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Vera_Pavlovna [14]
3 years ago
9

One circuit contains only an ac generator and a resistor, and the rms current in this circuit is IR. Another circuit contains on

ly an ac generator and a capacitor, and the rms current in this circuit is IC. The maximum, or peak, voltage of the generator is the same in both circuits and does not change. If the frequency of each generator is tripled, by what factor does the ratio IR/IC change?
Physics
1 answer:
snow_tiger [21]3 years ago
7 0

Answer:

1/3 times.

Explanation:

Let V₀ be the peak voltage .

IR ( rms )  = ( V₀ / √2 R )

R is value of resistance

IC =  ( V₀ ωC / √2  )

( 1 / ωC is reactance of capacitance in ac circuit  )

\frac{I_R}{I_C} =\frac{\frac{V_0}{\sqrt{2}R } }{\frac{V_0\omega C}{\sqrt{2} } }

= \frac{I_R}{I_C} = \frac{1}{\omega C R}

When frequency is tripled angular frequency will also be tripled

hence the ratio \frac{I_R}{I_C} becomes 1/3 times.

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The kinetic energy of a baseball is doubled by changing the speed of the baseball. By what factor has its speed changed? √ 2 v i
leonid [27]

Answer:

1/√2

Explanation:

Kinetic energy of a body is expressed as

KE = 1/2mv²

If it's KE is doubled

2KE = 1/2mv1² where v1 is the new speed when the kinetic energy is doubled

To know the value of the amount the velocity has changed, we will divide both equations

2KE/KE = (1/2mv²)/(1/2mv1²)

2 = v²/v1²

(v/v1)² = 2

v/v1 = √2

v1 = v/√2

v1 = 1/√2 × v

The new velocity has changed by

1/√2vinitial

7 0
3 years ago
A Tennis ball falls from a height 40m above the ground the ball rebounds
worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity <em>v</em> at time <em>t</em> before it hits the ground is

<em>v</em> = -<em>g t</em>

where <em>g</em> = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height <em>y</em> is

<em>y</em> = 40 m - 1/2 <em>g</em> <em>t</em>²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 <em>g</em> <em>t</em>²

==>  <em>t</em> = √(80/<em>g</em>) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

<em>v</em> = -<em>g</em> (√(80/<em>g</em>) s)

==>  <em>v</em> = -√(80<em>g</em>) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

<em>y</em> = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> to see how long it's airborne during this bounce:

0 = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

0 = <em>t</em> (14 m/s - 1/2 <em>g</em> <em>t</em>)

==>  <em>t</em> = 28/<em>g</em> s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

<em>y</em> = (14 m/s) (5 s - 2.86 s) - 1/2 <em>g</em> (5 s - 2.86 s)²

==>  (i) <em>y</em> ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But <em>y</em> will converge to 0 as <em>t</em> gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

<em>y</em> = (3.5 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> :

0 = (3.5 m/s) <em>t</em> - 1/2 <em>g t</em>²

0 = <em>t</em> (3.5 m/s - 1/2 <em>g</em> <em>t</em>)

==>  (iii) <em>t</em> = 7/<em>g</em> m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time <em>t</em> after 5.72 s (but before reaching the ground again) would be

<em>v</em> = 7 m/s - <em>g t</em>

At 6 s, the ball has velocity

(iv) <em>v</em> = 7 m/s - <em>g</em> (6 s - 5.72 s) ≈ 4.26 m/s

4 0
3 years ago
6)An electric field of 6 N/C points in the positive X direction. What is the electric flux through a surface that is 4 m2, if it
Salsk061 [2.6K]

Answer:

Flux is 21 Nm^2/C.

Explanation:

Electric field, E = 6 N/C along X axis

Electric filed vector, E = 6 i N/C

Area, A = 4 square meter

Area vector

\overrightarrow{A} = 4 (cos30 \widehat{i} + sin 30 \widehat{j})\\\\\overrightarrow{A} = 3.5 \widehat{i} + 2 \widehat{j}\\

The flux is given by

\phi= \overrightarrow{E}.\overrightarrow{A}\\\\\phi = 6 \widehat{i} .  \left (3.5 \widehat{i} + 2 \widehat{j}  \right )\\\\\phi = 21 Nm^2/C

3 0
2 years ago
The electric motor of a model train accelerates the train from rest to 0.685 m/s in 21.5 ms. The total mass of the train is 875
Natali [406]

Answer:

P=9.58 W

Explanation:

According to Newton's second law, and assuming friction force as zero:

F_m=m.a\\F_m=0.875kg*a

The acceleration is given by:

a=\frac{\Delta v}{t}\\a=\frac{0.685m/s}{21.5*10^{-3}s}\\\\a=31.9m/s^2

So the force exerted  by the motor is:

F_m=0.875kg*31.9m/s^2\\F_m=27.9N

The work done by the motor is given by:

W_m=F_m*d\\\\d=\frac{1}{2}*a*t^2\\d=\frac{1}{2}*31.9m/s^2*(21.5*10^{-3}s)^2\\\\d=7.37*10^{-3}m

W_m=27.9N*7.37*10^{-3}m\\W_m=0.206J

And finally, the power is given by:

P=\frac{W_m}{t}\\P=\frac{0.206J}{21.5*10^{-3}s}\\\\P=9.58W

8 0
3 years ago
The scientist who suggested that energy can be created under certain conditions was
tester [92]
Einstein hope this helped
7 0
3 years ago
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