Answer : (A) A charged object is brought near a neutral object without touching it.
Explanation:
Bodies can be charged by the method of conduction. By conduction the body acquires the same charge as on the charging body.
Charging the body can be understood by the following example of charging a paper cylinder by conduction. Make a paper cylinder by rolling a strip of paper on a pencil and then gently pulling out the pencil. Suspend the paper cylinder by a string tied to its center, now touch the paper cylinder with a glass rod rubbed with silk so it has a positive charge. Remove the class rod and the again bring it near to the paper cylinder.
The paper cylinder is repelled by the positively charged rod. And when you bring a negatively rubbed rod, next to the cylinder it will be attracted towards the rod. This means that the paper cylinder has acquired a positive charge as on the glass rod due to conduction.
Answer:
C
Explanation:
acceleration is a change in velocity which means a time rate change in velocity but velocity is the time rate change in displacement
Answer:
The final velocity is 28.14 m/s
Yes the angle of projection matters
Explanation:
Given;
initial velocity of the water balloon, u = 20 m/s
height of the building, h = 20 m
let the final speed of the ball when it hits the ground = v
The final speed is calculated as follows;
v² = u² + 2gh
v² = (20)² + 2(9.8)(20)
v² = 400 + 392
v² = 792
v = √792
v = 28.14 m/s
Yes the angle matters, if the balloon had been dropped at a certain angle, the final velocity would have been estimated using the following formula;

where;
θ is the angle of projection, which accounts for the vertical component of the velocity.
Answer:
Half
Explanation:
Given that:
- radial distance of satellite from the earth,

Now, if the satellite is moved to a distance 
<u>We have the mathematical expression for the potential energy fue to gravitational field as:</u>
...................(1)
where:

M = mass of earth
m = mass of satellite
R = radial distance of satellite
<u>Now from eq. (1) initially we have:</u>

<u>after the satellite is moved, we have:</u>



which is half of the initial condition.
Answer:
No.
Explanation:
Given the following :
Velocity (V) of ball = 5m/s
Radius = 1m
Can the ball reach the highest point of the circular track
of radius 1.0 m?
The highest point in the track could be considered as the diameter of the circle :
Radius = diameter / 2;
Diameter = (2 * Radius) = (2*1) = 2
Maximum height which the ball can reach :
Using the relation :
Kinetic Energy = Potential Energy
0.5mv^2 = mgh
0.5v^2 = gh
0.5(5^2) = 9.8h
0.5 * 25 = 9.8h
12.5 = 9.8h
h = 12.5 / 9.8
h = 1.2755
h = 1.26m
Therefore maximum height which can be reached is 1.26m.
Since h < Diameter