The momentum of the ball when it hits the ground is 4.89 kg.m/s.
The given parameters;
- <em>mass of the baseball, m = 0.145 kg</em>
- <em>height of fall of the ball, h = 58 m</em>
The final velocity of the ball when it hits the ground is calculated as follows;

The momentum of the ball when it hits the ground is calculated as follows;
P = mv
P = 0.145 x 33.72
P = 4.89 kg.m/s
Thus, the momentum of the ball when it hits the ground is 4.89 kg.m/s.
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Answer:
Explanation:
For lens A
object distance u = - 13.1 cm , focal length f = 6.19 cm
From lens formula
1/v - 1/u = 1/f
1 / v + 1/13.1 = 1/6.19
1/v = 1/6.19 - 1/13.1
= .16155 - .07633
= .08522
v = 11.7 3 cm
For lens B
object distance u = - ( 55.7 - 11.73) = - 43.97 cm , focal length f = 27.9 cm
From lens formula
1/v - 1/u = 1/f
1 / v + 1/43.97 = 1/27.9
1/v = 1/27.9 - 1/43.97
= .03584 - .022742
= .013098
v = 76.35 cm
Image will be formed 76.35 cm behind lens B .
magnification of lens system
= m₁ x m₂ , m₁ is magnification by lens A and m₂ is magnification by lens B
= (11.73 / 13.1) x (76.35 / 43.97)
= .8954 x 1.73
= 1.5547
size of image = total magnification x size of object
= 1.5547 x 6.47
= 10 cm approx. The first image will be real and inverted and second image will be erect with respect to object.
We will determine the wavelength through the relationship given by the distance between slits, this relationship is given under the function

Here,
m = Number of order bright fringe
= Wavelength
d = Distance between slits
Both distance are the same, then



Rearranging to find the second wavelength




Therefore the wavelength of the light coming from the second monochromatic light source is 550.3nm
Answer:
0.2
Explanation:
The given parameters are;
The acceleration of the train, a = 0.2·g
The mass of the person standing on the train = m
Let μ represent the coefficient of static friction, we have;
The force acting on the person, F = m × a = m × 0.2·g
The force of friction acting between the feet and the floor,
= m·g·μ
For the person not to slide we have;
The force acting on the person = The force of friction acting between the feet and the floor
F = 
∴ m × 0.2·g = m·g·μ
From which we get;
0.2 = μ
The coefficient of static friction that must exist between the feet and the floor if the person is not to slide, μ = 0.2.
Complete Question
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?
Answer:


Explanation:
From the question we are told that
Mass of ball 
Length of string 
Wind force 
Generally the equation for
is mathematically given as




Max angle =
Generally the equation for max Height
is mathematically given as



Generally the equation for Equilibrium Height
is mathematically given as


