The work done by the centripetal force during om complete revolution is 401.92 J.
<h3>What is centripetal force?</h3>
Centripetal force is a force that acts on a body undergoing a circular motion and is directed towards the center of the circle in which the body is moving.
To Calculate the work done by the centripetal force during one complete revolution, we use the formula below.
Formula:
- W = (mv²/r)2πr
- W = 2πmv²................... Equation 1
Where:
- W = Work done by the centripetal force
- m = mass of the ball
- v = velocity of the ball
- π = pie
From the question,
Given:
- m = 16 kg
- v = 2 m/s
- π = 3.14
Substitute these values into equation 1
Hence, The work done by the centripetal force during om complete revolution is 401.92 J.
Learn more about centripetal force here: brainly.com/question/20905151
Answer:
"A turbine takes the kinetic energy of a moving fluid, air in this case, and converts it to a rotary motion. As wind moves past the blades of a wind turbine, it moves or rotates the blades. These blades turn a generator."
Answer:
h = 2.49 [m]
Explanation:
In order to solve this problem we must use the definition of potential energy, which tells us that energy is equal to the product of mass by gravity by height.
The potential energy can be calculated by means of this equation:
Ep = m*g*h
where:
Ep = potential energy = 980 [J]
m = mass = 40 [kg]
g = gravity acceleration = 9.81 [m/s^2]
h = elevation [m]
Now replacing:
980 = 40*9.81*h
h = 2.49 [m]
1)
p = 2.4 * 10^5 Pa
T = 18° C + 273.15 = 291.15 k
r = 0.25 m => V = [4/3]π(r^3) = [4/3]π(0.25m)^3 = 0.06545 m^3 = 65.45 L
Use ideal gas equation: pV = nRT => n = pV / RT = [2.4*10^5 Pa * 0.06545 m^3] / [8.31 J/k*mol * 291.15k] = 6.492 mol
Avogadro number = 1 mol = 6.022 * 10^23 atoms
Number of atoms = 6.492 mol * 6.022 *10^23 atom/mol = 39.097 * 10^23 atoms = 3.91 * 10^24 atoms
2) Double atoms => double volume
V2 / V1 = r2 ^3 / r1/3
2 = r2 ^3 / r1 ^3 => r2 ^3 = 2* r1 ^3
r2 = [∛2]r1
The factor is ∛2
A) 
Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:
Parallel:
(1)
where
m is the mass
g = 9.8 m/s^2 the acceleration of gravity
is the coefficient of friction
R is the normal reaction
a is the acceleration
Perpendicular:
(2)
From (2) we find
And substituting into (1)
Solving for a,
B) 5.94 m/s
We can solve this part by using the suvat equation
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the displacement
Here we have
v = ?
u = 0 (it starts from rest)
s = 8.70 m
Solving for v,
