Using the law of conservation of angular momentum, we have
<span>I1 w1 = I2 w2 </span>
<span>ie., m1r^2/2 x w1 = ( m1r^2/2 + m2r^2 ) w2 </span>
<span>ie., new angular velocity w2 = m1 w1 / ( m1+ 2m2) = 125 x 3.1 / ( 125 + 2 x39.5 ) </span>
<span>= 1.8995 = 1.9 rad /sec ( nearly )</span>
Answer:
<em>a. t = 2.02 s
</em>
<em>b. d = 20.2 m</em>
Explanation:
<u>Horizontal Motion
</u>
If an object is thrown horizontally from a height h with a speed v, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.
The time the object takes to hit the ground can be calculated as follows:

The time does not depend on the initial speed.
The range or maximum horizontal distance traveled by the object can be calculated by the equation:

The man standing on the edge of the h=20 m cliff throws a rock with an initial horizontal speed of v=10 m/s.
a.
The time taken by the rock to reach the ground is:


t = 2.02 s
b.
The range is:

d = 20.2 m
The correct option is this: IT WOULD MOVE IN A CURVED CIRCULAR PATH.
Objects that are travelling in circular paths change directions all the time as they move round the circle, but they are prevented from moving off in a straight line by centripetal force. The centripetal force keeps pulling the objects towards the center of the circle. <span />
Answer:
4.04 s
Explanation:
h = vi + 1/2 a t ^2
HERE h = 80 m , vi = 0 , a =9.81 m/s^2
80 = 0 + 1/2 × 9.81 × t ^2
80 = 4.905 t^2
t^2 = 80/4.905
t ^2 = 16.30988
t = square root of 16.30988
t = 4.0385 s
t = 4.04 s
Answer:
The electric field outside the sphere will be
.
Explanation:
Given that,
Radius of solid sphere = R
Charge = q
According to figure,
Suppose r is the distance between the point P and center of sphere.
If
be the volume charge density,
Then, the charge will be,
.....(I)
Consider a Gaussian surface of radius r.
We need to calculate the electric field outside the sphere
Using formula of electric field


Put the value from equation (I)


Hence, The electric field outside the sphere will be
.