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3 years ago

15

Which of the following best models the changes in energy of a blackbody radiator?

2 answers:

Pani-rosa [81]3 years ago

8 0

We have that a blackbody radiator either constantly absorbs energy or constantly emits energy, depending on its surroundings. In this case, the energy is continuously and smoothly decreasing, thus it cannot be like B and C.
The energy loss or gain is also monotonous, it has the same direction; a radiator cannot gain energy at some point and then lose some. Hence, it does not resemble a wave either. The most appropriate model is the ramp. Energy is constantly emitted to surroundings and it decreases monotonically.

Katena32 [7]3 years ago

6 0

Answer:

D

Explanation:

summary of other answer

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The sound waves from a noisy jet travel from the air into water. Which property of the wave will not change?

notka56 [123]

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Hope this helps! :)

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3 years ago

A ball rolls over the edge of a platform with only a horizontal velocity. The height of the platform is 1.60m and the horizontal

AysviL [449]

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

$v=\sqrt{v_x^2+v_y^2}$ (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

$v_y^2=v_{oy}^2+2gh$ (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

$v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}$

vx is calculated by using the information about the horizontal range of the ball:

$R=v_o\sqrt{\frac{2h}{g}}$ (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

$v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}$

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

$v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}$

The velocity of the ball just before it touches the ground is 46.99 m/s

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USPshnik [31]

Answer:

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Explanation:

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The correct answer is B. it converts mechanical energy to electrical energy.

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Law Incorporation [45]

Yes, his velocity will decrease the further he slides.

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