If you put a object in front of the mirror why does it appear the opposite way

1 answer:

4 0

It is because that is how mirrors work, they reflect light, and since we see objects because we are seeing the light these objects reflect, what is reflected back by the mirror is what we see.

You might be interested in

Explain how a star's absorption spectrum and the concept of red shift help support the Big Bang Theory.

marta [7]

Answer:

Red shift supports the big bang theory. ... The light from distant galaxies is red shifted (this tells us the galaxies are moving away from us) and the further away the galaxy the greater the red shift (this tells us that the more distant the galaxy the faster it is moving). Constellations look like they are moving because earth is rotating on it's axis.

May I have brainliest, please?

4 0

2 years ago

A uniform cylindrical grindstone has a mass of 10 kg and a radius of 12 cm. (a) What is the rotational kinetic energy of the gri

Drupady [299]

Answer:

a) KE = 888.26J

b) N = 294.5 turns

Explanation:

For the kinetic energy:

$KE = I/2*\omega_o^2$

The inertia is:

$I=m/2*R^2=0.072kg.m^2$

So, the kinetic energy will be:

$KE = 888.26J$

Now, friction force is:

Ff = μ*N = 0.80*5N = 4N

The energy balance would be:

Kf - Ko = Wf where Kf=0; Ko = 888.26J; and Wf is the work done by friction force.

Wf = -Ff*d = -Ff*N*2*π*R where N is the amount of turns it gives.

Replacing these values into the energy balance:

0-888.26=-4*N*2*π*0.12

-888.26=-0.96*π*N

N=294.5 turns

6 0

3 years ago

A cyclical heat engine, operating between temperatures of 450º C and 150º C produces 4.00 MJ of work on a heat transfer of 5.00

gogolik [260]

Answer:

(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%

Explanation:

Using the formula that relate heat and work from the thermodynamic theory as: $W=Q=Q_{in}-Q_{out}$ solving to Q_out we get: $Q_{out}=Q_{in}-W=5(MJ)-4(MJ)=1(MJ)$ this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: $n=1-\frac{T_{Low} }{T_{High}}$ where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get: $n=1-\frac{423.15}{723.15} =0.415=41.5%$