Answer:
A) The acceleration is zero
<em>B) The total distance is 112 m</em>
Explanation:
<u>Velocity vs Time Graph</u>
It shows the behavior of the velocity as time increases. If the velocity increases, then the acceleration is positive, if the velocity decreases, the acceleration is negative, and if the velocity is constant, then the acceleration is zero.
The graph shows a horizontal line between points A and B. It means the velocity didn't change in that interval. Thus the acceleration in that zone is zero.
A. To calculate the acceleration, we use the formula:

Let's pick the extremes of the region AB: (0,8) and (12,8). The acceleration is:

This confirms the previous conclusion.
B. The distance covered by the body can be calculated as the area behind the graph. Since the velocity behaves differently after t=12 s, we'll split the total area into a rectangle and a triangle.
Area of rectangle= base*height=12 s * 8 m/s = 96 m
Area of triangle= base*height/2 = 4 s * 8 m/s /2= 16 m
The total distance is: 96 m + 16 m = 112 m
Answer:

Explanation:
As per energy conservation we know that the electrostatic potential energy of the charge system is equal to the initial kinetic energy of the alpha particle
So here we can write it as

now we know that


z = 79
here kinetic energy of the incident alpha particle is given as

now we have

now we have

Answer:
6.23x10^6Pa
Explanation:
Data obtained from the question include:
F (force) = 490N
r (radius) = 0.005m
A (area of the circlular heel) =?
P (pressure) =.?
First, we'll begin by calculating the area of the circlular heel. This is illustrated below:
Area of circle = πr^2
Area = 22/7 x (0.00)^2
Area = 7.86x10^-5m^2
Pressure is simply force per unit area. It represented mathematically as
Pressure = Force /Area
Pressure = 490/7.86x10^-5
Pressure = 6.23x10^6N/m2
Recall: 1N/m2 = 1Pa
Therefore, 6.23x10^6N/m2 = 6.23x10^6Pa
Therefore, the woman exert a pressure of 6.23x10^6Pa on the floor
Answer:
The kinetic energy lost in the collision is 48 J
Explanation:
Given;
mass of the first ball, m₁ = 2.0 kg
mass of the second ball, m₂ = 6.0 kg
initial speed of the first ball, u₁ = 12 m/s
initial speed of the second ball, u₂ = 4 m/s
let v be the final velocity of the two balls after the inelastic collision
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
2 x 12 + 6 x 4 = v(2 + 6)
48 = 8v
48 / 8 = v
v = 6 m/s
The initial kinetic energy of the balls is calculated as;
K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²
K.E₁ = 144 + 48
K.E₁ = 192 J
The final kinetic of the balls is calculated as;
K.E₂ = ¹/₂(m₁ + m₂)(v²)
K.E₂ = ¹/₂(2 + 6)(6²)
K.E₂ = ¹/₂(8)(6²)
K.E₂ = 144 J
The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J
Therefore, the kinetic energy lost in the collision is 48 J
According to another source this is what I got
<span>0.735 J ( Ep-potential energy, m-mass,g-gravitational acceleration = 9.81m/s², h-height; Ep = m * g * h; Ep = 0.0300 kg * 9.81 m/s² * 2.5 m )
</span>Hope it helps