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3 years ago

How does a comet change as it travels through space

1 answer:

5 0

Comets are like "dirty snowballs"; frozen gasses with dust and rocks in them. Each pass near the Sun causes the comet's nucleus to be exposed to intense sunlight, which causes some tiny fraction of the gas to evaporate and carry some of the dust and rock away into space. The gas and dust, near the Sun, cause the comet's "tail", and repeated passes cause dust and rock to spread out along most of the orbit of a comet. When the Earth enters one of these trails of old comet dust, we have meteor showers.

On rare occasions, comets break apart or even more rarely, crash into planets. In 1994, the comet Shoemaker-Levy 9 broke apart and then collided with the planet Jupiter.

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Which of the following usually occurs with a short circuit? a. All parts of the circuit will begin to carry higher amounts of cu

fgiga [73]

Answer:

Hi,

The correct answer option is; D. Most of the current will flow through one part of the circuit.

Explanation:

A short circuit is a low resistance path in an electric connection between two conductors supplying current in a circuit.

It happens when excess amounts of current flow in the power source through a 'short path'.

Short circuits occur at very high temperatures which is of course caused by the heat produced during dissipation.

An example of application of short circuit is arc welding, where heating is achieved through short circuit.

6 0

3 years ago

Which of the following could be the source of resistance in a household electric circuit?

Tpy6a [65]

D. all of these

all of these use electricity

Hope I helped!

8 0

3 years ago

Read 2 more answers

PLEASE HELP ME 45 POINTS

sergij07 [2.7K]

Answer:

a) We kindly invite you to see the explanation and the image attached below.

b) The acceleration of the masses is 4.203 meters per square second.

c) The tension force in the cord is 28.02 newtons.

d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.

e) The final speed of the system is 3.551 meters per second.

Explanation:

a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.

b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:

Mass A

$\Sigma F = T - m_{A}\cdot g = -m_{A}\cdot a$ (1)

Mass B

$\Sigma F = T - m_{B}\cdot g = m_{B}\cdot a$ (2)

Where:

$T$ - Tension force in the cord, measured in newtons.

$m_{A}$ , $m_{B}$ - Masses of blocks A and B, measured in kilograms.

$g$ - Gravitational acceleration, measured in meters per square second.

$a$ - Net acceleration of the each block, measured in meters per square second.

By subtracting (2) by (1), we get an expression for the acceleration of each mass:

$m_{B}\cdot a +m_{A}\cdot a = T-m_{B}\cdot g -T + m_{A}\cdot g$

$(m_{B}+m_{A})\cdot a = (m_{A}-m_{B})\cdot g$

$a = \frac{m_{A}-m_{B}}{m_{B}+m_{A}} \cdot g$

If we know that $m_{A} = 5\,kg$ , $m_{B} = 2\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then the acceleration of the masses is:

$a = \left(\frac{5\,kg-2\,kg}{5\,kg+2\,kg}\right) \cdot\left(9.807\,\frac{m}{s^{2}} \right)$

$a = 4.203\,\frac{m}{s^{2}}$

The acceleration of the masses is 4.203 meters per square second.

c) From (2) we get the following expression for the tension force in the cord:

$T = m_{B}\cdot (a+g)$

If we know that $m_{B} = 2\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ and $a = 4.203\,\frac{m}{s^{2}}$ , then the tension force in the cord:

$T = (2\,kg)\cdot \left(4.203\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}} \right)$

$T = 28.02\,N$

The tension force in the cord is 28.02 newtons.

d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:

$\Delta y = \frac{1}{2}\cdot a\cdot t^{2}$ (3)