Answer:
She stone hit ground 42.86 m far from base of cliff.
Explanation:
Initial height from ground = 40 meter.
We have equation of motion ,
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this the velocity of body in vertical direction = 0 m/s, acceleration = 9.8
, we need to calculate time when s = 40 meter.
Substituting
So it will take 2.86 seconds to reach ground.
So, stone travels horizontally at 15 m/s for 2.86 seconds.
Distance travelled = 15 x 2.86 = 42.86 m
So, the stone hit ground 42.86 m far from base of cliff.
Answer:
Explanation:
a ) F = (-kx + kx³/a²)
intensity of field
I = F / m
= (-kx + kx³/a²) / m
If U be potential function
- dU / dx = (-kx + kx³/a²) / m
U(x) = ∫ (kx - kx³/a²) / m dx
= k/m ( x²/2 - x⁴/4a²)
b )
For equilibrium points , U is either maximum or minimum .
dU / dx = x - 4x³/4a² = 0
x = ± a.
dU / dx = x - x³/a²
Again differentiating
d²U / dx² = 1 - 3x² / a²
Put the value of x = ± a.
we get
d²U / dx² = -2 ( negative )
So at x = ± a , potential energy U is maximum.
c )
U = k/m ( x²/2 - x⁴/4a²)
When x =0 , U = 0
When x= ± a.
U is maximum
So the shape of the U-x curve is like a bowl centered at x = 0
d ) Maximum potential energy
put x = a or -a in
U(max) = k/m ( x²/2 - x⁴/4a²)
= k/m ( a² / 2 - a⁴/4a²)
= k/m ( a² / 2 - a²/4)
a²k / 4m
This is the maximum total energy where kinetic energy is zero.
Answer:
K = 0.045 J
Explanation:
It is given that,
Mass of a ball, m = 10 g = 0.01 kg
Speed of the ball, v = 3 m/s
To find,
The energy of the ball.
Solution,
Due to the motion of the ball, it will have kinetic energy. It can be given by the formula as follows :

So, the ball will have a kinetic energy of 0.045 J.
<span>Solar panels convert light energy from sunlight into electricity energy , Metalloid is most likely used in solar panels The answer is : </span> A metalloid is used because it is a semiconductor and can become more conductive when more light shines on it. Metalloids are shiny,<span> semiconductive and they are brittle.</span>
6N I think I’m pretty sure