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Lostsunrise [7]
2 years ago
7

Which describes the relationship between photon energy and frequency?

Physics
2 answers:
nasty-shy [4]2 years ago
8 0
E = hf
E : photon energy
h : Plank's constant 6.63×10^-34
f : frequency

Hope it helped!
vovangra [49]2 years ago
8 0

high energy photons have high frequencies hope this helps brother

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Micah knows that a car had a change in velocity of 15 m/s. What does micah need to determine acceleration.
katrin2010 [14]
The time component is needed. The acceleration is the change of velocity divided by the time in when this change of velocity happens.
4 0
2 years ago
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A person carries a plank of wood 2.00 m long with one hand pushing down on it at one end with a force f1 and the other hand hold
slega [8]

Answer:


F₁ = 4,120.2 N


F₂ = 3,924N


Explanation:



1) Balance of angular momentum around the end where F₁ is applied.


F₂ × 0.5m - F₁ × 0 = mass × g × 1m


⇒ F2 × 0.5 m= 20 kg × 9.81 m/s² × 1 m = 1,962 N×m


F₂ = 196.2 Nm / 0.5m = 3,924 N


2) Balance of forces


F₁ - F₂ = mg


F₁ = F₂ + mg = 3,924N + 20kg (9.81 m/s²) = 4,120.2 N

4 0
3 years ago
What are earths two main motions called
andre [41]
Rotation and revolution
3 0
3 years ago
A photograph of you and your friends at your 8th birthday party.is it secondary or primary source and why?
Reptile [31]

Answer:

primary source

Explanation:

the explanation is in the image above

brainliest please

7 0
2 years ago
A V = 108-V source is connected in series with an R = 1.1-kΩ resistor and an L = 34-H inductor and the current is allowed to rea
soldi70 [24.7K]

Answer:

Explanation:

Given an RL circuit

A voltage source of.

V = 108V

A resistor of resistance

R = 1.1-kΩ = 1100 Ω

And inductor of inductance

L = 34 H

After he inductance has been fully charged, the switch is open and it connected to the resistor in their own circuit, so as to discharge the inductor

A. Time the inductor current will reduce to 12% of it's initial current

Let the initial charge current be Io

Then, final current is

I = 12% of Io

I = 0.12Io

I / Io = 0.12

The current in an inductor RL circuit is given as

I = Io ( 1—exp(-t/τ)

Where τ is time constant and it is given as

τ = L/R = 34/1100 = 0.03091A

So,

I = Io ( 1—exp(-t/τ))

I / Io = ( 1—exp(-t/τ))

Where I/Io = 0.12

0.12 = 1—exp(-t/τ)

0.12 — 1 = —exp(-t/τ)

-0.88 = -exp(-t/0.03091)

0.88 = exp(-t/0.03091)

Take In of both sides

In(0.88) = In(exp(-t/0.03091)

-0.12783 = -t/0.030901

t = -0.12783 × 0.030901

t = 3.95 × 10^-3 seconds

t = 3.95 ms

B. Energy stored in inductor is given as

U = ½Li²

So, the current at this time t = 3.95ms

I = Io ( 1—exp(-t/τ))

Where Io = V/R

Io = 108/1100 = 0.0982 A

Now,

I = Io ( 1—exp(-t/τ))

I = 0.0982(1 — exp(-3.95 × 10^-3 / 0.030901))

I = 0.0982(1—exp(-0.12783)

I = 0.0982 × 0.12

I = 0.01178

I = 11.78mA

Therefore,

U = ½Li²

U = ½ × 34 × 0.01178²

U = 2.36 × 10^-3 J

U = 2.36 mJ

8 0
2 years ago
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