Answer:
c. V = k Q1 * Q2 / R1 potential energy of Q1 and Q2 separated by R
V2 / V1 = (R1 / R2) = 1/4
V2 = V1 / 4
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.
Answer: First you must convert pound in kilogram, and feet in meter
Explanation:
To calculate momentum we use .
p=m*V
mass-m
speed-V
distance and time are used to calculate velocity(speed)
You are given :
mass- in pounds
for distance - in feet
before you do any calculation first you have to convert pounds in kilograms
and feet in meters.
Answer:
c. V = 2 m/s
Explanation:
Using the conservation of energy:

so:
Mgh = 
where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.
Also we know that:
V = WR
Where R is the radius of the disk, so:
W = V/R
Also, the moment of inertia of the disk is equal to:
I = 
I = 
I = 10 kg*m^2
so, we can write the initial equation as:
Mgh = 
Replacing the data:
(5kg)(9.8)(0.3m) = 
solving for V:
(5kg)(9.8)(0.3m) = 
V = 2 m/s
Let <em>F</em> be the magnitude of the force applied to the cart, <em>m</em> the mass of the cart, and <em>a</em> the acceleration it undergoes. After time <em>t</em>, the cart accelerates from rest <em>v</em>₀ = 0 to a final velocity <em>v</em>. By Newton's second law, the first push applies an acceleration of
<em>F</em> = <em>m a</em> → <em>a</em> = <em>F </em>/ <em>m</em>
so that the cart's final speed is
<em>v</em> = <em>v</em>₀ + <em>a</em> <em>t</em>
<em>v</em> = (<em>F</em> / <em>m</em>) <em>t</em>
<em />
If we force is halved, so is the accleration:
<em>a</em> = <em>F</em> / <em>m</em> → <em>a</em>/2 = <em>F</em> / (2<em>m</em>)
So, in order to get the cart up to the same speed <em>v</em> as before, you need to double the time interval <em>t</em> to 2<em>t</em>, since that would give
(<em>F</em> / (2<em>m</em>)) (2<em>t</em>) = (<em>F</em> / <em>m</em>) <em>t</em> = <em>v</em>