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Alexus [3.1K]
3 years ago
7

Which of the following is equivalent to Planks constant

Physics
1 answer:
Alinara [238K]3 years ago
8 0

Answer:

6.62607004 x 10^(-34)m²kg/s

Explanation:

This is the constant that shows the value of energy of a photon in relation to it's frequency.

Please let me know if you want this explained further!

Thanks!

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We are asked to solve for the index of refraction and the formula is n = c/v where "n" represents the index of refraction, "c" represents the speed of light in the vacuum while "v" represents the speed of another medium.
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c = 3 x 10^8 m/s
v = 2 x 10^8 m/s
n =?

Solving for n, we have the solution below:
n = 3x10^8 /  2x10^8
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The answer is 1.5 for the index of refraction.


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Why is it warmer near the equator and cooler near the poles?
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The doorbell transforms electrical energy into sound.

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3 years ago
Melvin is traveling south on I-95 at 29 m/s (65 mph) when a deer jumps into his path, 50 m ahead. a. If his reaction time is 0.1
aleksandr82 [10.1K]

Answer:

a. 5.22 meters

b. 2.9 seconds

c. No, Melvin does not hit the deer

Explanation:

The parameters with which Melvin is travelling are as follows;

The speed of Melvin's motion, u = 29 m/s

The distance from Melvin at which the deer jumps into the path = 50 m

a. Distance, d = Velocity, u × Time, t

The time it takes Melvin to react = 0.18 seconds

The distance, "d₁" Melvin travels before his foot hits the break = The velocity with which Melvin was traveling, "u" × The time duration it takes Melvin to hit the brakes, "t₁"

∴ d₁ = 29 m/s × 0.18 s = 5.22 m

The distance, Melvin travels before his foot hits the break = d₁ = 5.22 m

b. Melvin's acceleration after his foot hits the brakes, a = -10 m/s²

Therefore, we have;

The time it takes "t₂" it takes for him to come to a complete stop given as follows;

y = u + a × t₂

Where;

v = The final velocity after Melvin comes to a complete stop = 0 m/s

By substituting the known values, we have;

0 = 29 m/s + (-10 m/s²) × t₂ = 29 m/s - 10 m/s² × t₂

∴ 29 m/s = 10 m/s² × t₂

t₂ = (29 m/s)/(10 m/s²) = 2.9 s

The time it takes it takes for him to come to a complete stop = t₂ = 2.9 s

c. The distance, "d₂", Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop is given as follows;

v² = u² + 2·a·d₂

Therefore, we have;

0² = (29 m/s)² + 2 × (-10 m/s) × d₂ = (29 m/s)² - 2 × 10 m/s × d₂

∴  (29 m/s)² = 2 × 10 m/s × d₂

d₂ = ((29 m/s)²)/(2 × 10 m/s²) = (841 m²/s²)/(20 m/s²) = 42.05 m

The distance, Melvin reaches while accelerating (decelerating) at -10 m/s² to come to a complete stop = d₂ = 42.05 m

Given that d₂ = 42.05 m < 50 m (The distance separating Melvin's initial location and the deer, Melvin does not hit the deer.

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