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Temka [501]
3 years ago
12

Which object absorbs the most visible light

Physics
1 answer:
Papessa [141]3 years ago
4 0
The objects that look black are all tied for first place.
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A country is deciding what to do about pollution glven off by power plants.
DENIUS [597]

Answer:

option B is the correct answer

Explanation:

please follow me and Mark me brainliest please

3 0
2 years ago
A small glass bead has been charged to +20 nC. A small metal ball bearing 1.0 cm above the bead feels a 0.018 N downward electri
Alla [95]

Answer:

q=1\times10^{-8}C

Explanation:

Let the charge on the ball bearing is q.

charge on glass bead, Q = 20 nC = 20 x 10^-9 C

Force between them, F = 0.018 N

Distance between them, d = 1 cm = 0.01 m

By use of Coulomb's law in electrostatics

F=\frac{KQq}{d^{2}}

By substituting the values

0.018=\frac{9\times10^{9}\times20\times10^{-9}q}{0.01^{2}}

q=1\times10^{-8}C

Thus, the charge on the ball bearing is q=1\times10^{-8}C

7 0
2 years ago
Without friction, what is the mass of an ball accelerating at 1.8 m/sec2 to which an
Juli2301 [7.4K]

Answer:

<h2>23.33 kg </h2>

Explanation:

The mass of the object can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

a is the acceleration

From the question we have

m  = \frac{42}{1.8}   = 23.3333... \\

We have the final answer as

<h3>23.33 kg</h3>

Hope this helps you

4 0
2 years ago
Read 2 more answers
If a car is traveling on the highway at a constant velocity, the force that pushes the car forward must be A. equal to the weigh
Gala2k [10]

the correct answer is c

4 0
3 years ago
An object is 15 cm in front of a diverging lens with a
Rainbow [258]

A) See ray diagram in attachment (-6.0 cm)

By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f = -10 cm is the focal length (negative for a diverging lens)

p = 15 cm is the distance of the object from the lens

Solving for q,

\frac{1}{q}=\frac{1}{-10 cm}-\frac{1}{15 cm}=-0.167 cm^{-1}

q=\frac{1}{-0.167 cm^{-1}}=-6.0 cm

B) The image is upright

As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:

h_i = - h_o \frac{q}{p}

where h_i, h_o are the size of the image and of the object, respectively.

Since q < 0 and p > o, we have that h_i >0, which means that the image is upright.

C) The image is virtual

As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.

This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual

4 0
3 years ago
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