All categories

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>Answer;</h3>

A black T-shirt 

<h3>Explanation;</h3>

Materials can be either reflectors and absorbers of waves such as light waves, sound waves etc. 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 with smooth and shiny surfaces, while those with rough and opaque surfaces are poor reflectors of light. Additionally, these materials that are good reflectors are poor absorbers while those that are poor reflectors are good absorbers of light.
Dull surfaces are the best emitters and absorbers of light but the worst reflectors of light waves. White surfaces or polished surfaces are bad and worst emitters and absorbers of light waves respectively

You might be interested in

It is observed that the time for the ball to strike the ground at b is 2.5 s. determine the speed at which the ball was thrown.

Alika [10]

Ergrexgehtenhyrnsehtrsjyrrsjjrt

6 0

2 years ago

Kalea throws a baseball directly upward at time t = 0 at an initial speed of 13.7 m/s. How high h does the ball rise above its r

Trava [24]

Answer:

h = 9.57 seconds

Explanation:

It is given that,

Initial speed of Kalea, u = 13.7 m/s

At maximum height, v = 0

Let t is the time taken by the ball to reach its maximum point. It cane be calculated as :

$v=u-gt$

$u=gt$

$t=\dfrac{u}{g}$

$t=\dfrac{13.7}{9.8}$

t = 1.39 s

Let h is the height reached by the ball above its release point. It can be calculated using second equation of motion as :

$h=ut+\dfrac{1}{2}at^2$

Here, a = -g

$h=ut-\dfrac{1}{2}gt^2$

$h=13.7\times 1.39-\dfrac{1}{2}\times 9.8\times (1.39)^2$

h = 9.57 meters

So, the height attained by the ball above its release point is 9.57 meters. Hence, this is the required solution.

7 0

3 years ago

Consider two uniform solid spheres where both have the same diameter, but one has twice the mass of the other. how much larger i

egoroff_w [7]

The moment of inertia of the large sphere will be twice that of the smaller sphere.
The formula for the moment of inertia for a solid sphere is:
I = (2/5)mr^2
where
I = moment of inertia
m = mass
r = radius

Since both spheres have the same diameter, they also have the same radius, so the only change is their mass. And the moment of inertia is directly proportional to their mass as shown by the above formula. So the sphere with twice the mass will have twice the moment of inertia, or 2 times.

5 0

2 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$