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.
I'm not sure but I know u is 10^6
You need to know how much friction that object.
knives, axe ,blades and nails are examples of potential energy
or sharp objects
Answer:

Explanation:
The density of an object is given by

where
m is the mass of the object
V is its volume
In this problem,
m = 42 kg
V = 22 m^3
Substituting into the equation, we find the object's density:
