PHYSICS MIDDLE SCHOOL

The refractive index of water with respect to vacuum is 4/3 and refractive index of vacuum with respect to glass is 2/3. If the speed of light in glass is 2x 108m/s, find the speed of light in i) Vacuum ii) water

Answers

Answer 1

Answer:

As we know that

$\mu = (speed of light in air)/(speed of light in medium)$

now for the first part we know that

refractive index of vacuum with respect to glass is 2/3

speed of light in glass is $2 * 10^8 m/s$

now we will have

$(2)/(3) = (2* 10^8)/(c)$

$c = 3 * 10^8 m/s$

Part b)

Now similarly we can say

$\mu = (speed of light in air)/(speed of light in medium)$

here refractive index of water with respect to vacuum is 4/3

$(4)/(3) = (3 * 10^8)/(v)$

now we will have

$v = 2.25 * 10^8 m/s$

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If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?

Answers

Answer:50 $ms^(-1)$

Explanation:

Let $v_(a)$ be the airspeed.

Let $v_(w)$ be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If $v_(1)$ and $v_(2)$ are two perpendicular vectors,the resultant vector has the magnitude $\sqrt{|v|_(1)^(2)+|v|_(2)^(2)}$

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If 35 J of work was performed in 70 seconds, how much power was used to do this task?

Answers

$P = w * delta \: T$
P = Power
W = Work
Delta T = Change of Time

So:
P = 35 Joules
Delta T = 70 secs
W = ?

$P = 35 / 70 \n P = 0.5 \: watts$

* Joule/Sec = Watt

Answer: The power used to do this task was 0.5 Watts.

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Answers

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. A projectile is fired with an initial velocity of 113 m słatan angle of 60.0degrees above the horizontal from the top of a cliff 49.0 m high.
Calculate:
(a) the time to reach maximum height
(b) the maximum height above the base of the cliff reached by the
projectile
(c) thetotal time it is in the air
(d) the horizontal range of the projectile.

Answers

Answer:

a) 9.99 s

b) 538 m

c) 20.5 s

d) 1160 m

Explanation:

Given:

x₀ = 0 m

y₀ = 49.0 m

v₀ = 113 m/s

θ = 60.0°

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

a) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find t.

vᵧ = aᵧ t + v₀ᵧ

(0 m/s) = (-9.8 m/s²) t + (113 sin 60.0° m/s)

t ≈ 9.99 s

b) At the maximum height, the vertical velocity vᵧ = 0 m/s. Find y.

vᵧ² = v₀ᵧ² + 2aᵧ (y − y₀)

(0 m/s)² = (113 sin 60° m/s)² + 2 (-9.8 m/s²) (y − 49.0 m)

y ≈ 538 m

c) When the projectile lands, y = 0 m. Find t.

y = y₀ + v₀ᵧ t + ½ aᵧ t²

(0 m) = (49.0 m) + (113 sin 60° m/s) t + ½ (-9.8 m/s²) t²

You'll need to solve using quadratic formula:

t ≈ -0.489, 20.5

Since negative time doesn't apply here, t ≈ 20.5 s.

d) When the projectile lands, y = 0 m. Find x. (Use answer from part c).

x = x₀ + v₀ₓ t + ½ aₓ t²

x = (0 m) + (113 cos 60° m/s) (20.5 s) + ½ (0 m/s²) (20.5 s)²

x ≈ 1160 m

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