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.
Answer:
375 m.
Explanation:
From the question,
Work done by the frictional force = Kinetic energy of the object
F×d = 1/2m(v²-u²)..................... Equation 1
Where F = Force of friction, d = distance it slide before coming to rest, m = mass of the object, u = initial speed of the object, v = final speed of the object.
Make d the subject of the equation.
d = 1/2m(v²-u²)/F.................. Equation 2
Given: m = 60.0 kg, v = 0 m/s(coming to rest), u = 25 m/s, F = -50 N.
Note: If is negative because it tends to oppose the motion of the object.
Substitute into equation 2
d = 1/2(60)(0²-25²)/-50
d = 30(-625)/-50
d = -18750/-50
d = 375 m.
Hence the it will slide before coming to rest = 375 m
Answer;
Average speed = 47.5 km/hr
Explanation and solution;
Average Speed = Total distance /Total time
Total distance;
-In the first two hours travelling at 40km/hr, total distance traveled is 80km.
(40 × 2) = 80 km
-In next two hours distance covered is 110 km. (55 ×2)
Total distance = 110 + 80 = 190 km
Total time = 2 + 2 = 4 hours
Average Speed = Total distance /Total time
= 190/4
= 47.5 km/hr
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