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2 years ago

Study the interaction of the light rays with the mediums. Classify the medium in each image as being transparent or opaque.

Physics

2 answers:

Tresset [83]2 years ago

6 0

Answer:

a) transparent

b) transparent

c) opaque

Explanation:

In the first one, the light rays go completely through, so it is transparent.

The second one I'm not too sure about. It is refraction so it's going through a different material, but the fact that it went through makes me say transparent.

Last one, the light rays are reflecting off the surface so it's opaque.

Please feel free to correct me if I'm wrong. This is just my understanding

Talja [164]2 years ago

4 0

Answer:

a) transparent

b) transparent

c) opaque

Explanation:

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While watching a movie a spaceship explodes and there is a loud bang and flash of light. What is wrong with this scene? Explain

Ksenya-84 [330]

Well if the ship was in space their shouldn’t be a loud bang. Because you can’t hear anything in space

4 0

2 years ago

A 20 kg wagon is rolling to the right across a floor. A person attempts to catch and stop the crate and applies a force of 70 N,

Serggg [28]

Answer:

1.736m/s²

Explanation:

According to Newton's second law;

$\sum F_x = ma_x\\$

$Fm - Ff = ma_x\\$ where;

Fm is the moving force = 70.0N

Ff is the frictional force acting on the body

$Ff = \mu R\\Ff = \mu mg\\$

$\mu$ is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

a is the acceleration/deceleration

The equation becomes;

$Fm - Ff = ma_x\\Fm - \mu mg = ma\\$

Substitute the given parameters

$Fm - \mu mg = ma\\70 - 0.18(20)(9.8) = 20a\\70-35.28 = 20a\\34.72 = 20a\\a = \frac{34.72}{20}\\a = 1.736m/s^2\\$

Hence the deceleration rate of the wagon as it is caught is 1.736m/s²

7 0

2 years ago

What is a substance

Korolek [52]

Explanation:

A particular kind of matter with uniform properties..

8 0

2 years ago

Read 2 more answers

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

2 years ago

When a 3.0 kg mass is hung from a vertical massless spring, the spring is stretched 40 cm. What is the spring constant of the sp

Dmitry_Shevchenko [17]

Answer:

0.74 N/cm

Explanation:

The following data were obtained from the question:

Mass (m) = 3 Kg

Extention (e) = 40 cm

Spring constant (K) =?

Next, we shall determine the force exerted on the spring.

This can be obtained as follow:

Mass (m) = 3 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = mg

F = 3 × 9.8

F = 29.4 N

Finally, we shall determine the spring constant of the spring. This can be obtained as follow:

Extention (e) = 40 cm

Force (F) = 29.4 N

Spring constant (K) =?

F = Ke

29.4 = K × 40

Divide both side by 40

K = 29.4 / 40

K = 0.74 N/cm

Therefore, the spring constant of the spring is 0.74 N/cm

5 0

2 years ago