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cestrela7 [59]
3 years ago
10

If a certain shade of blue has a frequency of 7.33 x 10^14Hz what is the energy of exactly one photon of this light. Is this as

simple as multiplying the frequency by Planck's constant?
Physics
1 answer:
geniusboy [140]3 years ago
6 0
In calculating the energy of a photon of light, we need the relationship for energy and the frequency which is expressed as:
 
E=hv

where h is the Planck's constant (6.626 x 10-34 J s)and v is the frequency.
E = 6.626 x 10-34 J s (<span>7.33 x 10^14 /s) = 4.857 x 10^-19 J</span>
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Distance versus Displacement Worksheet
Umnica [9.8K]

when we find the distance we will add all the blocks so

distance = 6+6+4

distance = 14blocks

when we find the displacement we will add and minus too

As you can read he goes to the south 6 and to north 6 so he leave that place and back to the place again so the displacement is 0. and again he goes to the west 4 blocks so the displacement = <em><u>4blocks</u></em><em><u> </u></em><em><u>to</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>west</u></em>

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3 years ago
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What is the name of the hair-like structures on sponge cells that move back and forth to help move water, nutrients, and waste t
Dima020 [189]
I think it's B. Flagella

5 0
3 years ago
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If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle
ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0
2 years ago
The y-component of the force F which a person exerts on the handle of the box wrench is known to be 86 lb. Determine the x-compo
emmainna [20.7K]

Answer:

x-component of force is  38.18 lb where as magnitude of Force is 93.16

Explanation:

Fy of the force F exerted on the handle of the box wrench = 86 lb

Considering the triangle in Fig 1

magnitude of perpendicular = P =  12

magnitude of base = B = 5

using Pythagoras theorem

                        H= \sqrt{P^{2} + B^{2}}

                 H= \sqrt{12^{2} + 5^{2}}

            H=13\\\implies cos \theta = \frac{5}{13} \\\implies sin \theta = \frac{12}{13}\\

y-component of force is given given as:

                              86 = Fsin\theta\\F = 86 (\frac{13}{12})\\F = 93.16 lb\\F_{x} =F cos \theta\\F_{x} = 93.16 (\frac{5}{13})\\F_{x} =38.18 lb.

5 0
3 years ago
A particle moving along the x-axis has a position given by m, where t is measured in s. What is the magnitude of the acceleratio
8090 [49]

Question:

A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero

Answer:

24 m/s

Explanation:

Given:

x=(24t - 2.0t³)m

First find velocity function v(t):

v(t) = ẋ(t) = 24 - 2*3t²

v(t) = ẋ(t) = 24 - 6t²

Find the acceleration function a(t):

a(t) = Ẍ(t) = V(t) = -6*2t

a(t) = Ẍ(t) = V(t) = -12t

At acceleration = 0, take time as T in velocity function.

0 =v(T) = 24 - 6T²

Solve for T

T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2

Substitute -2 for t in acceleration function:

a(t) = a(T) = a(-2) = -12(-2) = 24 m/s

Acceleration = 24m/s

4 0
3 years ago
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