Answer:
Initial Kinetic energy of alpha particle is 9.45x10⁻¹³ J .
Explanation:
The distance at which the initial kinetic energy of the particle is equal to the potential energy is known as closest distance. As it is Rutherford scattering, so it is a coulomb potential energy.
Let K be the initial kinetic energy of alpha particle and r be the closest approach distance. So,
Initial Kinetic Energy = Coulomb Potential Energy
K = 
Here, k is constant, e is charge of electron and Z is the atomic number of silver.
Put 9x10⁹ N m²/C² for k, 1.6x10⁻¹⁹ C for e, 47 for Z and 22.9x10⁻¹⁵ m for r in the above equation.
K = 
K = 9.45x10⁻¹³ J
Explanation :
Using the law of conservation of energy
When two cars collide with each other then the momentum is same before collision and after collision but energy is changed after collision in form of heat and sound.
We know this collision is inelastic collision.
In Inelastic collision, when two objects collide with each other then the momentum is conserved but kinetic energy is not conserved.
The displacement is 2 m south
Explanation:
Distance and displacement are two different quantities:
- Distance is the total length of the path covered by an object during its motion, regardless of the direction. It is a scalar quantity
- Displacement is a vector connecting the initial position to the final position of motion of an object. The magnitude of the displacement is the distance in a straight line between the two points
For the car in this problem, the motion is:
10 m south
8 m north
Taking north as positive direction, we can describe the two parts of the motion as
m
Therefore, the final position of the car with respect to the original position is
which means 2 m south: so, the displacement of the car is 2 m south.
Learn more about distance and displacement:
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A star may form a black hole
once is true. After the marked supernova explosion, the star shrinks because it
sheds away the gaseous layers of the star. And especially if the star is large,
it will form into a black hole.
Answer:
(a) The work done is 0.05 J
(b) The force will stretch the spring by 3.8 cm
Explanation:
Given;
work done in stretching the spring from 30 cm to 45 cm, W = 3 J
extension of the spring, x = 45 cm - 30 cm = 15 cm = 0.15 m
The work done is given by;
W = ¹/₂kx²
where;
k is the force constant of the spring
k = 2W / x²
k = (2 x 3) / (0.15)²
k = 266.67 N/m
(a) the extension of the spring, x = 37 cm - 35 cm = 2 cm = 0.02 m
work done is given by;
W = ¹/₂kx²
W = ¹/₂ (266.67)(0.02)²
W = 0.05 J
(b) force = 10 N
natural length L = 30 cm
F = kx
x = F / k
x = 10 / 266.67
x = 0.0375 m
x = 3.75 cm = 3.8 cm
Thus a force of 10 N will stretch the spring by 3.8 cm
