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Darya [45]
3 years ago
7

What is the frequency of a wave that has a wavelength of 2000 meters and is traveling at 20 m/s?

Physics
1 answer:
Artist 52 [7]3 years ago
7 0

Answer:

freqiency=velocity/wavelength

ief=20/2000=0.01hz

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When light is reflected, the incident rays are bent and change direction.<br> True<br> False
Helga [31]

Answer: True

Explanation: When light is reflected off lets say a mirror it is bent and changes direction to bounce off of another wall or object. For example if you take a flash light and shine it into a mirror the light reflects into a different direction your welcome

8 0
3 years ago
Read 2 more answers
Two astronauts (each with mass 100 kg) are drifting together through space. They are connected to each other by a rope 5 m in le
Nana76 [90]

Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

600 J

Explanation:

m = Mass of astronauts = 100 kg

d = Diameter

r = Radius = \frac{d}{2}

v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s

The angular momentum of the system is 1000 kgm²/s

Rotational energy is given by

K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J

The rotational energy of the system is 400 J

There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s

Energy

E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J

The new energy will be 1000 J

Work done will be the change in the kinetic energy

W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J

The work done is 600 J

5 0
3 years ago
HELP ME PLEASE!!!!!!!!!
Stolb23 [73]

As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards

So here we know that

F = \frac{kq_1q_3}{d_{13}^2}

now from the above equation

F = \frac{(9\times 10^9)(2\times 10^{-6})(4 \times 10^{-6})}{0.5^2}

F = 0.288 N

so both of the charges will apply 0.288 N force on q3 charge along the line joining them

now the net force due to vector sum is given by

F_{net} = 2Fcos\theta

here we know that angle is

\theta = 37 degree

now we have

F_{net} = 2\times 0.288 cos37

F_{net} = 0.46 N

so net force on q3 is 0.46 N vertically upwards along +Y axis

6 0
3 years ago
The critical angle between a medium and air is 430. What is the speed of light in that medium?
vagabundo [1.1K]

Answer:

The speed of light is that medium is 281907786.2 m/s.

Explanation:

since the critical angle is Фc = 430, we know that the refractive index is given by:

n = 1/sin(Фc)

  = 1/sin(430)

  = 1.06

then if n is the refractive index of the medium and c is the speed of light, then the speed of light in the medium is given by:

v = c/n

  = (3×10^8)/(1.06)

  = 281907786.2 m/s

Therefore, the speed of light is that medium is 281907786.2 m/s.

8 0
3 years ago
You lay a mirror flat on the floor with one edge against a wall and aim a laser at the mirror. The ray reflects from the mirror
joja [24]

Answer:

the angle of incidence θ is 45.56 º

Explanation:

Given data

strikes the mirror before wall x = 30.7 cm

reflected ray strikes the wall y =  30.1 cm

to find out

the angle of incidence θ

solution

let us consider ray is strike at angle  θ so after strike on surface ray strike to wall at angle 90 - θ

we will apply here right angle triangle rule that is

tan( 90 - θ) = y /x

tan( 90 - θ)  = 30.1 / 30.7

90 - θ = tan^-1 (30.1/30.7)

90 - θ = 44.4345

θ = 45.56 º

the angle of incidence θ is 45.56 º

4 0
3 years ago
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