Answer:
2.40 x 10⁻¹³ C
Explanation:
= number of electrons = 6.25 x 10⁶
= charge on electron = - 1.6 x 10⁻¹⁹ C
= number of protons = 7.75 x 10⁶
= charge on proton = 1.6 x 10⁻¹⁹ C
Net charge is given as
Q = +
Q = (- 1.6 x 10⁻¹⁹) (6.25 x 10⁶) + (1.6 x 10⁻¹⁹) (7.75 x 10⁶)
Q = 2.40 x 10⁻¹³ C
Given:
(Initial velocity)u=20 m/s
At the maximum height the final velocity of the ball is 0.
Also since it is a free falling object the acceleration acting on the ball is due to gravity g.
Thus a=- 9.8 m/s^2
Now consider the equation
v^2-u^2= 2as
Where v is the final velocity which is measured in m/s
Where u is the initial velocity which is measured in m/s
a is the acceleration due to gravity measured in m/s^2
s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.
Substituting the given values in the above formula we get
0-(20x20)= 2 x- 9.8 x s
s= 400/19.6= 20.41m
Thus the maximum height attained is 20.41 m by the ball
Answer:
v = -14 m/s
Explanation:
Given that,
Initial location of the ball, X₁ = 10 m
Final position of the ball, X₂ = -25 m
Time taken to travel is, t = 2.5 s
The average velocity of the ball is given by the formula,
V = X₂ - X₁ / t m/s
Substituting the values in the above equation,
V = -25 - 10 / 2.5
= -14 m/s
The negative sign in the velocity indicates that ball rolls in the opposite direction.
Hence, the average velocity of the ball is v = -14 m/s
