All categories

2 years ago

What is the energy of a photon of blue light with a frequency of 6.53 × 10¹4 Hz?

Physics

1 answer:

shusha [124]2 years ago

3 0

Answer:

V = 6.53 × 10^14

h = 6.626 × 10^-34

E = hV

E = (6.626 × 10^-34) (6.53 × 10^14)

E = 4.33 × 10^-19

You might be interested in

What is the acceleration of a 4.5 kg mass pushed by a 15n force?<br> ANSWER ASAP PLEASE

nikitadnepr [17]

Answer: 3.33 m/s²

Explanation:

F = (m)(a)

15 N = (4.5 kg)(a)

15/4.9 = 3.33

a = 3.33 m/s²

3 0

3 years ago

A rower in a boat pushes the water using an oar.

brilliants [131]

Answer:

It's B, because of process of elimination:

A: The rower would be applying force to the oar not the other way around. So that's immediately out.

C: If the force of the oar applies to the water, the force of the water cannot apply to the oar.

D: Same as why A is incorrect.

So B is the only one left.

6 0

3 years ago

Read 2 more answers

A 10 kg rock is suspended 320 m above the ground.

alina1380 [7]

Potential energy<span> is energy which results from position or configuration. It is the energy stored in an object not moving. It is calculated by:

PE = mgh

where m is the mass, g is the gravitational acceleratin and h is the height.

PE = 10 ( 9.8) (320) = 31360 J = </span>3.1 x 104 J --------> OPTION 2

7 0

3 years ago

A bat at rest sends out ultrasonic sound waves at 46.2 kHz and receives them returned from an object moving directly away from i

murzikaleks [220]

Answer:

f" = 40779.61 Hz

Explanation:

From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;

from the Doppler effect equation, we can calculate the initial observed frequency as:

f' = f(1 - (v_o/v))

We are given;

f = 46.2 kHz = 46200 Hz

v_o = 21.8 m/s

v is speed of sound = 343 m/s

Thus;

f' = 46200(1 - (21/343))

f' = 43371.4285 Hz

In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;

Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;

f" = f'/(1 + (v_o/v))

f" = 43371.4285/(1 + (21.8/343))

f" = 40779.61 Hz

4 0

3 years ago

Water is entering the prism at a rate of A m^3/hr. The prism is empty at time 0. Express the depth d of the water in meters in t

zavuch27 [327]

This question is incomplete, the complete question is;

The picture shows a triangular prism. The end of prism are equilateral triangles with x meters. the other dimension of the prism is L meters

a) Find the volume V in terms of x and L

b) Water is entering the prism at a rate of A m³/hr. The prism is empty at time 0. Express the depth d of the water in meters in terms of A, the length of time t the water has been entering the trough, and the length L of the prism.

Answer:

a) the volume V in terms of x and L is ((√3/4)x²L) m³

b) required expression is (2/(3)^(1/u))√(At/L)

Explanation:

Given that;

form the question and image below;

triangular prism ends are equilateral triangle

side length = x meter

Dimension of the prism = L meter

Area of the equilateral triangle = √3/4 (side)² = √3/4 (x)² meter

Volume of the triangular prism = Area × height

= √3/4 (x)² × L

V = ((√3/4)x²L) m³

Therefore, the volume V in terms of x and L is ((√3/4)x²L) m³

Rate of water entering = A m³/hr

Depth of water tank = d meter

Time = t

Length of prism = L

now Rate of water entering is A m³/hr

dv/d = A [ V = ((√3/4)x²L) m³ ]

and

dv/dt = √3/4 [2x dx/dt ] L { L is constant }

A = √3/4 [2x dx/dt ] L

∫A dt = √3/2 [ Lx dx ] { Integrate both sides}

At = √3/2 × Lx × x²/2

x² = uAt / √3L { we find square root of both sides}

x = √( uAt / √3L )

x = (2/(3)^(1/u))√(At/L)

Therefore; required expression is (2/(3)^(1/u))√(At/L)

8 0

3 years ago