3 years ago

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Physics

1 answer:

Darina [25.2K]3 years ago

4 0

Resolving vectors question. See diagram enclosed.

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What is the<br> power of a light<br> bulb with 8V and<br> a current of 2A?

astra-53 [7]

Answer:

16 Watts

Explanation:

P = VI, where V is the voltage and I is the current (of the lightbulb)

V = 8 volts, I = 2 amps

P = 8 * 2 = 16 Watts

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3 years ago

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According to The Flash, The heroes top speed is Mach 3.3, or 2,500 miles per hour. How

Slav-nsk [51]

Answer:

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Explanation:

but maybe he can run it in five minutes

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3 years ago

A charged capacitor is connected to an ideal inductor to form an LC circuit with a frequency of oscillation f = 1.6 Hz. At time

IrinaK [193]

Answer:

$8.0\mu C$

Explanation:

We are given that

$f=1.6 Hz$

$q=3.0\mu C=3.0\times 10^{-6} C$

$1\mu C=10^{-6} C$

Current,I= $75\mu A=75\times 10^{-6} A$

$1\mu A=10^{-6} A$

We have to find the maximum charge of the capacitor.

Charge on the capacitor, $q=q_0cos\omega t$

$\omega=2\pi f=2\pi\times 1.6=3.2\pi rad/s$

$3\times 10^{-6}=q_0cos3.2\pi t$ ....(1)

$I=\frac{dq}{dt}=-q_0\omega sin\omega t$

$75\times 10^{-6}=-q_0(3.2\pi)sin3.2\pi t$ ....(2)

Equation (2) divided by equation (1)

$-3.2\pi tan3.2\pi t=\frac{75\times 10^{-6}}{3\times 10^{-6}}=25$

$tan3.2\pi t=-\frac{25}{3.2\pi}=-2.488$

$3.2\pi t=tan^{-1}(-2.488)=-1.188rad$

$q_0=\frac{q}{cos\omega t}=\frac{3\times 10^{-6}}{cos(-1.188)}=8.0\times 10^{-6}=8\mu C$

Hence, the maximum charge of the capacitor= $8.0\mu C$

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4 years ago

Why is it important to consider experimental error in all the empirical results presented?

NemiM [27]

Because when your running any experiment there will always be an experimental error so always be ready for it so that you can correct that error in your experiment

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3 years ago

A force F= (6.70 N)i + (5.60 N)j + (6.60 N)k act on a 9.80 kg mobile object that moves from an initial position of di = (9.10 m)

Mandarinka [93]

Answer:

W = 0.11J

P = 0.0186W

Explanation:

Please see attachment below.

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3 years ago