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yawa3891 [41]
3 years ago
14

The diagram shows the movement of air as a result of convection currents. At which point is the air at its highest density?

Physics
2 answers:
kozerog [31]3 years ago
5 0

Answer: X

Explanation:

In the given diagram, air moves as a result of convection currents. The air at the bottom, near the space heater gets heated up. As temperature rises, the density of air reduces. It becomes less dense and rises up. The least dense is at W. As the air rises, it gets cooled and becomes more dense and heavier and thus, sinks down. The highest density of air would be at X.

Gnesinka [82]3 years ago
3 0

X

Hot air rises = it's less dense than cold air = falls

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A 30 ohm resistor and a 20 ohm resistor are
Reptile [31]

The current that would pass through the 30 ohms resistor is 2 A.

<h3>What is electric current?</h3>

Electric current is the rate of flow of electric charge round a conductor.

To calculate the electric current that would pass through the 30 ohms resistor, we use the formula below

Formula:

  • I = V/Rt........... Equation 1

Where:

  • I = Electric current passing through the 30 ohms resistor
  • V = Voltage
  • Rt = Total or effective resistance of the resistors.

From the question,

Given:

  • V = 100 volts
  • Rt = (30+20) ohms (since both resistors are connected in series)
  • Rt = 50 ohms

Substitute these values into equation 1

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Hence, The current that would pass through the 30 ohms resistor is 2 A.

Learn more about electric current here: brainly.com/question/1100341

8 0
2 years ago
A person drops a brick from the top of a building. The height of the building is 400 m and the mass of the brick is 2.00 kg. Wh
DaniilM [7]
-- It takes the brick 8.9 seconds to reach the ground. 

-- At the instant of the "splat", it's falling at 89 m/s.

-- The mass doesn't matter. If not for air resistance, every object
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5 0
3 years ago
Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each o
zhannawk [14.2K]

Answer:

The magnitude of the charge on each sphere is 0.135 μC

Explanation:

Given that,

Mass = 1.0

Distance = 2.0 cm

Acceleration = 414 m/s²

We need to calculate the magnitude of charge

Using newton's second law

F= ma

a=\dfrac{F}{m}

Put the value of F

a=\dfrac{kq^2}{mr^2}

Put the value into the formula

414=\dfrac{9\times10^{9}\times q^2}{1.0\times10^{-3}\times(2.0\times10^{-2})^2}

q^2=\dfrac{414\times1.0\times10^{-3}\times(2.0\times10^{-2})^2}{9\times10^{9}}

q^2=1.84\times10^{-14}

q=0.135\times10^{-6}\ C

q=0.135\ \mu C

Hence, The magnitude of the charge on each sphere is 0.135μC.

7 0
3 years ago
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Oksanka [162]

Answer:

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\frac{dh}{dt}=0, \ \ \ \ h=constant\\\\h^2+l^2=b^2\\\\2h\frac{dh}{dt}+2l\frac{dl}{dt}=2b\frac{db}{dt}\\\\2\times5\times0+2\times12\times(-110)=2\times13\frac{db}{dt}\\\\\frac{db}{dt}=-101.54m/h

Hence, the rate at which the distance between the buses is changing when they are 13mi apart is 101.54m/h

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3 years ago
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the correct choice is

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3 years ago
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