C

Physics

1 answer:

maksim [4K]2 years ago

4 0

Answer:

Explanation:

a,f,,and g is correct because it is indicated that object moving forward

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A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-

SVETLANKA909090 [29]

The resonant frequency of a circuit is the frequency $\omega_0$ at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})$

Where

L = Inductance

C = Capacitance

Our values are given as

$C = 15*10^{-6}\mu F$

$L = 0.280*10^{-3}mH$

Replacing we have,

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})$

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{(15*10^{-6})(0.280*10^{-3})}})$

$f= 2455.81Hz$

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.

7 0

4 years ago

If 200 grams of water is to be heated from 24.0 degrees Celsius to 100.0 degrees Celsius to make a cup of tea, how much heat mus

Colt1911 [192]

Conveniently, some scientists have already figured out how much heat energy it takes to increase the temperature of one gram of water by exactly 1 degree Celsius. That is called the specific heat of water and it's value is 4.184J/g.

So simply multiply the specific heat with the number of grams and the number of degrees C like this:

$heat = 4.184J/g * 200g * 76^o C = 63,596.8J = 63.5968 kJ$