Answer:
Do not see a picture or graph but suspect it would show the golf ball falling faster and striking the ground slightly before the soccer ball.
Probably D: Soccer ball was affected by air resistance more than the golf ball.
Explanation:
Even though heavier, friction loss of the greater surface area soccer ball will counter pull of gravity more than the compact golf ball.
In a vacuum, (no friction) both objects fall at the same rate regardless of mass.
Answer:
Period of one vibration = 0.00215 second (Approx.)
Wavelength {Is speed of sound is 343 m/s] = 0.736 m (Approx.)
Explanation:
Given:
Frequency of wave = 466 Hz
Find:
Period of one vibration
Wavelength {Is speed of sound is 343 m/s]
Computation:
Period of one vibration = 1/F
Period of one vibration = 1 / 466
Period of one vibration = 0.00215 second (Approx.)
Wavelength = Velocity / Frequency
Wavelength {Is speed of sound is 343 m/s] = 343 / 466
Wavelength {Is speed of sound is 343 m/s] = 0.736 m (Approx.)
Explanation:
Given that,
Force with which a child hits a ball is 350 N
Time of contact is 0.12 s
We need to find the impulse received by the ball. The impulse delivered is given by :
So, the impulse is 42 N-m..
We know that he change in momentum is also equal to the impulse delivered.
So, impulse = 42 N-m and change in momentum =42 N-m.
Answer:
6957.04N
Explanation:
Using
vf2=vi2+2ad
But vf = 0 .
So convert 50km/hr to m/s, and you need to convert 61 cmto m
(50km/hr)*(1hr/3600s)*(1000m/km) = 13.9m/s
61cm * (1m/100cm) = .61m
So n
0 = (13.9m/s)^2 + 2a(.61m)
a = 158.11m/s^2
So
using F = ma
F = 44kg(158.11m/s^2) = 6957.04N
