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schepotkina [342]
3 years ago
5

If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the fr

equency is closest to If the voltage amplitude across an 8.50-nF capacitor is equal to 12.0 V when the current amplitude through it is 3.33 mA, the frequency is closest to 32.6 kHz. 32.6 MHz. 5.20 kHz. 5.20 MHz. 32.6 Hz.
Physics
1 answer:
Dmitriy789 [7]3 years ago
3 0

Answer:

Frequency will be equal to 5.20 kHz

So option (c) will be correct answer

Explanation:

We have given value of capacitance C=8.5nF=8.5\times 10^{-9}f

Potential difference across capacitor V = 12 volt

Current through capacitor i=3.33mA=3.33\times 10^{-3}A

Capacitive reactance will be equal to X_c=\frac{V}{i}=\frac{12}{3.33\times 10^{-3}A}=3603.60ohm

Capacitive reactance is equal to X_c=\frac{1}{\omega C}

3603.60=\frac{1}{\omega\times  8.5\times 10^{-9}}

\omega =32647.091rad/sec

2\pi f=32647.091

f=5198.98Hz

f = 5.20 kHz

So frequency will be equal to 5.20 kHz

So option (c) will be correct answer

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this is a 3 part questionOn vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85
mamaluj [8]

ANSWER:

(a) 1036 N

(b) -1036 N

(c) 2590 N

STEP-BY-STEP EXPLANATION:

Given:

Mc = 1400 kg

Mt = 560 kg

a = 1.85 m/s^2

(a)

Force by car on trailer:

\begin{gathered} F_c=m\cdot a \\ F_c=560\cdot1.85 \\ F_c=1036\text{ N} \end{gathered}

(b)

\begin{gathered} F_t=-F_c \\ F_t=-1036\text{ N} \end{gathered}

(c)

\begin{gathered} F_n=1400\cdot1.85 \\ F_n=2590\text{ N} \end{gathered}

3 0
1 year ago
Alkali metals _____.
Lyrx [107]
I think the answer is C
7 0
3 years ago
Spring #1 has a force constant of k, and spring #2 has a force constant of 2k. Both springs are attached to the ceiling. Identic
Gre4nikov [31]

Answer:

The ratio of the energy stored by spring #1 to that stored by spring #2 is 2:1

Explanation:

Let the weight that is hooked to two springs be w.

Spring#1:

Force constant= k

let x1 be the extension in spring#1

Therefore by balancing the forces, we get

Spring force= weight

⇒k·x1=w

⇒x1=w/k

Energy stored in a spring is given by \frac{1}{2}kx^{2} where k is the force constant and x is the extension in spring.

Therefore Energy stored in spring#1 is, \frac{1}{2}k(x1)^{2}

                                                              ⇒\frac{1}{2}k(\frac{w}{k})^{2}

                                                              ⇒\frac{w^{2}}{2k}

Spring #2:

Force constant= 2k

let x2 be the extension in spring#2

Therefore by balancing the forces, we get

Spring force= weight

⇒2k·x2=w

⇒x2=w/2k

Therefore Energy stored in spring#2 is, \frac{1}{2}2k(x2)^{2}

                                                              ⇒\frac{1}{2}2k(\frac{w}{2k})^{2}

                                                              ⇒\frac{w^{2}}{4k}

∴The ratio of the energy stored by spring #1 to that stored by spring #2 is \frac{\frac{w^{2}}{2k}}{\frac{w^{2}}{4k}}=2:1

4 0
3 years ago
Question:<br> the distance of the car can move by velocity 15 m/s for time 5s is :<br> Physics
tangare [24]

Answer:

c) 2.05

Explanation:

7 0
2 years ago
If a baseball travels a distance of 4 meters in 5 seconds, what is its average speed
Alisiya [41]

Answer:

speed=d/t.

s=4/5=0.8m/s.

4 0
3 years ago
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