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2 years ago

What does AC stand for in physics?

Physics

2 answers:

Oliga [24]2 years ago

6 0

Answer:

A. Alternating current

Svetradugi [14.3K]2 years ago

3 0

Answer:

Alternating current

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A motorboat accelerates uniformly from a velocity of 10.5 m/s to the west to a velocity of 6.5 m/s to the west. If its accelerat

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7.4m ............................

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3 years ago

What is less dense water or oil steel or water helium or air or oil or water

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Answer:

air because their is nothing contained within the air other than all the solutions that you have listed

Explanation:

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3 years ago

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

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2 years ago

Two water jets are emerging from a vessel at a height of 50 centimeters and 100 centimeters. If their horizontal velocities at t

givi [52]

For t1:

t1 = square root of 2h1 / g = square root of 2 * 0.5 / 9.8 = 0.319 sec

For t2:

t2 = sqaure root of 2h2 / g = square root of 2 * 1.0 / 9.8 = 0.451 sec

Wherein:
t = time(s) for the vertical movement
h= height
g = gravity (using the standard 9.8 m/sec measurement)

d1 = 1*0.319 = 0.319 m
d2 = 0.5 * 0.451 = 0.225 m

Where:

d = hor. distance

ratio = d1:d2
= 0.319 : 0.225
=3.19 : 2.25

The answer is 3.19 : 2.25

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3 years ago

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1. A psychologist exposes several lab animals to an enriched environment (toys to play with) and then compares their ability to

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Hypothesis: Enriched environments will have a positive effect and better performance of the experimental group of animals exposed to toys in comparison to the control group of animals in running in mazes.

Dependent Variable: Control Group

Independent Variable: Exposed/Experimental Group

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3 years ago

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