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2 years ago

12

What is the Index of refraction of a material of the speed of light in that materials is 2.251 x 10^8 m/s ? Use 3.00 x 10^8 m/s

for the speed of light in

1 answer:

Sladkaya [172]2 years ago

7 0

Answer:

1.33

Explanation:

$n = \frac{c}{v} \\ n = \frac{3 \times 10 {}^{8} }{2.251 \times {10}^{8} } \\ n = 1.33$

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Which shows the formula for converting from kelvins to degrees Celsius? °C = (9/5 × K) +32 °C = 5/9 × (K – 32) °C = K – 273 °C =

madam [21]

The freezing point of the water is 0 C , and it equals to 273 K

Then, To convert from Kelvins degrees to Celsius degrees we use the relation

$K = C + 273$

Also,

$C = K - 273$

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3 years ago

Read 2 more answers

A ticker timer makes 50 dots per second. When a body is pulled by a tap through the timer, the distance between the third and fo

fenix001 [56]

Answer:

The acceleration is 1 cm/s^2.

Explanation:

The acceleration is defined as the rate of change of velocity.

Here, initial velocity, u = 3/1 = 3 cm/s

final velocity, v = 4/1 = 4 cm/s

time, t = 1 s

Let the acceleration is a.

Use first equation of motion

v = u + at

4 = 3 + 1 x a

a = 1 cm/s^2

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2 years ago

A 10.0 g bullet moving at 300m/s is fired into a 1.00 kg block at rest. The bullet emerges (the bullet does not get embedded in

chubhunter [2.5K]

Answer:

v' = 1.5 m/s

Explanation:

given,

mass of the bullet, m = 10 g

initial speed of the bullet, v = 300 m/s

final speed of the bullet after collision, v' = 300/2 = 150 m/s

Mass of the block, M = 1 Kg

initial speed of the block, u = 0 m/s

velocity of the block after collision, u' = ?

using conservation of momentum

m v + Mu = m v' + M u'

0.01 x 300 + 0 = 0.01 x 150 + 1 x v'

v' = 0.01 x 150

v' = 1.5 m/s

Speed of the block after collision is equal to v' = 1.5 m/s

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2 years ago

The work energy principle states that the change in kinetic energy of an object is equal to the net work done on the object. If

sattari [20]

Answer:

$v=\sqrt{2gh}\ m/s$

Explanation:

From work energy theorem

Work done by all forces = Change in kinetic energy

Lets take

m= mass of object

h=height from the ground surface

initial velocity of object = 0 m/s

The final velocity of object is v

Work done by gravitational force = m g . h

The final kinetic energy = 1/2 m v²

So

Work done by all forces = Change in kinetic energy

m g h = 1/2 m v² - 0

v² = 2 g h

$v=\sqrt{2gh}\ m/s$

6 0

3 years ago

A 79.7 kg base runner begins his slide into second base while moving at a speed of 4.77 m/s. The coefficient of friction between

SashulF [63]

To solve this problem we will apply the concept related to the kinetic energy theorem. Said theorem states that the work done by the net force (sum of all forces) applied to a particle is equal to the change experienced by the kinetic energy of that particle. This is:

$\Delta W = \Delta KE$

$\Delta W = \frac{1}{2} mv^2$

Here,

m = mass

v = Velocity

Our values are given as,

$m = 79.7kg$

$v = 4.77m/s$

Replacing,

$\Delta W = \frac{1}{2} (79.7kg)(4.77m/s)^2$

$\Delta W = 907J$

Therefore the mechanical energy lost due to friction acting on the runner is 907J

6 0

3 years ago