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2 years ago

When the distance increases, the electrostatic force____; we call this relationship proportional

Physics

2 answers:

Ipatiy [6.2K]2 years ago

8 0

Answer:

See below

Explanation:

When the distance increases, the electrostatic force <u> DECREASES</u>; we call this relationship <u> INVERSELY</u> proportional

Alik [6]2 years ago

5 0

Answer:

I think decreases inversely

Explanation:

the third

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A majorette in the Rose Bowl Parade tosses a baton into the air with an initial angular velocity of 2.5 rev/s. If the baton unde

oksian1 [2.3K]

Answer:

14 rev

Explanation:

$w_{o}$ = initial angular velocity = 2.5 revs⁻¹

$w$ = final angular velocity = 0.8 revs⁻¹

$\alpha$ = Angular acceleration = - 0.2 revs⁻²

$\theta$ = Angular displacement

Using the equation

$w^{2} = w_{o}^{2} + 2 \alpha \theta\\0.8^{2} = 2.5^{2} + 2 (- 0.2) \theta\\ \theta = 14 rev$

So the number of revolutions are 14

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3 years ago

a 65 kg skater at rest on a frictionless rink throws a 2 kg ball, giving the ball a velocity of 7 m/s. What is the velocity of t

gayaneshka [121]

The answer to your question is 33

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3 years ago

How many DAYS occur between a new moon and a first quarter moon?

N76 [4]

Answer:

About 29.5

Explanation:

4 0

2 years ago

Read 2 more answers

What is a plasma?A.A superheated liquidB.A solid in which the particles vibrate very quicklyC.A gas in which the atoms have star

Mandarinka [93]

plasma is a superheated liquid

So, a would-be the correct option.

5 0

10 months ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

8 0

3 years ago