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3 years ago

How far is the Earth from the Sun?

Physics

2 answers:

a_sh-v [17]3 years ago

6 0

Answer:

93.531 million miles away❣️

Explanation:

ohaa [14]3 years ago

4 0

Answer:93.431 million mi

Explanation:

google lol

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Two round concentric metal wires lie on a tabletop, one inside the other. The inner wire has a diameter of 18.0 cm and carries a

Helen [10]

Answer:

Explanation:

The wires are in circular shape . They have common center .

magnetic field due to circular wire is given by the formula

B = $\frac{\mu_0\times I }{2r}$

where I is current , r is radius of the coil .

magnetic field due to inner wire

= $\frac{\mu_0\times 10 }{2\times.09}$

magnetic field due to outer wire

= $\frac{\mu_0\times I }{2\times.15}$

These should be equal and opposite so that by cancelling each other , they create zero field.

$\frac{\mu_0\times 10 }{2\times.09}$ = $\frac{\mu_0\times I }{2\times.15}$

I = 16.66 A

Direction of current should be in opposite direction ie anticlockwise when looking from above.

6 0

3 years ago

Why does the output of a microphone increase as the frequency of the sound waves which it receives increases

Sloan [31]

Answer:

See Explanation

Explanation:

The frequency of sound waves received by the microphone influences the output or pitch of the sound obtained from the microphone.

The higher the frequency of the sound received by the microphone, the higher the output of the microphone and vice versa. This is because, the higher the frequency of sound, the higher the oscillations produced and the greater the output of the microphone.

The rise and fall in the pitch of sound waves as the frequency of sound waves varies is called inflection.

7 0

3 years ago

You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

4 0

3 years ago

The graph shows two runners participating in a race.

Olin [163]

Answer:

'Daniela had a 5-meter head start, and Leonard caught up to her at 25 meters.'

Explanation:

hope that helps :)

6 0

3 years ago

Read 2 more answers

Doc Brown holds on to the end of the minute hand of the clock atop city hall. The tangential velocity of the minute hand is 0.41

alexdok [17]

The Professor's centripetal acceleration is 0.044 m/s²

Centripetal acceleration is the acceleration of an object moving in circular motion. It is usually directed towards the center of the rotation.

It is given by:

a = v²/r

where v is the velocity and r is the radius.

Given that the radius (r) = 4 m, velocity (v) = 0.419 m/s, hence:

a = v²/r = 0.419²/4 = 0.044 m/s²

The Professor's centripetal acceleration is 0.044 m/s²

Find out more at: brainly.com/question/6082363

3 0

2 years ago