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Strike441 [17]
3 years ago
5

Discuss how we are able to feel the sun's heat and see its brightness?

Physics
2 answers:
Anna007 [38]3 years ago
8 0

The sun makes so much energy that we are able to see the sun's brightness and feel its heat. Hydrogen fusion occurs in the Sun's core and the gas is heated by the energy that is released above the core. Eventually, the gas becomes less dense and the heat from below turns the gas buoyant

vaieri [72.5K]3 years ago
7 0

Answer:

Sun is bright because it is closer to the earth and the radiation that  reaches the earth's surface gives out the heat.

Explanation:

The thermonuclear fusion reactions are responsible for the heat and light emitted by any star.

As a result of the fusion reactions that occur in its core, sun's core is very very hot. This heat is radiated outward and after passing through several zones of the sun itself, the surface temperature of the Sun cools to about about 6000 K.

Sun radiates electromagnetic radiation that can travel through space and then penetrate the earth's atmosphere. Ozone layer which is present in the ozonosphere of the earth's atmosphere filters the UV rays coming from the sun. The radiation in the form of heat and light is absorbed by the particles of air in the earth's atmosphere, there by heating the surface.

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A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2,
Stella [2.4K]

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

Explanation:

The free-body  diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

  • the peripheral velocity that is directed downward (-V_y) along the y-axis
  • the linear velocity (V_x) that is directed along the x-axis

Now;

V_x = \frac{d}{dt}(12t^3+2) = 36 t^2

V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s

Also,

-V_y = R* \omega

where \omega(angular velocity) = \frac{d\theta}{dt} = \frac{d}{dt}(8t^4)

-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s  is   P =  104.04 \hat{i} -314.432 \hat{j}

3 0
3 years ago
An 820 N Marine in basic training climbs a 12.0-m vertical rope at a constant speed in 8.00 s. What is her power output
sweet-ann [11.9K]

Answer:

1230 W

Explanation:

P = \frac{W}{t} = \frac{Force * distance}{time} = \frac{820 N * 12.0 m}{8.00 s} = 1230 Watts

6 0
3 years ago
A cement truck of mass 14,000 kg moving 10m/s slams into a wall and comes to a halt in .2s. What is the force of impact on the t
skelet666 [1.2K]

And it’s x10 is 100000

6 0
3 years ago
Master of physics needed
Delicious77 [7]
Hey JayDilla, I get 1/3.  Here's how:
Kinetic energy due to linear motion is:
E_{linear}= \frac{1}{2}mv^2
where
v=r \omega
giving
E_{linear}= \frac{1}{2}mr^2 \omega ^2

The rotational part requires the moment of inertia of a solid cylinder
I_{cyl} =  \frac{1}{2}mr^2
Then the rotational kinetic energy is
E_{rot}= \frac{1}{2}I \omega ^2= \frac{1}{4}mr^2 \omega ^2
Adding the two types of energy and factoring out common terms gives
\frac{1}{2}mr^2 \omega ^2(1+ \frac{1}{2})
Here the "1" in the parenthesis is due to linear motion and the "1/2" is due to the rotational part.  Since this gives a total of 3/2 altogether, and the rotational part is due to a third of this (1/2), I say it's 1/3.

8 0
3 years ago
A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m
lbvjy [14]

The equation 15v_{i} + 2*0 = (15 + 2)v_{f} (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.  

The horizontal momentum is given by:

p_{i} = p_{f}

m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}

Where:

  • m₁: is the mass of the lab cart = 15 kg
  • m₂: is the <em>mass </em>of the object dropped = 2 kg
  • v_{1}_{i}: is the initial velocity of the<em> lab cart </em>
  • v_{2}_{i}: is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)
  • v_{1}_{f}: is the final velocity of the<em> lab cart </em>
  • v_{2}_{f}: is the <em>final velocity</em> of the <em>object </em>

Then, the horizontal momentum is:

15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

15v_{1}_{i} + 2*0 = v_{f}(15 + 2)

Therefore, the equation 15v_{i} + 2*0 = (15 + 2)v_{f} represents the horizontal momentum (option 3).

Learn more about linear momentum here:

  • brainly.com/question/2141713?referrer=searchResults
  • brainly.com/question/2400186?referrer=searchResults

I hope it helps you!            

4 0
3 years ago
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