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Norma-Jean [14]
2 years ago
8

3. What is the velocity of a wave that has a frequency of 750 Hz and a wavelength of 45.7 cm?

Physics
1 answer:
sweet-ann [11.9K]2 years ago
3 0

Explanation:

v=(750)(45.7)

v=34275

or

v=(750)(0.457)

v=342.75

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Tammy leaves the office, drives 26 km due
fredd [130]

Answer:

72.98 km

Explanation:

Her displacement is simply the distance from her final position to her initial position.

Now, I've drawn and attached a triangle diagram to depict this her movement.

Point O is her initial starting point.

Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.

From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.

Thus;

x² = 62² + 26² - 2(62 × 26)cos 120

x² = 4520 + 806

x² = 5326

x = √5326

x = 72.98 km

4 0
2 years ago
Arrange the following types of electromagnetic radiation in order of increasing (lowest to highest) frequency from left to right
algol13

Electromagnetic radiation in order of increasing (lowest to highest) frequency is Radio waves < Green visible light < UV radiation.

<h3>What is Frequency?</h3>

This is defined as the rate at which something occurs or is repeated over a particular period of time. The unit of frequency is referred to as Hertz.

Electromagnetic radiations with increasing frequency can be seen below:

Radio waves, microwaves, infrared waves, visible light, ultraviolet radiations, X rays, γ rays.

Read more about Frequency here brainly.com/question/254161

4 0
2 years ago
A student on an amusement park ride moves in circular path with a radius of 3.5 m once every 4.5 s. What is the tangential veloc
nata0808 [166]

Answer:

the tangential velocity of the student is 4.89 m/s.

Explanation:

Given;

the radius of the circular path, r = 3.5 m

duration of the motion, t = 4.5 s

let the student's tangential velocity = v

The tangential velocity of the student is calculated as follows;

v = \frac{2\pi r}{t} \\\\v = \frac{2 \pi \times 3.5}{4.5} \\\\v = 4.89 \ m/s

Therefore, the tangential velocity of the student is 4.89 m/s.

6 0
3 years ago
In the shadow of a tree with a dense, leafy canopy, one sees a number of light spots. Surprisingly, they all appear to be circul
Bad White [126]

The characteristics of the diffraction phenomenon allow to find the result for the shape of the points of light that you pass the tree is:

  • The shape of the dots is circular because it is in the range of far-field diffraction.

Diffraction is the phenomenon where the undulatory part of the light becomes evident, it is the interference of the waves that make up each ray of light, for this phenomenon to occur it must be fulfilled that the wavelength is of the order of the space where pass the light.

In the leafy tree it has many leaves, but there are spaces between them, some of these spaces are small and it fulfills the diffraction condition, therefore we see bright spots and not a continuous shadow.

Diffraction can be classified depending on the distance to the observer:

  • Near field or fresnel. In this case the distance from the observer is small and we can see the shape of the object that creates the diffraction.
  • Far field or Fraunhoger. In this case the distance between the obstacle (leaves) and the person is great, here the information on the shape of things is lost and we have two observable forms. Lines for the case of slits and circles for the case of objects with a closed shape.

In this case, the distance from the leaves to the observer is large, therefore we are in the case of far-field diffraction and since the edge of the leaves that forms the diffraction is closed, the observable shape is a circle.

In conclusion using the characteristics of the diffraction phenomenon we can find the result for the shape of the points of light that pass the tree is:

  • The shape of the dots is circular because it is in the range of far-field diffraction.

Learn more about diffraction here:  brainly.com/question/20140459

8 0
2 years ago
What is the magnitude of the acceleration vector which causes a particle to move from velocity −5i−2j m/s to −6i+ 7j m/s in 8 se
shusha [124]

Answer:

Acceleration, a=\dfrac{1}{8}(-i+9j)\ m/s^2

Explanation:

Initial velocity of a particle in vector form, u = (-5i - 2j) m/s

Final velocity of particle in vector form, v = (-6i + 7j) m/s

Time taken, t = 8 seconds

We need to find the magnitude of acceleration vector. The changing of velocity w.r.t time is called acceleration of a particle. It is given by :

a=\dfrac{v-u}{t}

a=\dfrac{(-6i+7j)\ m/s-(-5i-2j)\ m/s}{8\ s}    

a=\dfrac{(-i+9j)}{8\ s}\ m/s^2    

or

a=\dfrac{1}{8}(-i+9j)\ m/s^2

Hence, the value of acceleration vector is solved.

4 0
3 years ago
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