The work done by a rotating object can be calculated by the formula Work = Torque * angle.
This is analog to the work done by the linear motion where torque is analog to force and angle is analog to distance. This is Work = Force * distance.
An example will help you. Say that you want to calculate the work made by an engine that rotates a propeller with a torque of 1000 Newton*meter over 50 revolution.
The formula is Work = torque * angle.
Torque = 1000 N*m
Angle = [50 revolutions] * [2π radians/revolution] = 100π radians
=> Work = [1000 N*m] * [100π radians] = 100000π Joules ≈ 314159 Joules of work.
Magma forms by partial melting of upper mantle and crust. Partial melt means that only a fraction of the available material forms a melt, and that the remainder stays solid. The partial melt rises because of its lower density and ascends through he crust.
Answer:
a) 600 meters
b) between 0 and 10 seconds, and between 30 and 40 seconds.
c) the average of the magnitude of the velocity function is 15 m/s
Explanation:
a) In order to find the magnitude of the car's displacement in 40 seconds,we need to find the area under the curve (integral of the depicted velocity function) between 0 and 40 seconds. Since the area is that of a trapezoid, we can calculate it directly from geometry:
![Area \,\,Trapezoid=(\left[B+b]\,(H/2)\\displacement= \left[(40-0)+(30-10)\right] \,(20/2)=600\,\,m](https://tex.z-dn.net/?f=Area%20%5C%2C%5C%2CTrapezoid%3D%28%5Cleft%5BB%2Bb%5D%5C%2C%28H%2F2%29%5C%5Cdisplacement%3D%20%5Cleft%5B%2840-0%29%2B%2830-10%29%5Cright%5D%20%5C%2C%2820%2F2%29%3D600%5C%2C%5C%2Cm)
b) The car is accelerating when the velocity is changing, so we see that the velocity is changing (increasing) between 0 and 10 seconds, and we also see the velocity decreasing between 30 and 40 seconds.
Notice that between 10 and 30 seconds the velocity is constant (doesn't change) of magnitude 20 m/s, so in this section of the trip there is NO acceleration.
c) To calculate the average of a function that is changing over time, we do it through calculus, using the formula for average of a function:

Notice that the limits of integration for our case are 0 and 40 seconds, and that we have already calculated the area under the velocity function (the integral) in step a), so the average velocity becomes:

Answer:
Therefore the ratio of diameter of the copper to that of the tungsten is

Explanation:
Resistance: Resistance is defined to the ratio of voltage to the electricity.
The resistance of a wire is
- directly proportional to its length i.e

- inversely proportional to its cross section area i.e

Therefore

ρ is the resistivity.
The unit of resistance is ohm (Ω).
The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m
The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m
For copper:


......(1)
Again for tungsten:

........(2)
Given that
and 
Dividing the equation (1) and (2)

[since
and
]



Therefore the ratio of diameter of the copper to that of the tungsten is
