Answer:
Angle of projection from the horizontal will be
Explanation:
We have given that range of the projectile is 10 times the height of the projectile
Let the projectile is projected with velocity u at an angle
Range of the projectile is equal to
And height of the projectile is equal to
Now according to question range is 10 times of height
So
So angle of projection from the horizontal will be
(a) 1.01 cm
First of all, we need to find the spring constant of the spring.
The force initially applied to the spring is equal to the weight of the block hanging on it:
where m = 4.32 kg is the mass of the block and g = 9.8 m/s^2 is the acceleration of gravity.
When this force is applied, the spring stretches by
We can find the spring constant by using Hooke's law:
where k is the spring constant. Solving for k,
Later, the first object is removed and another object of mass
m' = 1.50 kg
is hung on the spring. The weight of this object is
So, if we use Hooke's law again, we can find the new stretching of the spring:
(b) 1.16 J
The work that must be done on the spring is equal to the elastic potential energy that would be stored in the spring, therefore:
where we have
k = 1448.6 N/m is the spring constant
x = 4.00 cm = 0.04 m is the new stretching
Solving the equation, we find the work that must be done by the external force:
Answer:
Explanation:
According to Coulomb's law, the magnitude of the electric force between two equals charges (q) is given by:
Here k is the coulomb constant and d is the distance between the charges. For two electrons we have: