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:
It is the carrier of genetic information.
Answer:
Please find the answer in the explanation.
Explanation:
Given that two porters are available to carry a long timber wood.out of them one is weak. how do you make less load to the weak one?
We can make the weak one to carry less load through two different ways or means.
First, if we can locate the centre of gravity and centre of mass of the long timbre wood, the week one can carry the other end away from the center of gravity and centre of mass.
Second, the strong porter can carry the long timbre wood almost to the fulcrum and allow the weak one to support at the other end. By doing this, the weak one will only carry light portion of the load.
Answer:
420 L
Explanation:
Applying Boyle's Law,
PV = P'V'.................... Equation 1
Where P = Initial pressure, P' = Final pressure, V = Initial volume, V' = Final volume.
make V' the subject of the equation
V' = PV/P'.................... Equation 2
From the question,
Given: P = 720 mmHg, V = 350 L, P' = 600 mmHg
Substitute these values into equation 2
V' = (720×350)/600
V' = 252000/600
V' = 420 L
