The acceleration of gravity is 9.8 m/s² . That means that a falling object
is always falling 9.8 m/s faster than it was falling 1 second earlier.
If an object is not slowed by air resistance, and has far enough to go
so that it's still falling after three whole seconds, then at the end of
three seconds it's falling at
(9.8 m/s²) x (3 sec) = 29.4 m/s
I think it’s A, I’m so sorry if I’m wrong.
if the velocity of the car reduces from 70km/h to 50km/h then the speed of the car will be equal to the speed of the lorry...
thus the relative velocity will be 0
Answer:
Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.
Explanation:
The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.
When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.
The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.
The velocity of the orbit is given by the relation,

Where
V - velocity of the orbit at a height h from the surface
R - Radius of the second object
G - Gravitational constant
h - height from the surface
The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.
Answer:
(a)106.4C
b)0.5676mm
Explanation:
(a)To get the charge that have passed through the starter then The current will be multiplied by the duration
I= current
t= time taken
Q= required charge
Q= I*t = 140*0.760 = 106.C
(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)
diameter of the conductor is 4.20 mm
But Radius= diameter/2= 4.20/2=
The radius of the conductor is 2.1mm, then if we convert to metre for consistency same then
radius of the conductor is 0.0021m.
We can now calculate the area of the conductor which is
A = π*r^2
= π*(0.0021)^2 = 13.85*10^-6 m^2
We can proceed to calculate the current density below
J = 140/13.85*10^-6 = 10108303A/m
According to the listed reference:
Where e= 1.6*10^-19
n= 8.46*10^28
Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 ) =0.0007468m/s=0 .7468 mm/s
Therefore , the distance traveled is:
x = v*t = 0.7468 * 0.760 = 0.5676mm