Answer:
Frequency of oscillation, f = 4 Hz
time period, T = 0.25 s
Angular frequency, 
Given:
Time taken to make one oscillation, T = 0.25 s
Solution:
Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:
f = 
f = 
Frequency of oscillation, f = 4 Hz
The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.
Therefore, time period, T = 0.25 s
Angular frequency of oscillation is given by:
If the gymnast mass were doubled, her height (h) from the top of the board would be as follows,
с Stay the same
Explanation:
- The Mass of an object or body does not affect the acceleration due to gravity in any kind of way.
- Light weight objects accelerate more slowly than the heavy objects because when the forces other than the gravity also plays a major role.
- Mass increases of a body when an object has higher velocity or the speed.
- The greater the force of gravity, it would give a direct impact on the object's acceleration; thus considering only a force, the heavier the object is, it would accelerate faster. But an acceleration depends upon the two factors which are force and mass.
- Newton's second law of motion states that the acceleration of an object is dependent upon the two factors which are, the net force of an object and the mass of the object.
Answer:
The distance the log has moved by the time Ernie reaches Bur is 1.33 m.
Explanation:
give information:
The log is 3.0 m long and has mass 20.0 kg.
Burt has mass 30.0 kg; Ernie has mass 40.0 kg
Ernie has mass 40.0 kg.
to find the distance, first, we have to calculate the center of mass
X = ∑ m x /∑m
= (20 x (3/2)) + (30 x 0) + (40 x 3)/ (20+30+40)
= 150/90
= 5/3
when Ernie walk, the center of the mass is
X = (70 x 0) + (20 x (3/2))/(70 + 20)
= 30/90
= 1/3
the distance of log moved = 5/3 - 4/3 = 1.33 m
Answer:
Explanation:
As in any sample you will have 75.8% of Cl-35 iosotopes and 24.3% of Cl-37 iosotopes you can get the average atomic mass as:
Answer
given,
I = 0.140 kg ·m²
decrease from 3.00 to 0.800 kg ·m²/s in 1.50 s.
a) 
τ = -1.467 N m
b) angle at which fly wheel will turn
θ = 20.35 rad
c) work done on the wheel
W = τ x θ
W = -1.467 x 20.35 rad
W = -29.86 J
d) average power of wheel
