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4 years ago

50 POINTS to whoever gets it right

Physics

1 answer:

sasho [114]4 years ago

6 0

D. The hydrogen, being a smaller atom, will take up less space than the oxygen. It will push the water down less.

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A parallel RLC circuit contains an inductor with a value of 400 microhenries and a capacitor with a value of 0.3 microfarads wha

ipn [44]

Answer:

The resonant frequency of this circuit is 14.5 kHz.

Explanation:

Given that,

Inductance of a parallel LCR circuit, $L=400\ \mu H=400\times 10^{-6}\ H$

Capacitance of parallel LCR circuit, $C=0.3\ \mu F=0.3\times 10^{-6}\ F$

At resonance the inductive reactance becomes equal to the capacitive reactance. The resonant frequency is given by :

$f=\dfrac{1}{2\pi \sqrt{LC} }$

$f=\dfrac{1}{2\pi \sqrt{400\times 10^{-6}\times 0.3\times 10^{-6}} }$

$f=14528.79\ Hz$

f = 14.5 kHz

So, the resonant frequency of this circuit is 14.5 kHz. Hence, this is the required solution.

4 0

3 years ago

(11) The speed of radio waves is 300 000 000 m/s.

Neko [114]

Answer:

Wavelength = 9.68 meters

Explanation:

Given the following data;

Speed = 300,000,000m/s

Frequency = 31 Megahertz to Hertz = 31 * 10⁶ Hz

To find the wavelength;

Wavelength = speed/frequency

Wavelength = 300,000,000/31,000,000

Wavelength = 9.68 meters

7 0

3 years ago

You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla

lesantik [10]

Answer:

When they are connected in series

The 50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

The power rating of the first bulb is $P_1 = 100 \ W$

The power rating of the second bulb is $P_2 = 50 \ W$

Generally the power rating of the first bulb is mathematically represented as

$P_1 = V^2 R$

Where $V$ is the normal household voltage which is constant for both bulbs

$R_1 = \frac{V^2}{P_1 }$

substituting values

$R_1 = \frac{V^2}{100}$

Thus the resistance of the second bulb would be evaluated as

$R_2 = \frac{V^2}{50}$

From the above calculation we see that

$R_2 > R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 1 = I^2_1 R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 2 = I^2_2 R_2$

Now given that they are connected in series which implies that the same current flow through them so

$I_1^2 = I_2^2$

This means that

$P \ \alpha \ R$

So when they are connected in series

$P_2 > P_1$

This means that the 50 W bulb glows more than the 100 \ W bulb

3 0

4 years ago

a positively charged body makes contact with a body. after a while, the charged body becomes neutralised. state 3 main condition

MrRa [10]

Answer:

A body will become positively charged when some electrons will come out from the body.Thus, positive charge is due to deficiency of electrons.

5 0

3 years ago

What part of the electromagnetic spectrum can be seen by humans

Amanda [17]

Answer:

Visible light

Explanation:

Electromagnetic spectrum is the classification of the electromagnetic waves according to their frequency/wavelength. In order from the shortest to the longest wavelength, we have

Gamma rays

X-rays

Ultraviolet

Visible light

Infrared

Microwaves

Radio waves

All these waves are invisible to human eye, except for the part referred as 'visible light'. The electromagnetic waves of this part of the spectrum are visible to human eye, and they appear as a different color depending on their wavelength. In particular, we have:

Violet: 380-450 nm

Blue: 450-495 nm

Green: 495-570 nm

Yellow: 570-590 nm

Orange: 590-620 nm

Red: 620-750 nm

4 0

3 years ago