N
Answer:
100N
Explanation:
Newton's third law of motion
For every action, there is an equal and opposite reaction.
Therefore 100N of force is exerted by the crate on student as a reaction to his action
True
The sample of the experiment is randomized in randomization.