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cestrela7 [59]
4 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]4 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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An escalator carries you from one level to the next in an airport terminal. The upper level is 4.4m above the lower level, and t
STALIN [3.7K]

Answer:

Part A W= 2772J

Part B = -2.772J

Explanation:

Work = force × distance

Part A when you move from lower to upper level

= mgh

= 63× 4.4×10

=2772J

=2.772KJ

Part B

When you move from upper to lower

The acceleration due to gravity becomes negative -10m/s^2

Therefore Work =-2772KJ

Work = -2.772KJ

8 0
3 years ago
Wat is the volume of a marble in mL?
rewona [7]
4/3 pi (radius)^3
If the radius is in cm the volume will be in mL
6 0
3 years ago
Two equally charged, 2.807 g spheres are placed with 3.711 cm between their centers. When released, each begins to accelerate at
statuscvo [17]
\begin{gathered} m=\text{ 2.807 g} \\ d=\text{ 3.711cm} \\ a=260.125\text{ m/s}^2 \\ m=\text{ mass of  both the spheres} \\ d=\text{ distance between the centers of sphere.} \\ a=\text{ acceleration of spheres.} \end{gathered}\begin{gathered} force\text{ due to the sphere having charge q, outside its surface is given by } \\ \vec{F}=\frac{1}{4\pi\epsilon_o}\frac{q_1q_2}{r^2}\hat{r} \\ q_1=charge\text{ on the source object.} \\ q_2=charge\text{ of the object in which we are observing the force.} \\ F=\text{ the force on the charged particle outside the sphere} \\ r=\text{ distance of the charged particle from the center of the sphere} \\ \hat{r}\text{= direction of the force acting on the charged particle} \end{gathered}\begin{gathered} from\text{ Newton's second law} \\ F=ma \\ F=\text{ force acting on the particle.} \\ m=\text{ mass of the object.} \\ a=\text{ acceleration of the object.} \end{gathered}\begin{gathered} from\text{ both the equation } \\ ma=\frac{1}{4\pi\epsilon_o}\frac{q_1q_{\frac{2}{}}}{r^2}\hat{r} \\ here\text{ q}_1\text{ and q}_2\text{ are the same, according to the question.} \end{gathered}\begin{gathered} converting\text{ all the values in s.i. unit} \\ m=2.807*10^{-3}kg \\ d=3.711*10^{-2}m \\ according\text{ to the question q}_1=\text{ q}_2 \\ value\text{ of }\frac{1}{4\pi\epsilon_o}=9*10^9\text{ Nm}^2\text{/C}^2 \\ now\text{ put all the values in the above equation } \\ 2.807*10^{-3}kg*260.125\text{ m/s\textasciicircum2}=9*10^9Nm^2\text{/C}^2*\frac{q^2}{3.711*10^{-2}m} \\  \end{gathered}\begin{gathered} by\text{ trasformation} \\ q=\sqrt{\frac{2.807*10^{-3}kg*260.125m/s^2*3.711*^10^{-2}m}{9*10^9Nm^2\text{/C}^2}} \\ by\text{ solving this we get } \\ q=17.3514*10^{-7}C \\ q=1.73514\text{ micro coulombs.} \end{gathered}Hence the correct answer is q= 1.73514 micro coulombs.
5 0
1 year ago
1. A concave mirror has a focal length of 1.50 meters. What is the radius of curvature of the mirror? An object is placed 4.00 m
matrenka [14]

1) 3.0 m

2) 2.40 m

Explanation:

1)

A concave mirror is a reflecting surface that causes the reflection of the rays of light coming to the mirror, producing an image of the object facing the mirror.

There are two types of mirror:

- Concave mirror: this is curved inward - as a result, the rays of light coming from the object are reflected back into a single point, called focal point

- Convex mirror: this is curved outward - as a result, the rays of light coming from the object are reflected back into diverging direction, not into a single point

For a curved mirror, the radius of curvature is twice the focal length:

R=2f

Where

R is the radius of curvature

f is the focal length

In this problem,

f = 1.50 m

So, the radius of curvature is

R=2(1.50)=3.0 m

2)

The distance of the image from the mirror can be found by using the mirror equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p is the distance of the object from the mirror

q is the distance of the image from the mirror

IN this problem we have:

f = 1.50 m is the focal length

p = 4.00 m is the distance of the object from the mirror

Solving for q, we  find:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{1.50}-\frac{1}{4.00}=0.416\\q=\frac{1}{0.416}=2.40 m

8 0
3 years ago
If you add together all of the forces exerted on a object and get a non zero value that is called
True [87]

Hello There!

<em>If you add together all of the forces exerted on a object and get a non zero value that is called the "NET FORCE" of the object</em>

3 0
3 years ago
Read 2 more answers
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