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3 years ago

Q3. What is the minim value of refractive index?

Physics

2 answers:

Airida [17]3 years ago

6 0

Answer:

Absolute refractive index is 1

Relative refractive index is lesser than 1.

Explanation:

Stels [109]3 years ago

6 0

Minimum value of Absolute refractive index is 1 and relative refractive index is lesser than 1.

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5 0

1 year ago

slader Question: A Model Rocket Is Launched Straight Upward With An Initial Speed Of 50m/s. Iit Accelerates With A Constant Upwa

xenn [34]

Answer:

Maximum height reached by the rocket is

$y_{max} = 308 m$

total time of the motion of rocket is given as

$T = 16.44 s$

Explanation:

Initial speed of the rocket is given as

$v_i = 50 m/s$

acceleration of the rocket is given as

$a = 2 m/s^2$

engine stops at height h = 150 m

so the final speed of the rocket at this height is given as

$v_f^2 - v_i^2 = 2 a d$

$v_f^2 - 50^2 = 2(2)(150)$

$v_f = 55.68 m/s$

so maximum height reached by the rocket is given as the height where its final speed becomes zero

so we will have

$v_f^2 - v_i^2 = 2 a d$

$0 - 55.68^2 = 2(-9.81)(y - 150)$

$y_{max} = 308 m$

Now the total time of the motion of rocket is given as

1) time to reach the height of 150 m

$v_f - v_i = at$

$55.68 - 50 = 2 t$

$t_1 = 2.84 s$

2) time to reach ground from this height

$\Delta y = v_y t + \frac{1}{2}gt^2$

$-150 = 55.68 t - \frac{1}{2}(9.81) t^2$

$t_2 = 13.6 s$

so total time of the motion of rocket is given as

$T = 13.6 + 2.84 = 16.44 s$

3 0

3 years ago

A car starts from rest and accelerated at a rate of 9m/s squared for 6.5 seconds. How much distance was covered by the car?

ki77a [65]

Answer:

58.5 meters

Explanation:

1. Find your formula. Use distance = speed x time for this problem

2. Plug in the given information. d (for distance) = 9m/s^2 * 6.5 s

3. Multiply number AND units. d = 58.5m

4. Check to make sure units & numbers make sense. In this case check that the answer is a lot bigger than what we stated with and that our units go with distance

5 0

2 years ago

Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball

Sidana [21]

Answer:

255.4 N/m

Explanation:

We can consider the system eyeball-attached to the musculature as a mass-spring system in simple harmonic motion, whose frequency of oscillation is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where in this case, we know:

f = 29 Hz is the frequency of oscillation

k is the spring constant, which is unknown

m = 7.7 g = 0.0077 kg is the mass of the eyeball

Solving the equation for k, we find the spring constant of the musculature attached to the eyeball:

$k=(2\pi f)^2 m=(2 \pi (29 Hz))^2 (0.0077 kg)=255.4 N/m$

6 0

3 years ago

The weight of a boy having a mass of 50 kg is_____ N

Elden [556K]

Answer: 490

Explanation:

5 0

3 years ago