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.
- Speed of the mobile = 250 m/s
- It starts decelerating at a rate of 3 m/s²
- Time travelled = 45s
- Velocity of mobile after 45 seconds
We can solve the above question using the three equations of motion which are:-
- v = u + at
- s = ut + 1/2 at²
- v² = u² + 2as
So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.
We are provided with,
- u = 250 m/s
- a = -3 m/s²
- t = 45 s
By using 1st equation of motion,
⇛ v = u + at
⇛ v = 250 + (-3)45
⇛ v = 250 - 135 m/s
⇛ v = 115 m/s
✤ <u>Final</u><u> </u><u>velocity</u><u> </u><u>of</u><u> </u><u>mobile</u><u> </u><u>=</u><u> </u><u>1</u><u>1</u><u>5</u><u> </u><u>m</u><u>/</u><u>s</u>
<u>━━━━━━━━━━━━━━━━━━━━</u>
Answer:
Ur answer is Stationary Front
Explanation:
Stationary Front is when a cold air mass and a warm air mass but are at a standstill the boundary is called Stationary Front.
Answer:
I don't know about these problems at all.
Explanation:
I don't know about physics at all
The power equals energy divided by the time
P=E/t
P=1200/60
P=20W
Hope you get it!
