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4 years ago

Under what circumstances will light display total internal reflection?

2 answers:

3 0

When light is passing from a denser medium to a lighter medium ( for example: from water to air)
 When the angle of refraction is greater than the critical angle( angle of incidence when the angle of refraction is perpendicular to the normal) of the denser medium.

Lapatulllka [165]4 years ago

3 0

Answer:

light passing from a low-velocity medium to a high-velocity medium at an incident angle larger than the critical angle will be totally reflected.

Explanation:

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Which of the following would be valid reason to NOT build the power plant?

Inessa [10]

Answer:

all of the above

Explanation:

because power plant are the ones that cause pollution,it can also destroy ecosystems for animals,and it cant be built anywhere.

6 0

3 years ago

Read 2 more answers

what velocity must a 1340kg car have in order to havw the same momentum as a 2680 kg truck traveling at a velocity of 15m/s to t

kykrilka [37]

Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.

Hope i helped
Have a good day :)

6 0

4 years ago

At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco

Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

$A_1*m=M*A_2$

$A_i =Area$

M,m = Counts per second

Our radios are given by

$r_1 = 11cm$

$R_2 = 20cm$

$m = 65cps$

Therefore replacing we have that,

$A_1*m=M*A_2$

$4\pi r_1^2*m = M * 4\pi R_2^2 M$

$r^2*m=MR^2$

$M = \frac{m*r^2}{R^2}$

$M = \frac{65*11^2}{20^2}$

$M = 19.6625cps$

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0

4 years ago

I WILL MARK YOU THE BRAINLIEST NO LINKS What goes were put them in the correct place

PIT_PIT [208]

Answer:

Fluid fricton goes to Static friction and sliding friction goes to rolling friction

Explanation:

6 0

3 years ago

Read 2 more answers

Please answer this question for me and explain why.

horsena [70]

Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass $m_2$ and the 4kg mass $m_1$ . If the tension in the string is $T$ then for the mass $m_2$

(1). $T-m_2g=-m_2a$ (the negative sign on the right side indicates that acceleration is downwards)

And for the mass $m_1$

(2). $T-m_1g =m_1a$ (the acceleration is upwards, hence the positive sign)

Solving for $T$ in the 2nd equation we get:

$T =m_1a+m_1g$ ,

and putting this into the 1st equation we get:

$m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g$

$\boxed{a= \dfrac{(m_2-m_1)}{(m_1+m_2)} g}$

Back to the question:

Using the formula for the acceleration we find

$a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g$

$a = \dfrac{g}{9},$

which is the acceleration that none of the given choices offer. Also, the acceleration of the two blocks is the same, because if it weren't, the difference in the instantaneous velocities of the objects would cause the string to break. Therefore, these two reasons make us decide that none of the choices are correct.

7 0

3 years ago