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2 years ago

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Which statement best defines constructive interference? O Energy reflects back toward the source of its power. O Two waves with

identical crests and troughs meet. thing O A wave appears to be at a standstill and vibrates in place O The trough of one wave matches the crest of another.

1 answer:

Brums [2.3K]2 years ago

4 0

Answer: two waves with identical crests and troughs meet

Explanation:

My teacher gave me the answer

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A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the

harina [27]

Answer:

$R=2.78\ \Omega$

Explanation:

Given that,

The current flowing in the circuit, I = 3 A

The power of the battery, P = 25 W

We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :

$P=I^2R$

Put all the values to find R.

$R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega$

So, the resistance is equal to $2.78\ \Omega$ .

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2 years ago

Science Net Forces. Could somebody help me?

tensa zangetsu [6.8K]

2) Unbalanced. Mike will push the box with a force of 20 N. The forces would be balanced if the box responded with 30 N.

3) Balanced. Both boys are pulling with the same force. Neither is winning.

4) Unbalanced. The rope will move with 10 N to the west. The teachers are winning.

5) Unbalanced. The kids are pulling 220 N to the east. The kids are winning.

6) Balanced. You and the dog are pulling with the same force.

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2 years ago

A sphere P, made of steel, has a weight of 10 N on Earth.

expeople1 [14]

An object’s mass is not determined by gravity, however weight is. With this information, you can figure out that Mars has less gravity acting on the object.

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3 years ago

A 4.8-kg block attached to a spring executes simple harmonic motion on a frictionless horizontal surface. At time t=0 s, the blo

garri49 [273]

Answer:

E) 80 N/m

Explanation:

Given;

mass of the block, m = 4.8 kg

displacement of the block, x = -0.5 m

velocity of the block, v = -0.8 m/s

acceleration of the block, a = 8.3 m/s²

From Newton's second law of motion;

F = ma

Also, from Hook's law;

F = -Kx

where;

k is the force constant

Thus, ma = -kx

k = -ma/x

k = -(4.8 x 8.3) / (-0.5)

k = 79.7 N/m

k ≅ 80 N/m

Therefore, the force constant of the spring is closest to 80 N/m

4 0

3 years ago

An extremely long wire laying parallel to the x -axis and passing through the origin carries a current of 250A running in the po

viktelen [127]

Answer:

$1.232\times 10^{-5}\ T.$

Explanation:

<u>Given:</u>

Current through the wire, passing through the origin, $I_1 = 250\ A.$

Current through the wire, passing through the y axis, $r_y=1.8\ m.$ , $I_2 = 50\ A.$

According to Ampere's circuital law, the line integral of magnetic field over a closed loop, called Amperian loop, is equal to $\mu_o$ times the net current threading the loop.

$\oint \vec B \cdot d\vec l=\mu_o I.$

In case of a circular loop, the directions of magnetic field and the line element $d\vec l$ , both are along the tangent of the loop at that point, therefore, $\vec B\cdot d\vec l = B\ dl$ .

$\oint \vec B \cdot d\vec l = \oint B\ dl = B\oint dl.\\$

$\oint dl$ is the circumference of the Amperian loop = $2\pi r$

Therefore,

$B\ 2\pi r=\mu_o I\\B=\dfrac{\mu_o I}{2\pi r}.$

It is the magnetic field due to a current carrying wire at a distance r from it.

For the first wire, passing through the origin:

Consider an Amperian loop of radius 3.510 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_1 = 3.510\ m.$

The magnetic field at the given point due to this wire is given by:

$B_1 = \dfrac{\mu_o I_1}{2\pi r_1}\\=\dfrac{4\pi \times 10^{-7}\times 250}{2\pi \times 3.510}=1.42\times 10^{-5}\ T.$

For the first wire, passing through the y-axis:

Consider an Amperian loop of radius (3.510+1.8) m = 5.310 m, concentric with the axis of the wire, such that it passes through the point where magnetic field is to be found, therefore, $r_2 = 5.310\ m.$

The magnetic field at the given point due to this wire is given by:

$B_2 = \dfrac{\mu_o I_2}{2\pi r_2}\\=\dfrac{4\pi \times 10^{-7}\times 50}{2\pi \times 5.310}=1.88\times 10^{-6}\ T.$

The directions of current in both the wires are opposite therefore, the directions of the magnetic field due to both the wires are also opposite to that of each other.

Thus, the net magnetic field at $r_r=-3.510\ m$ is given by

$B=B_1-B_2 = 1.42\times 10^{-5}-1.88\times 10^{-6}=1.232\times 10^{-5}\ T.$

6 0

3 years ago