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3 years ago

PLEASE ANSWER FIRST GETS BRAINLIEST

Physics

2 answers:

Ghella [55]3 years ago

8 0

Answer:

I think it's d. The coil gains an electical current

Explanation:

lesantik [10]3 years ago

4 0

Probably D...well, i think

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What does it mean to say that in any system the total energy score stays the<br> same

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a force sets an object in motion. when the force is multiplied by the time of its application we call the quantity impulse which changes the momentum of that object. what do we call the quantity force x (times) distance, and what quantity can this change?

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The following equation, N2 + 3 H2 —>2 NH3 ,describes a

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) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha

vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

$f= 466 Hz$

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s$

The angular frequency instead is given by

$\omega = 2\pi f$

And substituting

f = 466 Hz

We find

$\omega = 2\pi (466 Hz)=2926.5 rad/s$

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz$

While the angular frequency is given by:

$\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s$

(c) $4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s$

The minimum angular frequency of the light wave is

$\omega_1 = 2.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s$

The maximum angular frequency of the light wave is

$\omega_2 = 4.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s$

(d) $2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s$

In this case, the frequency is

$f=5.0 MHz = 5.0 \cdot 10^6 Hz$

So the period in this case is

$T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s$

While the angular frequency is given by

$\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s$

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3 years ago

Based on the principles of convection, conduction and thermal radiation, which scenario below is most similar to the following s

Ksivusya [100]

The situation (heat going through the ceiling) describes
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soaking through some material.

A). This is the one. Heat goes from from the marshmallow
to your hand by soaking through the wire. This is conduction too.

B). No. The heat in the room goes from the floor to the ceiling
because the warm air rises and carries it there. This is convection.

C). No. There's nothing for the heat to soak through between
the sun and the roof, and nothing that can move from the sun
to the roof and bring the heat with it. This is radiation.

D). No. Cold water sinks from the surface to the bottom because
warm water rose from the bottom to the surface, taking heat with it.
This is convection.

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