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3 years ago

What happens to a sound wave as air temperature decreases?

Physics

1 answer:

weqwewe [10]3 years ago

8 0

Answer:

Explanation:

goes down as the temperature decreases, and vice versa. Sound's frequency is independent of temperature, while its speed is directly proportional to temperature. Heat, like sound, is a form of kinetic energy. Molecules at higher temperatures have more energy, thus they can vibrate faster. Since the molecules vibrate faster, sound waves can travel more quickly. ... This is faster than 331 meters per second, which is the speed of sound in air at freezing temperatures.

hope this helps!!!!

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An capacitor consists of two large parallel plates of area A separated by a very small distance d. This capacitor is connected t

scoundrel [369]

Answer:

Will be doubled.

Explanation:

For a capacitor of parallel plates of area A, separated by a distance d, such that the charges in the plates are Q and -Q, the capacitance is written as:

$C = \frac{Q}{V} = e_0\frac{A}{d}$

where e₀ is a constant, the electric permittivity.

Now we can isolate V, the potential difference between the plates as:

$V = \frac{Q}{e_0} *\frac{d}{A}$

Now, notice that the separation between the plates is in the numerator.

Thus, if we double the distance we will get a new potential difference V', such that:

$V' = \frac{Q}{e_0} *\frac{2d}{A} = 2*( \frac{Q}{e_0} *\frac{d}{A}) = 2*V\\V' = 2*V$

So, if we double the distance between the plates, the potential difference will also be doubled.

6 0

3 years ago

What makes a planet different from other celestial bodies?.

Serggg [28]

Answer:

It's more habitable.

Explanation:

The atmosphere, calculated to equations, are a lot more pulled down.

5 0

3 years ago

Which circuit offers the greater resistance to the battery, two bulbs in series or two bulbs in parallel?

dolphi86 [110]

The resistance of two things in series is the SUM of their individual resistances. So the resistance of two bulbs in series is <u><em>double</em></u> the resistance of one bulb.

(If they're in parallel, their combined resistance is <u><em>1/2</em></u> the resistance of one bulb.)

So two bulbs <em>in series</em> is the greater resistance. <em>(a) </em>

7 0

3 years ago

Michael is biking on a trail and is accelerating at a rate of 1.2 m/s/s for 15 seconds. He began this part of his ride with a ve

Serhud [2]

Answer:

Michael's final velocity is 19.62 m/s.

Explanation:

We can find the final velocity of Michael by using the following kinematic equation:

$v_{f} = v_{0} + at$ (1)

Where:

$v_{f}$ : is the final velocity =?

$v_{0}$ : is the initial velocity = 1.62 m/s

a: is the acceleration = 1.2 m/s²

t: is the time = 15 s

By entering the above values into equation (1) we have:

$v_{f} = 1.62 m/s + 1.2 m/s^{2}*15 s$

$v_{f} = 19.62 m/s$

Therefore, Michael's final velocity is 19.62 m/s.

I hope it helps you!

7 0

3 years ago

I NEED HELP ASAP! BRAINIEST TO THE CORRECT ANSWER. HELP ME NOW!

sergij07 [2.7K]

Answer:

$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

V = 12 V
P = 24 W

$\implies \mathsf{24=\frac{12^2}{R} }$

$\implies \mathsf{24R=12^2 }$

$\implies \mathsf{24R=144 }$

$\boxed{\mathfrak{Power(P)=\frac{Voltage(V)^2}{Resistance(R)} }}$

Power (P)= 100 W

<em>these lights operate at the usual 240 volts direct from the main electricity supply. Therefore,</em>

V = 240 V

$\implies \mathsf{100=\frac{240^2}{R} }$

<em>R and 100 can interchange places</em>

$\implies \mathsf{R=\frac{240^2}{100} }$

$\implies \mathsf{R=\frac{57600}{100} }$

<u></u>

By Ohm's Law:

$\boxed{\mathsf{Voltage(V)=Current(I) \times Resistance(R)}}$

=> 240 = I × 576

=> I = 0.417 A

I don't know it's resistance,... so sorry

The brightness of the bulb in series is <em><u>less than</u></em> when they're placed individually.

For bulbs in series their resistance gets added to form the equivalent resistance of the two bulbs.

Their resistances are nothing but mere numbers and the sum of two numbers(positive of course) is greater than the numbers.

So, the effective resistance of some bulbs in series <u>is more</u> than the individual resistance.

And

<em>Brightness, i. e., Power</em>

$\boxed{\mathfrak{Power \propto \frac{1}{Resistance} }}$

If resistance increases, Power decreases.

Here, the effective resistance was for sure larger, therefore resistance was increasing, hence power decreased taking brightness along with it.

3 0

3 years ago