Answer:
245 m
Explanation:
v = at + v₀
50.0 m/s = a (9.8 s) + 0 m/s
a = 5.10 m/s²
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (9.8 s) + ½ (5.10 m/s²) (9.8 s)²
x = 245 m
First, we have a change in the velocity from 85 to 164 m/s in 10 sec.
Then, we calculate the <u>acceleration </u>as:
Hence we need to calculate the velocity of the space vehicle at t = 2 sec using the first equation of motion:
Then, using the second equation of motion to calculate the distance:
Choices 'B'; and 'D' both begin with the correct words.
But they should end with the equation
R = V / I
Answer:
(1) 0.333 Hz
(2) 4 sec
(3) 2 sec, 0.5 Hz
Explanation:
(1) We have given time period of pendulum is 3 sec
So T = 3 sec
Frequency will be equal to 
(2) Frequency of the pendulum is given f = 0.25 Hz
Time period is equal to 
(3) It is given that a pendulum makes 10 back and forth swings in 20 seconds
So time taken to complete 1 back and forth swings = 
So time period T = 2 sec
Frequency will be equal to 
Answer:
The puck moves a vertical height of 2.6 cm before stopping
Explanation:
As the puck is accelerated by the spring, the kinetic energy of the puck equals the elastic potential energy of the spring.
So, 1/2mv² = 1/2kx² where m = mass of puck = 39.2 g = 0.0392 g, v = velocity of puck, k = spring constant = 59 N/m and x = compression of spring = 1.3 cm = 0.013 cm.
Now, since the puck has an initial velocity, v before it slides up the inclined surface, its loss in kinetic energy equals its gain in potential energy before it stops. So
1/2mv² = mgh where h = vertical height puck moves and g = acceleration due to gravity = 9.8 m/s².
Substituting the kinetic energy of the puck for the potential energy of the spring, we have
1/2kx² = mgh
h = kx²/2mg
= 59 N/m × (0.013 m)²/(0.0392 kg × 9.8 m/s²)
= 0.009971 Nm/0.38416 N
= 0.0259 m
= 2.59 cm
≅ 2.6 cm
So the puck moves a vertical height of 2.6 cm before stopping
