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Ivanshal [37]
2 years ago
10

When the angle of incidence equals the angle of reflection it is called the law of ​

Physics
2 answers:
BartSMP [9]2 years ago
8 0

The angle of incidence ALWAYS equals the angle of reflection.

THAT'S the law of reflection.

Reptile [31]2 years ago
6 0

The law of reflection

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What are two ways in which the suns energy can be captured and used?
My name is Ann [436]

The oldest way ... the way we've been using as long as we've been
walking on the Earth ... has been to use plants.  Plants sit out in the
sun all day, capturing its energy and using it to make chemical compounds. 
Then we come along, cut the plants down, and eat them.  Our bodies
rip the chemical compounds apart and suck the solar energy out of them,
and then we use the energy to walk around, sing, and play video games.  

Another way to capture the sun's energy is to build a dam across a creek
or a river, so that the water can't flow past it.  You see, it was the sun's
energy that evaporated the water from the ocean and lifted it high into
the sky, giving it a lot of potential energy.  The rain falls on high ground,
up in the mountains, so the water still has most of that potential energy
as it drizzles down the river to the ocean.  If we catch it on its way, we
can use some of that potential energy to turn wheels, grind our grain,
turn our hydroelectric turbines to get electrical energy ... all kinds of jobs. 

A modern, recent new way to capture some of the sun's energy is to use
photovoltaic cells.  Those are the flat blue things that you see on roofs
everywhere.  When the sun shines on them, they convert some of its
energy into electrical energy.  We use some of what they produce, and
we store the rest in giant batteries, to use when the sun is not there.
 
7 0
3 years ago
Read 2 more answers
A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 34 m/s when a 60
german

Answer:

34 m/s

Explanation:

m = Mass of glider with person = 680 kg

v = Velocity of glider with person = 34 m/s

m_1 = Mass of glider without person = 680-60 kg

v_1 = Gliders speed just after the skydiver lets go

m_2 = Mass of person = 60 kg

v_2 = Velcotiy of person = 34 m/s

As the linear momentum of the system is conserved

m_1v_1+m_2v_2=mv\\\Rightarrow v_1=\dfrac{mv-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{680\times 34-60\times 34}{680-60}\\\Rightarrow v_1=34\ m/s

The gliders speed just after the skydiver lets go is 34 m/s

3 0
3 years ago
Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
3 years ago
Ms. Sparkle bought 12 cans of diet soda. Each can
Anit [1.1K]
4.2 liters..... there are 1,000 mL in a liter and there is a total of 4200 mL in this case which is divided by 1000 which gives you 4.2 liters.
6 0
2 years ago
A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function
Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$v(t)=\sin (\omega t) * \cos ^2(\omega t)

To get the position equation we just need to integrate the above equation:

$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t

$\mathrm{u}=\cos (\omega \mathrm{t})

Then:

$d u=-\sin (\omega t) d t

\Rightarrow d t=-d u / \sin (\omega t)

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$f(t)=\frac{\cos ^3(\omega t)}{3}+C

To find the value of C, we use the fact that f(0) = 0.

$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0

C = -1 / 3

Then the position function is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

4 0
1 year ago
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