Answer:
Explanation:
Given
length of window 
time Frame for which rock can be seen is 
Suppose h is height above which rock is dropped
Time taken to cover 
so using equation of motion

where y=displacement
u=initial velocity
a=acceleration
t=time
time taken to travel h is

Subtract 1 and 2 we get


and from equation 
so 

and 
so 



substitute the value of
in equation 2


Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
Answer:
Time, t = 0.87 seconds
Explanation:
Given that,
Initial velocity of the object, u = 4.3 m/s
The coefficient of kinetic friction between horizontal tabletop and the object is 0.5
We need to find the time taken by the object for the object to come to rest i.e. final velocity will be 0.
Using first equation of motion to find it as :

a is the acceleration, here, 


So, the time taken by the object to come at rest is 0.87 seconds. Hence, this is the required solution.