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Savatey [412]
3 years ago
8

If you quadruple the temperature of a black body, by what factor will the total energy radiated per second per square meter incr

ease?
Physics
1 answer:
Anastaziya [24]3 years ago
8 0
Radiant heat transfer is proportional to the 4-th power of absolute temperature.
Therefore if the temperature is quadrupled, the radiant heat energy will increase by a factor of
4⁴ = 256

Answer: 256
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What do all wetlands have in common?
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They are all coverd in water 24/7 they never clear up
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For a star with a parallax angle of 1/2 of an at arcsecond, what will be its distance in parsec?
avanturin [10]

(1 parsec) is the distance at which an object has a parallax of 1 arcsecond. The distance is about 3.26 light years.

Another way to understand it is:  The distance from which the Earth's orbit appears 1 arcsecond across.

For a parallax angle of 1/2 arcsecond, the distance is <em>2 parsecs </em>(about 6.52 light years).

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8 0
3 years ago
A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward
AleksandrR [38]

Answer:

(a)  104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.  

The component of F perpendicular to the ramp is Fy.

(a)  

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.  

Resolve F into its x-component from Pythagorean theorem:  

Fx=Fcos30°

Solve for F:  

F= Fx/cos30°  

Substitute for Fx from given data:  

Fx=90 N/cos30°

   =104 N

(b) Resolve r into its y-component from Pythagorean theorem:

     Fy = Fsin 30°

   Substitute for F from part (a):

     Fy = (104 N) (sin 30°)  

          = 52 N  

5 0
3 years ago
You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo
jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

\Delta E_k=W=W_g-W_p

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J

hence, the change in Er is about 1.52J times the initial rotational energy

5 0
3 years ago
Read 2 more answers
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