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3 years ago

When a light ray is reflected from a surface, the ratio of the angle of incidence to the angle of reflection is

Physics

1 answer:

balandron [24]3 years ago

7 0

Answer:

equal to 1

Explanation:

When any ray of light falls or strikes on a surface, the ray is reflected back and the angle of reflection is equal to angle of the incidence, where both the angles are always measured from the path of light that is normal to surface of the point at which the ray of light strikes the surface. This is called as the laws of reflection of light.

And therefore, the ratio of the angle of incidence to the angle of reflection is always equal to 1.

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If the momentum of a 1000 kg car travelling at 10 m/s was transferred completely to a 20.0 kg traffic barrier, what would the fi

Alenkinab [10]

Answer:

500 m/s

Explanation:

Momentum, p is a product of mass and velocity. From law of conservation of momentum, the initial momentum equals final momentum

$m_1v_1=m_2v_2\\1000*10=20v_2\\v_2=10000/20=500 m/s$

Here m and v represent mass and velocity respectively and subscripts 1 and 2 represent car and barrier respectively

Therefore, the velocity of barrier will be 500 m/s

6 0

3 years ago

Which units are used to measure both velocity and speed? Check all that apply. m/s d/t km/h mph lph

GaryK [48]

The units used to measure speed and velocity are as follows: m/s, Km/h, and mph.

5 0

4 years ago

Read 2 more answers

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

3 years ago

Really need help! ;(<br> Could you please explain it? :) <br> GIVING 10 POINTS

ale4655 [162]

Answer:b

Explanation: If you look at the line on the graph, you can see that it is going downward, meaning it has a negative slope, and choice b is the only one that has a negative slope

4 0

2 years ago

Read 2 more answers

A 50-foot flagpole is at the entrance of a building that is 300 feet tall. If the length of the flagpole's shadow is 30 feet at

natka813 [3]

First you will want to sketch out both of the situations. It should be two sketches, one for the flagpole and one for the building.

To solve this, you will want to create a proportion.

Flagpole height/flagpole shadow=building height/building shadow

Therefore, it should look like this:
50/30= 300/x

Solve for x:
50x=9,000
X= 180 feet

YOUR ANSWER IS C.

7 0

3 years ago