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snow_lady [41]
3 years ago
6

It was a cold, sunny day and the ground was covered with fresh snow. Maria wore her black sweatshirt outside to play in the snow

. She knew that black would _________ light waves and help keep her warm.
refract


absorb


diffract


reflect
Physics
2 answers:
Sergeeva-Olga [200]3 years ago
7 0

Answer: The correct answer is absorb.

Explanation:

In the given problem, It was a cold, sunny day and the ground was covered with fresh snow. Maria wore her black sweatshirt outside to play in the snow.

A black color is a good absorber while the white color is good reflector.

Black objects absorb all the incident radiations which falls on them regardless of the frequencies of the radiations while the white reflects all the incidents radiations completely.

Therefore, she knew that black would absorb light waves and help keep her warm.

Sonbull [250]3 years ago
5 0

It was a cold, sunny day and the ground was covered with fresh snow. Maria wore her black sweatshirt outside to play in the snow. She knew that black would absorb light waves and help keep her warm

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Two particles with oppositely signed charges are held a fixed distance apart. The charges are equal in magnitude and they exert
damaskus [11]

Answer:

the force will decrease to 3/4 of its original value.

Explanation:

The initial electric force between the two charges is:

F = k \frac{q\cdot q}{r^2}

where

k is the Coulomb's constant

q is the magnitude of each charge

r is their separation

Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of

q+\frac{q}{2}=\frac{3}{2}q

while the other charge will be

q-\frac{q}{2}=\frac{q}{2}

So, the new force will be

F' = k \frac{(\frac{q}{2})\cdot (\frac{3}{2}q)}{r^2}=\frac{3}{4} (k\frac{q\cdot q}{r^2})=\frac{3}{4}F

So, the force will decrease to 3/4 of its original value.

6 0
3 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem 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 is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

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A)6.15 cm to the left of the lens

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3 years ago
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SVEN [57.7K]

Answer:

d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

Explanation:

For this case  we can use the second law of Newton given by:

\sum F = ma

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Where N represent the normal force, \mu_k the kinetic friction coeffient and a the acceleration.

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F_f = ma

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d = \frac{v^2_i}{2a}= \frac{(20m/s)^2}{2* 3.43 m/s^2}=58.309m

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The answer is B friction force
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3 years ago
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