<u>Answer</u>
1.26 rad/s
the first step is to find the frequency of the wheel.
The wheel rotates at rate of 5 revolutions in 25 seconds. The frequency is the number of revolutions per second.
frequency (f) = 5/25
= 1/5
= 0.2 Hz
Angular velocity = 2πf
= 2×π×0.2
= 0.4π rad/s or
= 1.26 rad/s
Answer:
The time taken for the race is 17.20 s.
Explanation:
It is given in the problem that a 62.0 kg sprinter starts a race with an acceleration of 1.44 meter per second square.The initial speed of the sprinter is zero as it starts from the rest.
Calculate the final speed of the sprinter.
The expression for the equation of the motion is as follows;

Here, u is the initial speed, v is the final speed, a is the acceleration and s is the distance.
Put u= 0, s=30 m and
.


Calculate time taken to cover 30 m distance.
The expression for the equation of motion is as follows;

Put u= 0, s=30 m and
.

t=6.45 s
Calculate the time taken to complete his race.
T= t+t'
Here, t is the time taken to cover 30 m distance and t' is the time taken to cover 100 m distance.

Put s= 30 m,
and s'= 100 m.

T= 17.20 s
Therefore, the time taken for the race is 17.20 s.
Answer:
The planet will move from east to west for a couple of months in the night sky.
Explanation:
Retrograde motion is an optical effect due to the fact that Earth rotates more quickly than the planet that apparently has a retrograde motion in the sky.
For example, Saturn has a slower speed in its orbit around the Sun. That means that the Earth will pass it, and that will give the effect that the planet is moving backward. That same scenario can be seen between two cars on a highway, the faster car will see the slower car when it passes as it is moving for a short fragment of time in backward.
Remember that the planets in the night sky move from west to east, in the case of a planet with retrograde motion, it will move from east to west for a couple of months.
20/45=0.4*100= 44.4 so the answer is..................................................
Answer: 44.4%
Answer:
Option c) are perpendicular to the electric field
Explanation:
Equipotential surfaces are perpendicular to the electric field. the electric field lines are projected outwards from the equipotential surface, i.e., the lines of the electric field are at 90
to the equipotential surface.
Equipotential surface are those surfaces that have the same potential at any point on the surface. Thus the potential difference at any point on the surface is zero due to same potential.
Any charge particle on this surface will move in a perpendicular direction to the Coulombian force. No work is done by the force on a particle moving on an equipotential surface.