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IceJOKER [234]
3 years ago
7

Visible light travels at a speed 3.0 × 108 of m/s. If red light has a wavelength of 6.5 × 10–7 m, what the frequency of this lig

ht? 2.2 × 10–15 HZ 2.0 × 102 HZ 3.0 × 108 HZ 4.6 × 1014 HZ
Physics
1 answer:
zlopas [31]3 years ago
5 0

Answer:

4.6\cdot 10^{14} Hz

Explanation:

The frequency of an electromagnetic wave is given by:

f=\frac{c}{\lambda}

where

c=3.0\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

Red light has a wavelength of

\lambda=6.5\cdot 10^{-7} m

So by using the formula above we can calculate its frequency:

f=\frac{3.0\cdot 10^8 m/s}{6.5\cdot 10^{-7} m}=4.6\cdot 10^{14} Hz

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dusya [7]

C cost (I need points)

5 0
3 years ago
I need answers and solvings to these questions​
den301095 [7]

1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)

2) The angular acceleration is C) 15.7 rad/s^2

3) The frequency of the oscillation is C) 1.6 Hz

4) The period of vibration is B) 0.6 s

5) The diameter of the nozzle is A) 5.0 mm

6) The force that must be applied is B) 266.7 N

Explanation:

1)

The period of a simple pendulum is given by

T=2\pi \sqrt{\frac{L}{g}}

where

T is the period

L is the length of the pendulum

g is the acceleration of gravity

From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is

B) III. only (the length of the pendulum)

2)

The angular acceleration of the rotating disc is given by the equation

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f is the final angular velocity

\omega_i is the initial angular velocity

t is the time elapsed

For the compact disc in this problem we have:

\omega_i = 0 (since it starts from rest)

\omega_f = 300 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=31.4 rad/s is the final angular velocity

t = 2 s

Substituting, we find

\alpha = \frac{31.4-0}{2}=15.7 rad/s^2

3)

For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation

a=-\omega^2 x

where

a is the acceleration

x is the displacement

\omega is the angular frequency of the system

For the oscillator in this problem, we have the following relationship

a=-100 x

which implies that

\omega^2 = 100

And so

\omega = \sqrt{100}=10 rad/s

Also, the angular frequency is related to the frequency f by

f=\frac{\omega}{2\pi}

Therefore, the frequency of this simple harmonic oscillator is

f=\frac{10}{2\pi}=1.6 Hz

4)

When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write

mg=kx

where

m is the mass

g=9.8 m/s^2 is the acceleration of gravity

k is the spring constant

x = 8.0 cm = 0.08 m is the stretching of the spring

We can re-arrange the equation as

\frac{k}{m}=\frac{g}{x}=\frac{9.8}{0.08}=122.5

The angular frequency of the spring is given by

\omega=\sqrt{\frac{k}{m}}=\sqrt{122.5}=11.1 Hz

And therefore, its period is

T=\frac{2\pi}{\omega}=\frac{2\pi}{11.1}=0.6 s

5)

According to the equation of continuity, the volume flow rate must remain constant, so we can write

A_1 v_1 = A_2 v_2

where

A_1 = \pi r_1^2 is the cross-sectional area of the hose, with r_1 = 5 mm being the radius of the hose

v_1 = 4 m/s is the speed of the petrol in the hose

A_2 = \pi r_2^2 is the cross-sectional area of the nozzle, with r_2 being the radius of the nozzle

v_2 = 16 m/s is the speed in the nozzle

Solving for r_2, we find the radius of the nozzle:

\pi r_1^2 v_1 = \pi r_2^2 v_2\\r_2 = r_1 \sqrt{\frac{v_1}{v_2}}=(5)\sqrt{\frac{4}{16}}=2.5 mm

So, the diameter of the nozzle will be

d_2 = 2r_2 = 2(2.5)=5.0 mm

6)

According to the Pascal principle, the pressure on the two pistons is the same, so we can write

\frac{F_1}{A_1}=\frac{F_2}{A_2}

where

F_1 is the force that must be applied to the small piston

A_1 = \pi r_1^2 is the area of the first piston, with r_1= 2 cm being its radius

F_2 = mg = (1500 kg)(9.8 m/s^2)=14700 N is the force applied on the bigger piston (the weight of the car)

A_2 = \pi r_2^2 is the area of the bigger piston, with r_2= 15 cm being its radius

Solving for F_1, we find

F_1 = \frac{F_2A_1}{A_2}=\frac{F_2 \pi r_1^2}{\pi r_2^2}=\frac{(14700)(2)^2}{(15)^2}=261 N

So, the closest answer is B) 266.7 N.

Learn more about pressure:

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#LearnwithBrainly

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3 years ago
An artist wants to create a metal sculpture using a mold so that his artwork can be readily mass produced. He wants his sculptur
lukranit [14]

Answer:NO

Explanation:

No the mold should not be of the same size as that of sculpture because the material from which molds is made may shrink or expand depending upon its properties .

For example grey cast iron shrinks on cooling.

We need to make mold bigger in general so that if there is a need of finishing it can be done easily without altering the size of sculpture.

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3 years ago
Which type of relationship is formed when a megabat eats a fig and drops the
Alina [70]

The type of relationship formed when a megabat eats a fig and drops the seeds in a new location is COMENSALISM. It is an ecological interaction.

<h3>What is commensalism?</h3>

Commensalism is a type of ecological interaction in which one organism benefits (in this case, the tree)  and the other organism neither benefits nor harm (the megabat).

Mutualism is a type of ecological association in which both organisms benefit from such interaction.

Conversely, parasitism is a type of ecological interaction in which one organism benefits and the other organism is harmed.

Learn more about commensalism here:

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2 years ago
Time shifting occurs when _______.
Svet_ta [14]
The answer is C, individuals copy works to view at a later time.
4 0
3 years ago
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