Answer:
50 m
Explanation:
Acceleration= force/mass
3000/3000=1m/s^-2
Applying equation of motion:
V^2=U^2+2as; V is final velocity, u is initial velocity, a is acceleration and s is the distance covered.
0=10^2 -2*1s;
Solve for s
Answer:
0.47 m
Explanation:
= Number of vibrations = 37
= total time taken = 33 s
= time period of each vibration
frequency of vibration is given as
Hz
= distance traveled along the rope = 421 cm = 4.21 m
= time taken to travel the distance = 8 s
= speed of the wave
Speed of the wave is given as

= wavelength of the harmonic wave
wavelength of the harmonic wave is given as

Answer:


Explanation:
= Initial momentum of the pin = 13 kg m/s
= Initial momentum of the ball = 18 kg m/s
= Momentum of the ball after hit
= Angle ball makes with the horizontal after hitting the pin
= Angle the pin makes with the horizotal after getting hit by the ball
Momentum in the x direction

Momentum in the y direction


The pin's resultant velocity is 

The pin's resultant direction is
below the horizontal or to the right.
Answer:
a. 13.7 s b. 6913.5 m
Explanation:
a. How much time before being directly overhead should the box be dropped?
Since the box falls under gravity we use the equation
y = ut - 1/2gt² where y = height of plane above ocean = 919 m, u = initial vertical velocity of airplane = 0 m/s, g = acceleration due to gravity = -9.8 m/s² and t = time it takes the airplane to be directly overhead.
So,
y = ut - 1/2gt²
y = 0 × t - 1/2gt²
y = 0 - 1/2gt²
y = - 1/2gt²
t² = -2y/g
t = √(-2y/g)
So, t = √(-2 × 919 m/-9.8 m/s²)
t = √(-1838 m/-9.8 m/s²)
t = √(187.551 m²/s²)
t = 13.69 s
t ≅ 13.7 s
So, the box should be dropped 13.69 s before being directly overhead.
b. What is the horizontal distance between the plane and the victims when the box is dropped?
The horizontal distance x between plane and victims, x = speed of plane × time it takes for box to drop = 505 m/s × 13.69 s = 6913.45 m ≅ 6913.5 m