Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
Answer:
137.2 in pounds and in Newton's it's 588.399
Answer:

Explanation:
Given that,
The current flowing in the circuit, I = 3 A
The power of the battery, P = 25 W
We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :

Put all the values to find R.

So, the resistance is equal to
.
Answer:
Dont know if this is right but i say C
Explanation:
Answer:
option C
Explanation:
The correct answer is option C
When the driver takes the sharp right turn the door will exert rightward pressure on the driver.
When the driver takes the sudden right turn the tendency of the body is to be in the straight line by the vehicle moves in the circular path so, as the vehicle turns it applies a rightward force on you.
The pushing of the door to you because of the centripetal force acting on the car due to sudden sharp turn.