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s344n2d4d5 [400]
4 years ago
8

As the wavelength increases, the frequency (2 points) decreases and energy decreases. increases and energy increases. decreases

and energy increases. increases and energy decreases
Physics
2 answers:
lora16 [44]4 years ago
4 0

Answer:

I TOOK THE TEST! the answer is INCREASES AND ENERGY INCREASES!

frozen [14]4 years ago
3 0
Bohr's equation for the change in energy is
\Delta E= \frac{hc}{\lambda}
where
h = Planck's constant
c == the velocity of light
λ = wavelength.

The velocity is related to wavelength and frequency, f, by
c = fλ

Let us examine the given answers on the basis of the given equations.

a. As λ increases, f decreases and ΔE decreases.
     TRUE

b. As λ increases, f increases and ΔE increases.
    FALSE

c. As λ increases, f increases and ΔE decreases.
    FALSE

Answer: 
As the wavelength increases, the frequency decreases and energy decreases.

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what do radio waves transfer between a cell phone and a cell phone tower? A.Sound. B.Energy. C. the medium D. air particles
sergeinik [125]
The answer would be energy, because all wave emitters give off at least one type of energy....
8 0
4 years ago
How much time is required for a car to reach a speed of 100 m/s if it starts from rest and its average acceleration is 12 m/s2?
gtnhenbr [62]

Answer:

t = 8.3s

Explanation:

V = 100 m/s U = 0 m/s t = ? a = 12 m/s2

V = U + at

100 = 0 + 12 × t = 100 = 12t

12t = 100

t = 100/12 = 25/3

t = 8.3s

7 0
3 years ago
Read 2 more answers
What objects have less resistance
Leto [7]

Answer:

Free fall and air resistance

5 0
3 years ago
Gas stored in a tank at 273 k has a pressure of 388kpa. The safe limit for the pressure is 825kpa at whst temperature Will gas r
vagabundo [1.1K]

Answer:

The answer to your question is  T2 = 580.5 °K

Explanation:

Data

Temperature 1 = T1 = 273°K

Pressure 1 = P1 = 388 kPa

Pressure 2 = P2 = 825 kPa

Temperature 2 = ?

Process

1.- Use the Gay-Lussac law

                  P1/T1 = P2/T2

-Solve for T2

                  T2 = P2T1/P1

-Substitution

                  T2 = (825)(273) / 388

-Simplification

                   T2 = 225225 / 388

-Result

                    T2 = 580.5 °K

6 0
4 years ago
To maintain a constant speed, the force provided by a car's engine must equal the drag force plus the force of friction of the r
bogdanovich [222]

Answer:

Toyota Camry

     F_d = 51.852 N

     F_d = 100.042 N

Hummer H2

     F_d = 412.0888 N

     F_d = 8351.755 N

Explanation:

Given:

- The density of air p_air = 1.2 kg/m^3

- The drag force equals car's engine force.

Find:

- What are the drag forces in newtons at 80 km/h and 105 km/h for a Toyota Camry? (Drag area = 0.70 m2 and drag coefficient = 0.28.)

- What are the drag forces in newtons at 80 km/h and at 105 km/h for a Hummer H2? (Drag area = 2.44 m2 and drag coefficient = 0.57.)

Solution:

- The formula for drag force is given as follows:

                                 F_d = 0.5*C_d*p_air*A*V^2

Where,

A : The drag Area  m^2

C_d: The drag coefficient

V: Velocity  m/s

a)  Toyota Camry

                  C_d = 0.28 , A = 0.70 m^2 , V = 80 km/h = 22.222 m/s

Then compute the F_d drag force:

                  F_d = 0.5*0.28*1.2*0.70*(22.22)^2

                  F_d = 51.852 N

                 C_d = 0.28 , A = 0.70 m^2 , V = 105 km/h = 29.1667 m/s

Then compute the F_d drag force:

                  F_d = 0.5*0.28*1.2*0.70*(29.1667)^2

                  F_d = 100.042 N

b)   Hummer H2

                 C_d = 0.57 , A = 2.44 m^2 , V = 80 km/h = 22.222 m/s

Then compute the F_d drag force:

                  F_d = 0.5*0.57*1.2*2.44*(22.22)^2

                  F_d = 412.0888 N

                 C_d = 0.28 , A = 0.70 m^2 , V = 105 km/h = 29.1667 m/s

Then compute the F_d drag force:

                  F_d = 0.5*0.57*1.2*2.44*(29.1667)^2

                  F_d = 8351.755 N

4 0
3 years ago
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