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2 years ago

8. What is the frequency of green light waves that have a wavelength of 5.2 x 10-7 m.? The speed of light is 3.0 x 108 m/s

Physics

1 answer:

o-na [289]2 years ago

4 0

Answer:

$f=5.76\times 10^{14}\ Hz$

Explanation:

We need to find the frequency of green light having wavelength o $5.2\times 10^{-7}\ m$ . It can be calculated as follows :

$c=f\lambda\\\\f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{5.2\times 10^{-7}}\\\\f=5.76\times 10^{14}\ Hz$

So, the required frequency of green light is equal to $5.76\times 10^{14}\ Hz$ .

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Mrrafil [7]

For a simple harmonic motion energy is given with:
$E=\frac{1}{2}kA^2$
Where k is a constant that depends on the type of the wave you are looking at and A is amplitude.
Let's calculate the energy of the wave using two different amplitudes given in the problem:
$E_1=\frac{1}{2}k(1)^2=\frac{1}{2}k\\E_2=\frac{1}{2}k(2)^2=\frac{4}{2}k=2k\\$
We can see that energy associated with the wave is 4 times smaller when we decrease its amplitude by half. So the answer should be C.

6 0

3 years ago

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A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.

Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0

3 years ago

A kayaker needs to paddle north across a 100-m-wide harbor. The tide is going out, creating a tidal current that flows to the ea

Savatey [412]

Answer:

$41.81^{\circ}$

Explanation:

The tidal current flows to the east at 2.0 m/s and the speed of the kayaker is 3.0 m/s.

Let Vector $\overrightarrow{OA}$ is the tidal current velocity as shown in the diagram.

In order to travel straight across the harbor, the vector addition of both the velocities (i.e the resultant velocity, $\vec {R}$ must be in the north direction.

Let $\overrightarrow{AB}$ is the speed of the kayaker having angle \theta measured north of east as shown in the figure.

For the resultant velocity in the north direction, the tail of the vector $\overrightarrow {OA}$ and head of the vector $\overrightarrow{AB}$ must lie on the north-south line.