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3 years ago

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An empty parallel plate capacitor is connected between the terminals of a 18.8-V battery and charges up. The capacitor is then d

isconnected from the battery, and the spacing between the capacitor plates is doubled. As a result of this change, what is the new voltage between the plates of the capacitor

1 answer:

Basile [38]3 years ago

7 0

Answer:

$p.d' = 37.6 V$

Explanation:

From the question we are told that:

Potential difference $p.d=18.8V$

New Capacitor $C_1=C_2/2$

Generally the equation for Capacitor capacitance is mathematically given by

$C=\frac{eA}{d}$

Generally the equation for New p.d' is mathematically given by

$C_2V=C_1*p.d'$

$p.d' = 2V$

$p.d'= 2 * 18.8$

$p.d' = 37.6 V$

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Two boys are at the top of a waterslide at Seven Peaks Water Park. One boy (boy A) slips off the top of the tower and falls unob

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Answer: boy B, same

Explanation:

Given

One boy trips off the waterslide while the other starts sliding down

As there is no horizontal velocity, both boys have to travel the same vertical distance.

Their starting vertical velocity is zero and they need to travel the same vertical distance. Therefore, both boys splash water with the same velocity.

The time taken by boy B is more than boy A as boy B will travel some horizontal distance due to slide which will increase its time to reach the bottom.

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2 years ago

Careful measurements have been made of Olympic sprinters in the 100-meter dash. A quite realistic model is that the sprinter's v

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Answer:

a.

$\displaystyle a(0 )=8.133\ m/s^2$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=0.52\ m/s^2$

b. $\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. $t=9.9 \ sec$

Explanation:

Modeling With Functions

Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by

$\displaystyle V(t)=a(1-e^{bt})$

For Carl Lewis's run at the 1987 World Championships, the values of a and b are

$\displaystyle a=11.81\ ,\ b=-0.6887$

Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is

$\displaystyle V(t)=11.81(1-e^{-0.6887t})$

a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?

To compute the accelerations, we must find the function for a as the derivative of v

$\displaystyle a(t)=\frac{dv}{dt}=11.81(0.6887\ e^{0.6887t})$

$\displaystyle a(t)=8.133547\ e^{-0.6887t}$

For t=0

$\displaystyle a(0)=8.133547\ e^o$

$\displaystyle a(0 )=8.133\ m/s^2$

For t=2

$\displaystyle a(2)=8.133547\ e^{-0.6887\times 2}$

$\displaystyle a(2)=2.05\ m/s^2$

$\displaystyle a(4)=8.133547\ e^{-0.6887\times 4}$

$\displaystyle a(4)=0.52\ m/s^2$

b. Find an expression for the distance traveled at time t.

The distance is the integral of the velocity, thus

$\displaystyle X(t)=\int v(t)dt \int 11.81(1-e^{-0.6887t})dt=11.81(t+\frac{e^{-0.6887t}}{0.6887})+C$

$\displaystyle X(t)=11.81(t+1.45201\ e^{-0.6887t})+C$

To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates

$\displaystyle x(0)=0=>11.81\times1.45201+C=0$

Solving for C

$\displaystyle c=-17.1482\approx -17.15$

Now we complete the equation for the distance

$\displaystyle X(t)=11.81(t+1.45\ e^{-0.6887t})-17.15$

c. Find the time Lewis needed to sprint 100.0 m.

The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.

We are required to find the time at which the distance is 100 m, thus

$\displaystyle X(t)=100=>11.81(t+1.45\ e^{-0.6887t})-17.15=100$

Rearranging

$\displaystyle t+1.45\ e^{-0.6887t}=9.92$

We define an auxiliary function f(t) to help us find the value of t.

$\displaystyle f(t)=t+1.45\ e^{-0.687t}-9.92$

Let's try for t=9 sec

$\displaystyle f(9)=9+1.45\ e^{-0.687\times 9}-9.92=-0.92$

Now with t=9.9 sec

$\displaystyle f(9.9)=9.9+1.45\ e^{-0.687\times 9.9}-9.92=-0.0184$

That was a real close guess. One more to be sure for t=10 sec

$\displaystyle f(10)=10+1.45\ e^{-0.687\times 10}-9.92=0.081$

The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).

$At t\approx 9.9\ sec, \text{ Lewis sprinted 100 m}$

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Uma massa de 500 Kg desloca-se com velocidade 58 km por hora. Calcule o módulo de sua quantidade por movimento

Simora [160]

The momentum of the object is 8050 kg m/s

Explanation:

The momentum of an object is defined as

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v = 58 km/h is its velocity

Converting the velocity into m/s,

$v=58 \frac{km}{h}\cdot \frac{1000 m/km}{3600 s/h}=16.1 m/s$

Therefore now we can find the momentum of the object:

$p=(500)(16.1)=8050 kg m/s$

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3 years ago

The North Star is located in the Big Dipper. True or False

harina [27]

Answer:

False

Explanation:

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3 years ago

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The peak value of an alternating current in a 1500-W device is 6.4 A. What is the rms voltage across it

horrorfan [7]

Answer:

6787.5 V

Explanation:

From the question,

P = IV..................... Equation 1

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make V the subject of the equation

V = P/I................. Equation 2

Given: P = 1500 W, I = 6.4/√2 = 4.525 A

Substitute these values into equation 2

V = 1500(4.525)

V = 6787.5 V

Hence the rms voltage = 6787.5 V

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3 years ago