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kirza4 [7]
3 years ago
12

Pierce conducts an experiment in which waves collide in a way that the energy increases. What has occurred? refraction reflectio

n constructive interference destructive interference
Physics
2 answers:
Nadya [2.5K]3 years ago
4 0

Answer:

constructive interference.

Explanation:

Vitek1552 [10]3 years ago
3 0

Answer:

constructive interference.

Explanation:

Electromagnetic waves is a propagating medium used in all communications device to transmit data (messages) from the device of the sender to the device of the receiver.

Generally, the most commonly used electromagnetic wave technology in telecommunications is radio waves.

Radio waves can be defined as an electromagnetic wave that has its frequency ranging from 30 GHz to 300 GHz and its wavelength between 1mm and 3000m.

In this scenario, an experiment is conducted by Pierce in which waves collide in a way that the energy increases. Thus, this is an example of constructive interference i.e constructive interference has occurred.

An interference occurs when two pair of waves such as sound and light waves pass through each other i.e crossing each other's path while traveling through the same medium or space.

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A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward
AleksandrR [38]

Answer:

(a)  104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.  

The component of F perpendicular to the ramp is Fy.

(a)  

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.  

Resolve F into its x-component from Pythagorean theorem:  

Fx=Fcos30°

Solve for F:  

F= Fx/cos30°  

Substitute for Fx from given data:  

Fx=90 N/cos30°

   =104 N

(b) Resolve r into its y-component from Pythagorean theorem:

     Fy = Fsin 30°

   Substitute for F from part (a):

     Fy = (104 N) (sin 30°)  

          = 52 N  

5 0
3 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by b) is at 36 ft. from the wall is the following:

\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s

Explanation:

The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)

\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Product rule

\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt} Chain rule

\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}

\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)

In here we can identify b=36\, ft, l=39 and \frac{db}{dt}=8\,ft/s.

The result is then

\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s

3 0
3 years ago
What’s the Answer for 5-8?
Dafna1 [17]

6. B. 4 electrons

7. Carbon is in everything and it's a compound that all organisms need

8. D. 3

9. Nitrogen gas is almost 80% of the atmosphere

10. Diatomic molecules compose 2 atoms

3 0
3 years ago
What is the purpose of the background research step of the scientific method?
Lilit [14]
Background research is used to familiarize yourself with the topic or problem you're conducting your experiment on and help you form an educated guess explaining or solving it that will become your hypothesis. 
3 0
3 years ago
A sled travels 15 meters down a slope inclined at 30 degrees with a horizontal. What is the horizontal displacement of the sled?
Arte-miy333 [17]

Length of the slope is given as

L = 15 m

also the inclination is given as

\theta = 30 degree

now the horizontal displacement is let say "x"

now from geometry we can say

\frac{x}{L} = cos30

x = L cos30

now substitute all values in it

x = 15 * cos30

x = 12.99 m

<em>so it will displace horizontally by 12.99 m</em>

6 0
4 years ago
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