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:
12.5 m/s
Explanation:
The motion of the hammer is a free fall motion, so a uniformly accelerated motion, therefore we can use the following suvat equation:

Where, taking downward as positive direction, we have:
s = 8 m is the displacement of the hammer
u = 0 is the initial velocity (it is dropped from rest)
v is the final velocity
is the acceleration of gravity
Solving the equation for v, we find the final velocity:

So, the final speed is 12.5 m/s.
Answer:
5x10^-3
Explanation:
Hooke's Law states that the force needed to compress or extend a spring is directly proportional to the distance you stretch it.
Hooke's Law can be represented as
<h3> F = kx, </h3>
<em>where F is the force </em>
<em> k is the spring constant</em>
<em> x is the extension of the material </em>
<em />
Plug values in the equation
Step 1 find the original extension
0.045 = (400)x
x = 1.125x 10^-4 m d
Step 2 find the new extension
0.045+2 = 400(x)
2.045 = 400x
x = 5.1125x10^-3
Step 3 subtract the new extension with original
Total extension of the spring = 5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3
Answer:

Explanation:
The relationship between the linear distance covered by an object and its angular displacement is given by the following formula:
s = rθ
where,
s = distance traveled on road = ?
r radius of tires = diameter/2 = 2.2 m/2 = 1.1 m
θ = angular displacement = (60 rev)(2π rad/1 rev) = 377 rad
Therefore,
