Answer:
Explanation:
Given that,
The radius of a flywheel, r = 0.3 m
Angular acceleration of a flywheel,
We need to find the magnitude of the tangential acceleration after 2.00 s of acceleration.
The relation between the tangential and angular acceleration is given by :
So, the required magnitude of tangential acceleration is .
Explanation :
It is given that, the driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.
We know that acceleration is defined as the rate of change of velocity.
Where
dv is the change in velocity, dv = 0 - 60 m/s = -60 m/s
dt is the change in time, dt = 40 s - 30 s = 10 s
So,
From the graph it is clear that, from 30 s to 40 s the car is decelerating. So, at every second within this time the value of acceleration will be same i.e. .
Answer:
Explanation:
The electrostatic potential energy for pair of charge is given by
U=1/4π∈₀×(q₁q₂/r)
Hence for a system of three charges the electrostatic potential energy can be found by adding up the potential energy for all possible pairs or charges.For three equal charges on the corners of an equilateral triangle,the electrostatic potential energy is given by:
U=1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)
U=3×1/4π∈₀×(q²/r)
Substitute given values
So