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2 years ago

9

What is the index of refraction of the glass?

1 answer:

Serggg [28]2 years ago

4 0

You don't need to do any calculations, you are already given that n=1.5
n is index of refraction. it's 1.5

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How does friction affect the distance of an object?

stiv31 [10]

It slows the object down so it cannot move well and evetually the object cannot be pushed and farther

4 0

2 years ago

Two bodies of specific heats S1 and S2 having the same heat capacities are combined to form a single composite body. What is the

Dafna11 [192]

$\qquad\qquad\huge\underline{{\sf Answer}}♨$

Heat capacity of body 1 :

$\qquad \sf \dashrightarrow \:m_1s_1$

Heat capacity of body 2 :

$\qquad \sf \dashrightarrow \:m_2s_2$

it's given that, the the head capacities of both the objects are equal. I.e

$\qquad \sf \dashrightarrow \:m_1s_1 = m_2s_2$

$\qquad \sf \dashrightarrow \:m_1 = \dfrac{m_2s_2}{s_1}$

Now, consider specific heat of composite body be s'

According to given relation :

$\qquad \sf \dashrightarrow \:(m_1 + m_2) s' = m_1s_1 + m_2s_2$

$\qquad \sf \dashrightarrow \:s' = \dfrac{ m_1s_1 + m_2s_2}{m_1 + m_2}$

$\qquad \sf \dashrightarrow \:s' = \dfrac{ m_2s_2+ m_2s_2}{ \frac{m_2s_2}{s_1} + m_2 }$

[ since, $m_2s_2 = m_1s_1$ ]

$\qquad \sf \dashrightarrow \:s' = \dfrac{ 2m_2s_2}{ m_2(\frac{s_2}{s_1} + 1)}$

$\qquad \sf \dashrightarrow \:s' = \dfrac{ 2 \cancel{m_2}s_2}{ \cancel{m_2}(\frac{s_2}{s_1} + 1)}$

$\qquad \sf \dashrightarrow \:s' = \dfrac{ 2 s_2}{ (\frac{s_2 + s_1}{s_1} )}$

$\qquad \sf \dashrightarrow \: s' = \dfrac{2s_1s_2}{s_1 + s_2}$

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6 0

2 years ago

Read 2 more answers

A cheetah starts from rest and accelerated at 8.7 m/s^2 for 3s. How far did the cheetah go in that time

nika2105 [10]

Answer:

cheetah goes 52.2 m in that time.

Explanation:

4 0

2 years ago

Hii! help asap. i’ll give brainliest thanks!

oksian1 [2.3K]

Answer:

a i believe

Explanation:

7 0

2 years ago

A 1.0 meter long rod is held horizontally in the east-west direction on a distant planet and is dropped from a height of 1.23 m.

enyata [817]

Answer:

The induced emf between two end is $34.02 \times 10^{-5}$ V

Explanation:

Given:

Length of rod $l = 1$ m

Height $h = 1.23$ m

Magnetic field $B = 6.93 \times 10^{-5}$ T

For finding induced emf,

$\epsilon = Blv$

Where $v =$ velocity of rod,

For finding the velocity of rod.

From kinematics equation,

$v^{2} = v_{o} ^{2} + 2gh$

Where $v_{o} =$ initial velocity, $g = 9.8 \frac{m}{s^{2} }$

$v = \sqrt{2gh}$

$v = \sqrt{2 \times 9.8 \times 1.23}$

$v = 4.91$ $\frac{m}{s}$

Put the velocity in above equation,

$\epsilon = 6.93 \times 10^{-5} \times 1 \times 4.91$

$\epsilon = 34.02 \times 10^{-5}$ V

Therefore, the induced emf between two end is $34.02 \times 10^{-5}$ V

6 0

2 years ago