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STALIN [3.7K]
3 years ago
8

The speed of an electromagnetic wave is a constant, 3.0 × 108 m/s. The wavelength of a wave is 0.3 meters. What is the frequency

? A. 9.0 × 108 Hz B. 1.0 × 109 Hz C. 1.0 × 108 Hz D. 9.0 × 107 Hz
Physics
1 answer:
marusya05 [52]3 years ago
5 0

Answer: option B. 1.0 × 10⁹ Hz


Explanation:


The speed, s, of a wave, equals the product of its frequency, ν, times its wavelength, λ:

  • s = νλ.

As the question states, the speed of an electromagnetic wave is a constant, c, equal to 3.0 × 10⁸ m/s.


Substituting this constant in the equation for the speed of the wave, you get:

  • c =  νλ.

From that equation, you can solve for the frequency to show the inverse realation of frequency and wavelength:

  • ν =  c / λ

Now, you just have to substitute values and compute:

  • ν =  c / λ = 3.0 × 10⁸ m/s / 0.3m = 10. × 10⁸ Hz = 1.0 × 10⁹ Hz
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Five difference between elastic collision and inelastic collision?​
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Answer:

Elastic Collision

Inelastic Collision

The total kinetic energy is conserved. The total kinetic energy of the bodies at the beginning and the end of the collision is different.

Momentum does not change. Momentum changes.

No conversion of energy takes place. Kinetic energy is changed into other energy such as sound or heat energy.

Highly unlikely in the real world as there is almost always a change in energy. This is the normal form of collision in the real world.

An example of this can be swinging balls or a spacecraft flying near a planet but not getting affected by its gravity in the end.

8 0
3 years ago
Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an
joja [24]

Answer:

Approximately 9.7\; \rm m \cdot s^{-2}.

Explanation:

Assuming that there is no other force on this vehicle, the 16000\; \rm N force from the road would be the only force on this vehicle. The net force would then be equal to this 16000\; \rm N\! force. The size of the net force would be 16000\; \rm N\!\!.

Let m denote the mass of this vehicle and let \Sigma F denote the net force on this vehicle.  

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, a, would be:

\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}.

For this vehicle, \Sigma F = 16000\; \rm N whereas m = 1650\; \rm kg. The acceleration of this vehicle would be:

\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}.

8 0
3 years ago
What is the total momentum of a 30 kg object traveling left at 3 m/s and a 50 kg object traveling at 2 m/s to the right?
madam [21]

Answer:

there are 25 kg objective travelling at 2m/s to the right.

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2 years ago
"a wind shift from the south or southwest to the northwest is commonly associated with the passage of which type of front
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If i wouldve know i would tell you sorry
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3 years ago
An object is thrown from the ground with an initial velocity of 30 m/s. What is the velocity at the point 25 m above the ground?
dsp73

Answer:

It's a pretty simple suvat linear projectile motion question, using the following equation and plugging in your values it's a pretty trivial calculation.

V^2=U^2+2*a*x

V=0 (as it is at max height)

U=30ms^-1 (initial speed)

a=-g /-9.8ms^-2 (as it is moving against gravity)

x is the variable you want to calculate (height)

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x=-30^2/2*-9.8

x=45.92m

3 0
3 years ago
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