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2 years ago

How does black paint interact with a light wave?

Physics

1 answer:

KATRIN_1 [288]2 years ago

8 0

Answer:

Light rays that come from a source such as the sun reflect off items and enter our eye. ... While black objects absorb the energy from all colors and become hot, the objects gradually release some of that energy back into the air around it

Explanation:

Hi Army Pls Mark brainliest

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4 years ago

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3 years ago

A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 686 K, and r

steposvetlana [31]

Answer:

ΔS = 3.09 J/K

Explanation:

Entropy is defined as the magnitude of disorder in a system. Mathematically, entropy change is given as, the heat absorbed or released by a system divided by the change in absolute temperature. So, the formula for entropy change is as follows:

ΔS = ΔQ/ΔT

where,

ΔS = Change in Entropy of Metal Wire = ?

ΔQ = Heat Conducted through Wire = 1216 J

ΔT = Difference in Temperature = 686 K - 292 K = 394 K

Therefore,

ΔS = 1216 J/394 K

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3 years ago

The 20 g bullet is traveling at 400 m/s when it becomes embedded in the 2 -kg stationary block. Determine the distance the block

Marina CMI [18]

Answer:

The block+bullet system moves 4 m before being stopped by the frictional force.

Explanation:

Using the law of conservation of llinear momentum and the work energy theorem, we can obtain this.

According to Newton's second law of motion

Momentum before collision = Momentum after collision

Momentum before collision = (0.02×400) + 0 (stationary block)

Momentum before collision = 8 kgm/s

Momentum after collision = (2+0.02)v

8 = 2.02v

v = 3.96 m/s.

According to the work-energy theorem,

The kinetic energy of the block+bullet system = work done by Friction to stop the motion of the block+bullet system

Kinetic energy = (1/2)(2.02)(3.96²) = 15.84 J

Work done by the frictional force = F × (distance moved by the force)

F = μmg = 0.2(2.02)(9.8) = 3.96 N

3.96d = 15.84

d = (15.84/3.96) = 4 m

5 0

3 years ago

The student in item 1 moves the box up a ramp inclined at 12 degrees with the horizontal. If the box starts from rest at the bot

Alik [6]

Answer:

Acceleration up the ramp = -2.745 m/ $s^{2}$

Explanation:

Given

box is pulled at angle Θ = $25^{o}$

Force applied F = 185N

coefficient of friction, μ $_{k}$ = 0.27

mass of the box m = 35 kg

We know that,

acceleration due to gravity g = 9.8 m/ $s^{2}$

horizontal component $F_{x}$ = F cosΘ = 185 * cos $25^{o}$ =167.67

vertical component $F_{y}$ = F sinΘ =185*sin $25^{o}$ = 78.18

Another vertical component is due to gravity $F_{g}$ , force in given by

$F_{g}$ = mg

= 35 x 9.8

= 343.35 N

Normal force $F_{n}$ = $F_{g}$ - $F_{y}$

= 343.35 - 78.18

= 265.17 N

Frictional force $F_{k}$ = $F_{n}$ * μ $_{k}$

= 265.17 * 0.27

= 71.596 N

To find acceleration, we know that,

force = mass x acceleration

acceleration = $\frac{force}{mass}$

Here force is the summation of frictional force and horizontal component of the applied force. These force act in opposite directions.

force = $F_{k}$ - $F_{x}$

= 71.596 - 167.67

= -96.074

acceleration = $\frac{-96.074}{35}$

= -2.745 m/ $s^{2}$

6 0

3 years ago