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3 years ago

5

When light moves from air into glass at an angle, is the light refracted toward or away from the normal?

1 answer:

Whitepunk [10]3 years ago

6 0

Answer: at an angle !!

Explanation:

light refracts at an angle

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According to the Second Law of Thermodynamics, a. any process during which the entropy of the universe increases will be product

Iteru [2.4K]

Answer:

The correct option is;

a. Any process in which the entropy of the universe increases will be product-favored

Explanation:

According to the second law of thermodynamics, the change in entropy of a closed system with time is always positive. That is the entropy of the entire universe, considered as an isolated system, always increases with time, hence the entropy change in the universe will always be positive.

$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} >0$

Therefore, any process in which the entropy of the universe increases will be product favored.

4 0

3 years ago

A smooth circular hoop with a radius of 0.800 m is placed flat on the floor. A 0.300-kg particle slides around the inside edge o

Marrrta [24]

Answer:

a)11.25 J

b)Number of revolution = 1

Explanation:

Given that

Radius ,r= 0.8 m

m= 0.3 kg

Initial speed ,u= 10 m/s

final speed ,v= 5 m/s

a)

Initial energy

$KE_i=\dfrac{1}{2}mu^2$

$KE_i=\dfrac{1}{2}0.3\times 10^2$

KEi= 15 J

Final kinetic energy

$KE_f=\dfrac{1}{2}mv^2$

$KE_f=\dfrac{1}{2}0.3\times 5^2$

KEf=3.75 J

The energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J

b)

The minimum value to complete the circular arc

$V=\sqrt{r.g}$

Now by putting the values

$V=\sqrt{0.8\times 10}$

V= 2.82 m/s

So kinetic energy KE

$KE=\dfrac{1}{2}mV^2$

$KE=\dfrac{1}{2}0.3\times 2.82^2$

KE=1.19 J

ΔKE= KEi - KE

ΔKE= 15- 1.19 J

ΔKE=13.80 J

The minimum energy required to complete 2 revolutions = 2 x 11.25 J

= 22.5 J

Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.

Number of revolution = 1

4 0

3 years ago

Which year was pluto no longer considered a planet?.

katrin2010 [14]

2006, i hope this helps

6 0

2 years ago

A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni

mina [271]

Answer:

Induced emf, $\epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}$

Explanation:

The varying magnetic field with time t is given by according to equation as :

$B=B_{max}e^{-t/\tau}$

Where

$B_{max}\ and\ t$ are constant

Let $\epsilon$ is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

$\epsilon=-\dfrac{d\phi}{dt}$

$\epsilon=-\dfrac{d(BA)}{dt}$

$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}$

$\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}$

So, the induced emf in the loop as a function of time is $A\dfrac{B_{max}e^{-t/\tau}}{\tau}$ . Hence, this is the required solution.

7 0

3 years ago

An astronaut weighs 863 N on Earth. What is the astronaut’s mass?

Alecsey [184]

86.3 pounds.... I think

6 0

4 years ago

Read 2 more answers