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3 years ago

What are three examples of constructive forces

Physics

1 answer:

Olin [163]3 years ago

5 0

The three main constructive forces are crustal deformation, volcanic eruptions, and deposition of sediment.

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Compare the charges and masses of<br> protons, neutrons, and electrons.

dlinn [17]

Answer:

Protons have a positive charge

neutrons have a neutral charge

Electrons have a negative charge

Explanation:

Protons and electrons balance each other out.

6 0

3 years ago

Which of the following phrases defines the angle of refraction?

Sophie [7]

<h2>Answer:The angle between the refracted ray and the normal.</h2>

Explanation:

Refraction is defined as the phenomenon of bending or changing course of a light ray when it passes from one medium to another medium.

The new ray formed after the bending at the surface of new medium is called the refracted ray.

The angle between the refracted ray and the normal of the point on the surface of the new medium where the ray enters is called angle of refraction.

Refer the attachment for instance.

In short,it is the angle between the refracted ray and the normal.

4 0

4 years ago

What is centripental force?

agasfer [191]

Answer:

Centripetal force is the force that is necessary to keep an object moving in a curved path and that is directed inward towards the center of rotation.

Explanation:

Definition of centripetal force:

Centripetal force is the force that is necessary to keep an object moving in a curved path and that is directed inward towards the center of rotation.

Example of centripetal force

A string on the end of which a stone is whirled about exerts a centripetal force on the stone.

The diagram is shown below

Where

The centripetal forces acting towards the centre C that is $\vec {AC}$

and the direction is from A to C.

And the stone is moving in a circular motion with center as C.

7 0

3 years ago

A 1.70 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

Nikitich [7]

Resistance = ρ * (L/A) and Rf = Ri * ([1 + α * (Tf – Ti)]
ρ = Resistivity L = length in meters A = cross sectional area in m^2 α = temperature coefficient of resistivity
L = 1.50 m Area = π * r^2 r = d/2 = 0.25 cm = 2.5 * 10^-3 m Area = π * (2.5 * 10^-3)^2

The cylindrical rod is similar to a resistor. Since the current is decreasing, the resistance must be increasing. This means the resistance is increasing as the temperature increases. Resistance = Voltage ÷ Current At 20˚, R = 15 ÷ 18.5 At 92˚, R = 15 ÷ 17.2

Now you know the resistance at the two temperatures. Let’s determine the resistivity at the two temperatures. Resistance = ρ * (L/A) ρ = Resistance * (A/L)
At 20˚, ρ = (15 ÷ 18.5) * [π * (2.5 * 10^-3)^2] ÷ 1.5 = At 92˚, ρ = (15 ÷ 17.2) * [π * (2.5 * 10^-3)^2] ÷ 1.5 =
Now you know the resistivity at the two temperatures. Let’s determine the temperature coefficient of resistivity for the material of the rod.
Rf = Ri * ([1 + α * (Tf – Ti)] Rf = 15 ÷ 17.2, Ri = 15 ÷ 18.5, Tf = 92˚, Ti = 20˚
15 ÷ 17.2 = 15 ÷ 18.5 * [1 + α * (92 – 20)] Multiply both sides by (18.5 ÷ 15) (18.5 ÷ 15) * (15 ÷ 17.2) = 1 + α * 72 Subtract 1 from both sides (18.5 ÷ 15) * (15 ÷ 17.2) – 1 = α * 72 Divide both sides by 72 α = 1.05 * 10^-3

4 0

3 years ago

A loudspeaker has a circular opening with a radius of 0.1158 m. The electrical power needed to operate the speaker is 27.0 W. Th

Vilka [71]

Answer:

electrical power is converted by the speaker into sound power 17.25 %

Explanation:

given data

radius = 0.1158 m

power needed = 27.0 W

average sound intensity = 25.6 W/m²

solution

we know sound intensity that is

I = $\frac{power}{area}$ ...................1

power = intensity × area

Power = 25.6 × 4 ×π×r²

Power = 25.6 × 4 ×π×0.1158²

power = 4.313 W

so here electric power % into sound power

= $\frac{4.313}{25}$ × 100

= 17.25 %

3 0

3 years ago