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valkas [14]
3 years ago
14

A capacitor with a very large capacitance is in series with a capacitor that has a very small capacitance. what can we say about

the capacitance of the series combination? question 1 options:
a.it is slightly smaller than the capacitance of the large capacitor.

b.it is slightly larger than the capacitance of the small capacitor.

c.it is slightly smaller than the capacitance of the small capacitor.

d.it is impossible to say anything about the equivalent series capacitance without knowing the actual numeric values of the individual capacitors in the combination.

e.it is slightly larger than the capacitance of the large capacitor.
Physics
1 answer:
Misha Larkins [42]3 years ago
5 0
<span>A capacitor with a very large capacitance is in series with a capacitor
that has a very small capacitance.

The capacitance of the series combination is slightly smaller than the
capacitance of the small capacitor. (choice-C)

The capacitance of a series combination is

             1 / (1/A + 1/B + 1/C + 1/D + .....) .

If you wisk, fold, knead, and mash that expression for a while,
you find that for only two capacitors in series, (or 2 resistors or
two inductors in parallel), the combination is   

             (product of the 2 individuals) / (sum of the individuals)  .

In this problem, we have a humongous one and a tiny one.
Let's call them  1000  and  1 .
Then the series combination is

           (1000 x 1) / (1000 + 1)

        =       (1000) / (1001)

        =         0.999 000 999 . . . 

which is smaller than the smaller individual.

It'll always be that way.   </span>
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A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m. If the ball revolves twice every second,
Aloiza [94]

Answer:

19.74 N

Explanation:

mass of ball (m) = 0.25 kg

radius (r) = 0.5 m

time (t) = 2 revolutions per seconds = 1/2 = 0.5 second per revolution

find the tension in the string

tension (T) = \frac{mv^{2} }{r}

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tension now becomes (T) = \frac{m}{r} x (\frac{2πr}{t})^{2}

tension (T) = \frac{4π^{2}rm }{t^{2} }

  • now substituting the values of mass (m), time (t) and radius (r) into the equation above we have

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4 0
3 years ago
A 75 kg football player is gliding forward across very smooth ice at 4.6 m/s. He throws a 0.47 kg football straight forward. A)
lord [1]

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4.53482 m/s

4.506 m/s

Explanation:

m_1 = Mass of player = 75 kg

v_1 = Initial velocity of player = 4.6 m/s

m_2 = Mass of ball = 0.47 kg

v_1 = Initial velocity of ball = 15 m/s

The linear momentum of the system is conserved

(m_1+m_2)v_1=m_1v+m_2v_2\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2v_2}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times 15}{75}\\\Rightarrow v=4.53482\ m/s

The player's speed is 4.53482 m/s

In the second case the equation of momentum is

(m_1+m_2)v_1=m_1v+m_2(v_2+v_1)\\\Rightarrow v=\dfrac{(m_1+m_2)v_1-m_2(v_2+v_1)}{m_1}\\\Rightarrow v=\dfrac{(75+0.47)4.6-0.47\times (15+4.6)}{75}\\\Rightarrow v=4.506\ m/s

The player's speed is 4.506 m/s

4 0
4 years ago
A stationary sub uses sonar to send a 1.18x10^3 hertz sound wave through ocean water. The reflected sound wave from the flat oce
vampirchik [111]

Answer:

a) v = 1524.7 m/s

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Explanation:

a) First, in order to calculate the speed of the sound wave, you take into account that the velocity is constant, then, you use the following formula:

v=\frac{d}{t}

d: distance traveled by the sound wave, which is twice the distance to the ocean bottom = 2*324 m = 648 m

t: time that sound wave takes to return to the sub = 0.425

v=\frac{648m}{0.425s}=1524.7\frac{m}{s}

hence, the speed of the sound wave is 1524.7 m/s

b) Next, with the value of the velocity of the wave you can calculate the wavelength of the wave, by using the following formula:

v=\lambda f\\\\\lambda=\frac{v}{f}

f: frequency = 1.18*10^3 Hz

\lambda=\frac{1524.7m/s}{1.18*10^3s^{-1}}=1.29m

And the period is:

T=\frac{1}{f}=\frac{1}{1.18*10^3s^{-1}}=8.47*10^{-4}s

hence, the wavelength and period of the sound wave is, respectively, 1.29m and 8.47*10^-4 s

5 0
3 years ago
A laser beam strikes a plane's surface with an angle of52.
Vadim26 [7]

Answer:

Angle between incident ray and reflected ray will be 104°

Explanation:

We have given angle of incidence = 52 °

From law of reflection angle of incidence will be equal to angle of reflection

So angle of reflection will be also 52°

We have to find the angle between incident ray and reflected ray

As the incidence angle and reflected angle both is from normal of the surface and opposite to each other

So angle between incident ray and reflected ray will be 52°+52° = 104°

7 0
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