This instrument is called a spring scale.
Working of a Half wave rectifier
The diode is connected in series with the secondary of the transformer and the load resistance RL. The primary of the transformer is being connected to the ac supply mains. The ac voltage across the secondary winding changes polarities after every half cycle of the input wave.
Given that force is applied at an angle of 30 degree below the horizontal
So let say force applied if F
now its two components are given as


Now the normal force on the block is given as



now the friction force on the cart is given as



now if cart moves with constant speed then net force on cart must be zero
so now we have




so the force must be 199.2 N
This problems a perfect application for this acceleration formula:
Distance = (1/2) (acceleration) (time)² .
During the speeding-up half: 1,600 meters = (1/2) (1.3 m/s²) T²
During the slowing-down half: 1,600 meters = (1/2) (1.3 m/s²) T²
Pick either half, and divide each side by 0.65 m/s²:
T² = (1600 m) / (0.65 m/s²)
T = square root of (1600 / 0.65) seconds
Time for the total trip between the stations is double that time.
T = 2 √(1600/0.65) = <em>99.2 seconds</em> (rounded)
Answer:
time rising = 34 / 9.8 = 3.47 sec
total time in air = 2 * 3.47 sec = 6.94 sec
(time rising must equal time falling)
R = 17 m/s * 6.94 s = 118 m
Can also use range formula
R = v^2 sin (2 theta) / g
tan theta = 34 / 17 = 2
theta = 63.4 deg
2 theta = 126.9 deg
sin 126.9 = .8
v^2 = 17^2 + 34^2 = 1445 m^2/s^2
R = 1445 * .8 / 9.8 = 118 m agreeing with answer found above