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inysia [295]
3 years ago
15

Charcoal appears black in color because it

Physics
2 answers:
steposvetlana [31]3 years ago
4 0
C. This is the idea of "black body radiation". Charcoal/carbon is a "perfect black body". Absorbs all radiation. There's a whole host of stuff about this in physics, including, I think, Planck's (Nobel Prize winner) black body radiation theory.
Natalka [10]3 years ago
3 0
I think the answer is C but correct me if im wrong hope this helps :)
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You calculate that the maximum static friction is 420N. If you apply 500N of force, will the couch slide?
Valentin [98]
Assuming that the entirety of the force is opposite the static friction, then the couch should begin sliding as the applied force overcomes the couch's static friction with what it's making contact with.
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Give details of aids in brief and explain how to protect yourself and prevent.
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8 0
3 years ago
Does light have mass?
Lapatulllka [165]
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5 0
3 years ago
Read 2 more answers
A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho
d1i1m1o1n [39]

Answer:

Part a)

KE_r = 8 J

Part b)

v = 3.64 m/s

Part c)

KE_f = 12.7 J

Part d)

v = 2.9 m/s

Explanation:

As we know that moment of inertia of hollow sphere is given as

I = \frac{2}{3}mR^2

here we know that

I = 0.0484 kg m^2

R = 0.200 m

now we have

0.0484 = \frac{2}{3}m(0.200)^2

m = 1.815 kg

now we know that total Kinetic energy is given as

KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2

20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2

20 = 1.5125 v^2

v = 3.64 m/s

Part a)

Now initial rotational kinetic energy is given as

KE_r = \frac{1}{2}I(\frac{v}{R})^2

KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2

KE_r = 8 J

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

mgh = (KE_i) - KE_f

1.815(9.8)(0.900sin27.1) = 20- KE_f

7.30 = 20 - KE_f

KE_f = 12.7 J

Part d)

Now we know that

\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7

\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7

1.5125 v^2 = 12.7

v = 2.9 m/s

8 0
3 years ago
An electrical motor spins at a constant 2695.0 rpm. If the armature radius is 7.165 cm, what is the acceleration of the edge of
djyliett [7]

Answer:

Acceleration will be 5706.77rad/sec^2

So option (D) will be correct answer

Explanation:

We have given angular speed of the electrical motor \omega =2695rpm

We have to change this angular speed in rad/sec for further calculation

So \omega =2695rpm=2695\times \frac{2\pi }{60}=282.2197rad/sec

Armature radius is given r = 7.165 cm = 0.07165 m

We have to find the acceleration of edge of motor

Acceleration is given by a=\omega ^2r=282.2197^2\times 0.07165=5706.77rad/sec^2

So acceleration will be 5706.77rad/sec^2

So option (D) will be correct answer

5 0
4 years ago
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