An example of a solution is

The light emitted by an argon laser light includes light with a wavelength of 488 nm. what is the frequency of this light? 6.15

Paraphin [41]

Frequency and wavelength are inversely related. That is, when frequency increases, the value of the wavelength decreases and vice versa. The constant of proportionality of the relation is the speed of light that has a value of 3.0x10^8 m/s. We calculate as follows:

wavelength = speed of light /frequency
488x10^-9 = 3.8x10^8 / frequency
frequency = 6.15x10^14 /s = 6.15x10^14 Hz ------> OPTION 3

3 0

4 years ago

The Gulf Stream off the east coast of the United States can flow at a rapid 3.9 m/s to the north. A ship in this current has a c

Alex Ar [27]

Answer:

72.54 degree west of south

Explanation:

flow = 3.9 m/s north

speed = 11 m/s

to find out

point due west from the current position

solution

we know here water is flowing north and ship must go south at an equal rate so that the velocities cancel and the ship just goes west

so it become like triangle with 3.3 point down and the hypotenuse is 11

so by triangle

hypotenuse ×cos(angle) = adjacent side

11 ×cos(angle) = 3.3

cos(angle) = 0.3

angle = 72.54 degree west of south

3 0

3 years ago

Read 2 more answers

A student throws a 0.22 kg rock horizontally at 20.0 m/s from 10.0 m above the ground. Find the initial kinetic energy of the ro

LekaFEV [45]

Answer:

44J

Explanation:

Given parameters:

Mass of rock = 0.22kg

Initial velocity = 20m/s

Distance moved = 10m

Unknown:

Initial kinetic energy of the rock = ?

Solution:

To solve this problem, we need to understand that kinetic energy is the energy due to the motion of a body.

It is mathematically expressed as;

Kinetic energy = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

Kinetic energy = $\frac{1}{2}$ x 0.22 x 20² = 44J

6 0

3 years ago

To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeof

BlackZzzverrR [31]

<h2>Answer: 26,8 s</h2>

Explanation:

If we are talking about an acceleration at a constant rate , we are dealing with constant acceleration, hence we can use the following equations:

${V_{f}}^{2}={V_{o}}^{2}+2ad$ (1)

$V_{f}=V_{o}+at$ (2)

Where:

$V_{f}$ is the final velocity of the plane (the takeoff velocity in this case)

$V_{o}=0$ is the initial velocity of the plane (we know it is zero because it starts from rest)

$a=5m/s^{2}$ is the constant acceleration of the plane to reach the takeoff velocity

$d=1800m$ is the distance of the runway

$t$ is the time

Knowing this, let's begin with (1):

${V_{f}}^{2}=0+2(5m/s^{2})(1800m)$ (3)

${V_{f}}^{2}=18000m^{2}/s^{2}$ (4)

$V_{f}=134.164 m/s$ (5)

Substituting (5) in (2):

$134.164 m/s=0+(5m/s^{2})t$ (6)

Finding $t$ :

$t=26.8 s$ This is the time needed to take off

6 0

3 years ago

Two particles, each with charge 55.3 nC, are located on the y axis at y 24.9 cm and y -24.9cm (a) Find the vector electric field

ser-zykov [4K]

Answer:

Ex = kq 2x / ∛ (x² + y²)² and Ex = 2008 N / C

Explanation:

a) The electric field is a vector quantity, so we must find the field for each particle and add them vectorially, as the whole process is on the X axis,

The equation for the electric field produced by a point charge is

E = k q / r²

With r the distance between the point charge and the positive test charge

We look for each electric field

Particle 1. Located at y = 24.9 m, let's use Pythagoras' theorem to find the distance

r² = x² + y²

E1 = k q / (x² + y²)

Particle 2. located at x = -24.9 m

r² = x² + y²

E2 = k q / (x² + y²)

We can see that the two fields are equal since the particles have the same charge and coordinate it and that is squared.

In the attached one we can see that the Y components of the electric fields created by each particle are always the same and it is canceled, so we only have to add the X components of the electric fields. Let's use Pythagoras' theorem to find

Let's measure the angle from axis X

cos θ = CA / H = x / (x2 + y2) ½

E1x = E1 cos θ

E2x = = E1 cos θ

The resulting field

Ey = 0

Ex = E1x + E2x 2 E1x

Ex = 2 k q / (x² + y²) cos θ) = 2 k q / (x² + y²) x / √(x² + x²)

Ex = kq 2x / ∛ (x² + y²)²

b) For this part we substitute the numerical values

Ex = 8.99 10⁹ 55.3 10⁻⁹ x / (x² + 0.249 2) ³/₂

Ex = 497.15 x / (x² + 0.062) ³/₂

Point where can the value of the electric field x = 38.1 cm = 0.381 m

Ex = 497.15 0.381 / (0.381² + 0.062) ³/₂

Ex = 497.15 0.381 / (0.1452 + 0.062) 3/2 = 189.41 / 0.2072 3/2

Ex= 189.41 /0.0943

Ex = 2008 N / C

c) E = 1.00 kN / C = 1000 N / C

To solve this part we must find x in the equation

Ex = 497.15 x / (x² + 0.062) ³/₂

Let's use some arithmetic

Ex / 497.15 = x / (x² + 0.062) ³/₂

[Ex / 497.15] ²/₃ = [x / (x² + 0.062) 3/2] ²/₃

∛[Ex / 497.15]² = (∛x²) / (x² + 0.062) (1)

The roots of this equation are the solution to the problem,

For Ex = 1.00 kN / C = 1000 N / C

[Ex / 497.15] 2/3 = 1000 / 497.15) 2/3 = 1,312

1.312 = (∛x² ) / (x² + 0.062)

1.312 (x² + 0.062) = ∛x²

1.312 X² - ∛x² + 1.312 0.062 = 0

1.312 X² - ∛x² + 0.0813 = 0

We need used computer

4 0

3 years ago