21. The total resistance of a 6-ohm resistor and a second 6-ohm resistor in parallel is Question 21 options:

a)

37 ohms.

b)

20 ohms.

c)

12 ohms.

d)

6 ohms.

e)

3 ohms.

22. A graduation cap is thrown straight up by a graduate. It rises vertically. Which one of the following statements describes the energy transformation of the cap as it rises? Neglect air resistance. Question 22 options:

a)	Both the potential energy and the total energy of the cap increase.

b)	The total energy of the cap increases.

c)	The kinetic energy decreases and the potential energy increases.

d)	The kinetic energy increases and the potential energy decreases.

e)	Both the kinetic energy and the potential energy of the cap remain the same.

23. A 36 g golf ball is projected upwards from a level field with an initial vertical component of 29.4 m/s and a horizontal component of 9.8 m/s. Frictional forces are considered negligible. What is the total time that the ball remains in the air?

Question 23 options:

a)

288 s

b)

3.00 s

c)

12.0 s

d)

6.00 s

13. A ball is tossed Northward with a speed of 6.00 m/s at an angle of 60.0 above the horizontal. At its highest point, what are its velocity v and its acceleration a? Question 13 options:

a)	v = 3.00 m/s North, a = 0

b)	v = 3.00 m/s North, a = 9.80 m/s2 down

c)	v = 0, a = 9.80 m/s2 down

d)	v = 0, a = 0

14. The coefficient of static and kinetic frictions between a 6.0 kg box and a desk are 0.10 and 0.15, respectively. What is the net force on the box when a 10 N horizontal force is applied to the box?

Question 14 options:

a)

5.9 N

b)

2.2 N

c)

8.8 N

d)

zero

Solve and explain each question

Given . . Initial Speed = V. = 6 . Angle = 0 = 60" above horizontal. . At maximum height, v= ? , a= ? Salution: - This is an case of projectile motion, . As you com see that, at maximum height of the Hmax projectile, velocity component is directed along horizontal . . vertical component is zero at max, height. . Gravitational acceleration is acting downward always . Due to gravitational acceleration, ball will be moving along the path after max . height . # Speed :- horizontal component of velocity is always constant because acceleration along horizontal dirn is zero. Vx = V. coso = 6x ( 08 60 = 67 - Vx = 3 mys It is constant and some every where along the path of the projectile.. vertical component of velocity at max height is zero . Vy = D (2 ". Resultant velocity will be :- V = d ( vaj + ( Vy ) 2 = 1 (3)2+ 0 = 132 V = 3 m along horizontal. Speed at maximum height. # acceleration ?- . Horizontal component = ax = 0 . vertical component = qy = 9180 m ( down ) F a = ( ax ) ? + ( ag ) 2 ay a = of o+ ( 9 80 ) 2 a = 9.80 . as 9x= 0 so net acceleration is along vertical component. 9 = 9180 12 ( down ) 4 acceleration . Ans