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Marrrta [24]
3 years ago
7

Does the gas have a well defined structure

Physics
2 answers:
dem82 [27]3 years ago
6 0

What gas? Gas is an invisible form so I don't see how it would have a "defined structure".

NeTakaya3 years ago
4 0
No gas does not have a structure.
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I need 1, 2 and 3 <br><br> Please help!
svetoff [14.1K]

1)Kenetic Energy is defined as energy which a body possesses by virtue of being in motion. 2)KE) is KE = 0.5 x mv2. Here m stands for mass, the measure of how much matter is in an object, and v stands for the velocity of the object, or the rate at which the object changes its position..

And I hope this helped :)

7 0
2 years ago
The rate at which the earth's surface loses heat and keeps earth temperature at which living things can survive
agasfer [191]
<span>I believe it's insulation.</span>
5 0
3 years ago
Can someone help me with this please?
Slav-nsk [51]

Answer:

C2, C1, C4, C5 and C6 are in parallel. Therefore, we use the formula Cp = C1 + C2 + ....

Cp = C2 + C1 + C4 + C5 + C6 = ( 7 * 10 ^-3) + (18 * 10^-6) + (0.8F) + (200 * 10^-3 F) + (750 * 10^-6) = 1.008F

Now, Cp will become one capacitor and it will be aligned with C3, therefore it will now become a circuit in series.

We use the formula: 1/Cs = 1/C1 + 1/C2 + .... + ....1/Cn

Thus,

1/Cs = 1/C3 + 1/Cp

1/Cs = 1/(14 * 10^-3 F) + 1/(1.008F)

Cs = 1.4 * 10 ^-2 or if we do not round too much it will give exactly 0.0138 F

So the answer should be a)

8 0
3 years ago
An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400Hz. What is the distance from the center of
Gnoma [55]

Length of the pipe = 0.39 m

Third harmonic frequency = 1400 Hz

For the third harmonic:

Wavelength = \frac{2L}{3}

The center of the open pipe will host a node and the nearest anti - node from the center will be at the 0.25 × wavelength

Distance from center  = 0.25 × wavelength

Distance = 0.25 x \frac{2L}{3}

Plugging the value of the length of the pipe (L) = 0.39 m = 39 cm

Distance = 0.25 x \frac{2 \times 39}{3}

Distance from the center to the nearest anti - node = 6.5 cm

Hence, the nearest distance to the anti - node from the center = 6.5 cm

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7 0
3 years ago
A 5.5kg radio is pushed across the table. If the acceleration is 5m/s to the right, find the net force exerted on the radio.
Sholpan [36]

Answer:

Net force exerted on the radio is 27.5 Newton.

Given:

Mass = 5.5 kg

Acceleration = 5 \frac{m}{s^{2} }

To find:

Force exerted on the radio = ?

Formula used:

F = ma

Where F = net force

m = mass

a = acceleration

Solution:

According to Newton's second law of motion,

F = ma

Where F = net force

m = mass

a = acceleration

F = 5.5 × 5

F = 27.5 Newton

Hence, Net force exerted on the radio is 27.5 Newton.


4 0
3 years ago
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