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garri49 [273]
3 years ago
10

I need help with this answer

Physics
1 answer:
iren2701 [21]3 years ago
7 0
It’s oxygen i’m pretty sure
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In each cylinder, draw the piston and the gas particles based on the temperature shown.
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Harrizon [31]
Ohms law = v= Ir

V= 0.02 x 4000 = 80v
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A capacitor C is fully charged by connecting it to battery of V Volt. Then it is disconnected from battery. If the separation be
vodomira [7]

Answer:

Explanation:

i )

When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2

So charged stored in it will remain unchanged .

ii )

Potential difference = charge / capacitance

in the first case potential difference = Q / C

in the second case potential difference = Q / 2C

So potential difference becomes half .

iii ) electric field = potential diff / plate separation

in the first case electric field = Q / (d x C )

in the second case electric field = 2 Q / (d x 2C)

= Q / (d  x C )

So electric field remains unchanged .

iv)

energy stored in first case = Q² / 2C

In the second case energy stored = Q² / 2x2C

so energy stored becomes half .

4 0
4 years ago
A pelican flying along a horizontal path drops
Ludmilka [50]

Answer:

The initial speed of the pelican is 8.81 m/s.

Explanation:

Given;

height of the pelican, h = 5.0 m

horizontal distance, X = 8.9 m

The time of flight is given by;

t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2*5}{9.81} } \\\\t = 1.01 \ s

The initial horizontal speed of the pelican is given by;

X = vₓt

vₓ = X / t

vₓ = 8.9 / 1.01

vₓ = 8.81 m/s

Therefore, the initial speed of the pelican is 8.81 m/s.

8 0
3 years ago
If a 0.08 kg cell phone falls off a table at 15 m/s, then what is its kinetic energy right before it hits the ground?
Mariana [72]

The kinetic energy of the phone right before it hits the ground is 9J.

<h3>Kinetic energy of the phone</h3>

The kinetic energy of the phone right before it hits the ground is calculated as follows;

K.E = ¹/₂mv²

where;

  • m is mass of the phone
  • v is velocity of the phone

K.E = ¹/₂(0.08)(15)²

K.E = 9 J

Thus, the kinetic energy of the phone right before it hits the ground is 9J.

Learn more about kinetic energy here: brainly.com/question/25959744

#SPJ1

7 0
2 years ago
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