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DerKrebs [107]
3 years ago
13

26. The equation E=hf describes the energy of each photon in a beam of light. If Planck’s constant, h, were larger, would photon

s of light of the same frequency be more energetic or less energetic. Be sure to explain using at least 3 content related sentences.
Physics
1 answer:
Elenna [48]3 years ago
6 0
The equation of the energy of a photon is E=h*f.

If we increase the Planck's constant h, the energy would increase. 

For example, lets double the value of Planck's constant and name it H:

H=2*h. Now lets put that into the equation for energy that we will call E₂:

E₂=H*f=2*h*f=2*E. 

So we can clearly see that E₂=2*E or that if we double Planck's constant, the energy also doubles. 
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hammer [34]
33 mins hope it’s right
8 0
3 years ago
A force did 80 j of work on an object in 4 m how big was the force
yanalaym [24]

Work = (force) x (distance)

80 J = (force) x (4 m)

Force = (80 J) / (4 m)  =  20 N

That's IF the force was in the same direction as the 4m of motion.
If the force was kind of slanted, then it had to be stronger, and
it had a component of 20N in the direction of the motion.

3 0
3 years ago
Which of the following must be true about the object labeled X in the circuit below?
Alex787 [66]

the answer is c and if I help you  thank me
4 0
3 years ago
A 5.00-A current runs through a 12-gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5 * 1028 free ele
evablogger [386]

Answer:

a)n= 3.125 x 10^{19 electrons.

b)J= 1.515 x 10^{6 A/m²

c)V_{d =1.114 x 10^{4m/s

d) see explanation

Explanation:

Current 'I' = 5A =>5C/s

diameter 'd'= 2.05 x 10^{-3 m

radius 'r' = d/2 => 1.025 x 10^{-3 m

no. of electrons 'n'= 8.5 x 10^{28}

a) the amount of electrons pass through the light bulb each second can be determined by:

I= Q/t

Q= I x t => 5 x 1

Q= 5C

As we know that: Q= ne

where e is the charge of electron i.e 1.6 x 10^{-19C

n= Q/e => 5/ 1.6 x 10^{-19

n= 3.125 x 10^{19 electrons.

b)  the current density 'J' in the wire is given by

J= I/A => I/πr²

J= 5 / (3.14 x (1.025x 10^{-3)²)

J= 1.515 x 10^{6 A/m²

c) The typical speed'V_{d' of an electron is given by:

V_{d = \frac{J}{n|q|}

    =1.515 x 10^{6 / 8.5 x 10^{28} x |-1.6 x 10^{-19|

V_{d =1.114 x 10^{4m/s

d) According to these equations,

J= I/A

V_{d = \frac{J}{n|q|} =\frac{I}{nA|q|}

If you were to use wire of twice the diameter, the current density and drift speed will change

Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.

Also drift velocity will decrease as it is inversely proportional to the area

 

5 0
2 years ago
Read 2 more answers
The net external force on a golf cart is 411 N north of tha cart has a total mass of 281 kg what is tha cart acceleration
Alisiya [41]

1.46m/s²

Explanation:

Given parameters:

Mass of cart = 281kg

Net external force = 411N

Unknown:

Acceleration of the cart = ?

Solution:

According to newton's second law " the net force on a car is the product of its mass and acceleration";

            Force = mass x acceleration

Since acceleration of the cart is unknown;

     Acceleration = \frac{force}{mass} = \frac{411}{281}

  Acceleration = 1.46m/s²

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

8 0
3 years ago
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