Picture? I may be able to answer if you have a chart or some kind of graph as a referral to the question
Answer:
Explanation:
Given
mass of block 
at 
displacement is 
velocity 
acceleration 
suppose
is the general equation of SHM
where A=amplitude
=natural frequency of oscillation
therefore velocity and acceleration is given by


for t=0



divide 1 and 3 we get


Now square and 1 and 2 we get



electric potential energy..electrons get the the kinetic energy from the voltage applied across the conductor
Answer:
Scenario A, B and E is True.
Explanation:
Scenario A) True. Removing carbon dioxide from atmosphere decreases greenhouse effect of atmosphere. Thus, temperature rise decreases.
Scenario B) True. The more evaporation creates the more greenhouse effect. Therefore, temperature rise increases.
Scenario C) False. Removing carbon dioxide from atmosphere decreases greenhouse effect of atmosphere. Thus, temperature rise decreases.
Scenario D) False. The more evaporation creates the more greenhouse effect. Therefore, temperature rise increases.
Scenario E) True. If reflected radiation increases from Earth, temperature rise of the Earth will decrease. Ice cover increases reflectivity which leads temperature level decrease.
Scenario F) False. If reflected radiation increases from Earth, temperature rise of the Earth will decrease. Ice cover increases reflectivity which leads temperature level decrease.
Answer:
The frequency of the phonograph record is 0.2 Hz
Explanation:
The frequency of an object moving in uniform circular motion is the number of completed cycles the object makes in a specified time period
The given parameters of the phonograph record are;
The radius of the record = 0.15 m
The number of times the phonograph record rotates, n = 18 times
The time it takes the phonograph record to rotate the 18 times, t = 90 seconds
The frequency of the phonograph record, f = (The number of times the phonograph record rotates) ÷ (The time it takes the phonograph record to rotate the 18 times)
∴ The frequency of the phonograph record, f = n/t = 18/(90 s) = 0.2 Hz
The frequency of the phonograph record = 0.2 Hz.