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nekit [7.7K]
3 years ago
14

A sphere of uniform density with mass 21 kg and radius 0.8 m is spinning, making one complete revolution every 0.8 s. the center

of mass of the sphere has a speed of 8 m/s. (a) what is the rotational kinetic energy of the sphere?
Physics
1 answer:
kompoz [17]3 years ago
8 0
Using the equation;
TE = 1/2mv^2(1+2); where k = 2/5 for a solid sphere; V is the velocity, and m is the mass.
Total energy = 0.5 × 21 × 8² (7/5)
                    = 940.8 J
The rotational kinetic energy of the sphere is 940.8 J

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Amount of work done by a rotating object
Oduvanchick [21]
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.

 
5 0
2 years ago
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What current flows through a 2.56-cm-diameter rod of pure silicon that is 20.0 cm long, when 1.00 ✕ 103 V is applied to it? (Suc
vfiekz [6]

Answer:

Current, I = 0.0011 A

Explanation:

It is given that,

Diameter of rod, d = 2.56 cm

Radius of rod, r = 1.28 cm = 0.0128 m

The resistivity of the pure silicon, \rho=2300\ \Omega-m

Length of rod, l = 20 cm = 0.2 m

Voltage, V=1\times 10^3\ V

The resistivity of the rod is given by :

R=\rho\dfrac{L}{A}

R=2300\ \Omega-m\dfrac{0.2\ m}{\pi (0.0128\ m)^2}

R = 893692.30 ohms

Current flowing in the rod is calculated using Ohm's law as :

V = I R

I=\dfrac{V}{R}

I=\dfrac{10^3\ V}{893692.30\ \Omega}

I = 0.0011 A

So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.

6 0
3 years ago
Using the set-up seen here, Ms. Garcia places a golf ball between the globe and the flashlight. Turning off the lights in the ro
Julli [10]
This is a solar Eclipse.  
7 0
3 years ago
A charge of 7.2 × 10-5 C is placed in an electric field with a strength of 4.8 × 105 StartFraction N over C EndFraction. If the
barxatty [35]

Answer:

2.2 meters

Explanation:

Potential energy, PE created by a charge, q at a radius r from the charge source, Q,  is expressed as:

KE=\frac{kQq}{r}\     \ \ \ \ \ \ ...i

k is Coulomb's constant.

#The electric field,E at radius r is expressed as:

E=\frac{kQ}{r^2}\ \ \ \ \ \ \ \ \ \ ...ii

From i and ii, we have:

KE=Eqr

r=(KE)/Eq

#Substitute actual values in our equation:

r=\frac{75J}{(7.2\times 10^{-5}C)(4.8\times 10^5 V/m)}\\\\=2.1701\approx2.2\ m

Hence, the distance between the charge and the source of the electric field is 2.2 meters

7 0
2 years ago
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Starting with the definition 1.00 in. = 2.54 cm, find the number of kilometers in 8.00 mi .
velikii [3]
Use Factor-Label Method:

8miles 63360 inches
---------- X --------------------- X
1 1 mile

2.54cm 1 meter
X ------------ X ---------------- X
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1 km
----------------- = 12.87 km
1000meters


8 miles = 12.87 km
8 0
3 years ago
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