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3 years ago

Which object absorbs the most visible light?

Physics

2 answers:

klio [65]3 years ago

4 0

Answer:

C. a black t-shirt

Explanation:

wariber [46]3 years ago

3 0

<h3><u>Answer;</u></h3>

<em>A black T-shirt </em>

<h3><u>Explanation;</u></h3>

<em><u>Materials can be either reflectors and absorbers of waves such as light waves, sound waves etc.</u></em> Reflectors are those materials that reflects waves, such that the waves bounces back on hitting the surface of those materials.
Examples of such materials are those <em><u>with smooth and shiny surfaces, while those with rough and opaque surfaces are poor reflectors of light. </u></em>Additionally, these materials that are good reflectors are poor absorbers while those that are poor reflectors are good absorbers of light.
<u><em>Dull surfaces are the best emitters and absorbers of light but the worst reflectors of light waves.</em></u> White surfaces or polished surfaces are bad and worst emitters and absorbers of light waves respectively

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If a trapezoidal channel has side slopes of 1:1, hydraulic depth is 5 feet, the bottom width is 8 feet, flow is 2,312 cubic feet

Evgesh-ka [11]

Answer:

$S = \dfrac{1}{2.5}$

Explanation:

given,

side slope = 1 : 1

hydraulic depth(y) = 5 ft

bottom width (b)= 8 ft

x = 1

Q = 2,312 ft³/s

n = 0.013

slope of channel = ?

$R = \dfrac{A}{P}$

$R = \dfrac{y(b + xy)}{b+2y\sqrt{1+x^2}}$

$R = \dfrac{5(8+ 5)}{8+2\times 5\sqrt{1+1^2}}$

R = 4.69 m

using manning's equation

$Q = \dfrac{1}{n}AR^{2/3} S^{1/2}$

$2312= \dfrac{1}{0.013}\times (5(8+5))\times 4.69^{2/3} S^{1/2}$

$2312=14009.37\times S^{1/2}$

S = 0.406

$S = \dfrac{1}{2.5}$

5 0

3 years ago

INTEGRATED SCIENCE-8th grade science <br> Please help report cards are due for me at dec.18

serious [3.7K]

Answer:

Our drinking water comes from lakes, rivers and groundwater. For most Americans, the water then flows from intake points to a treatment plant, a storage tank, and then to our houses through various pipe systems. A typical water treatment process.

Explanation:

7 0

3 years ago

A kangaroo can jump straight up to a height of 2.0 m. What is its takeoff speed

12345 [234]

7.17m/s glad I could help

5 0

3 years ago

An experiment is done to compare the initial speed of projectiles fired from high-performance catapults. The catapults are place

anzhelika [568]

Answer:1.084

Explanation:

Given

mass of Pendulum M=10 kg

mass of bullet m=5.5 gm

velocity of bullet u

After collision let say velocity is v

conserving momentum we get

$mu=(M+m)v$

$v=\frac{m}{M+m}\times u$

Conserving Energy for Pendulum

Kinetic Energy=Potential Energy

$\frac{(M+m)v^2}{2}=(M+m)gh$

here $h=L(1-\cos \theta )$ from diagram

therefore

$v=\sqrt{2gL(1-\cos \theta )}$

initial velocity in terms of v

$u=\frac{M+m}{m}\times \sqrt{2gL(1-\cos \theta )}$

For first case $\theta =6.8^{\circ}$

$u_1=\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}$

for second case $\theta =11.4^{\circ}$

$u_2=\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}$

Therefore $\frac{u_1}{u_2}=\frac{\frac{M+m_1}{m_1}\times \sqrt{2gL(1-\cos 6.8)}}{\frac{M+m_2}{m_2}\times \sqrt{2gL(1-\cos 11.4)}}$

$\frac{u_1}{u_2}=\frac{1819.181\times 0.0838}{1001\times 0.1404}$

$\frac{u_1}{u_2}=1.084$

i.e. $\frac{v_1}{v_2}=1.084$

4 0

3 years ago

A ball is thrown at 60 degrees and lands in 18.5 seconds what is the velocity of the ball at the start

n200080 [17]

Answer:

the initial velocity of the ball is 104.67 m/s.

Explanation:

Given;

angle of projection, θ = 60⁰

time of flight, T = 18.5 s

let the initial velocity of the ball, = u

The time of flight is given as;

$T = \frac{2u\times sin(\theta)}{g} \\\\2u\times sin(\theta) = Tg\\\\u = \frac{Tg}{2\times sin(\theta)} \\\\u = \frac{18.5 \times 9.8}{2\times sin(60^0)} \\\\u = 104.67 \ m/s$

Therefore, the initial velocity of the ball is 104.67 m/s.

3 0

3 years ago