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Alexeev081 [22]
3 years ago
12

How does a comet change as it travels through space

Physics
1 answer:
AfilCa [17]3 years ago
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. 

<span>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.</span>
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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)

Where:

a - Net acceleration, measured in meters per square second.

t - Time, measured in seconds.

\Delta y - Covered distance, measured in meters.

If we know that a = 4.203\,\frac{m}{s^{2}} and \Delta y = 1.5\,m, then the time taken by the system is:

t = \sqrt{\frac{2\cdot \Delta y}{a} }

t = \sqrt{\frac{2\cdot (1.5\,m)}{4.203\,\frac{m}{s^{2}} } }

t \approx 0.845\,s

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

e) The final speed of the system is calculated by the following formula:

v = a\cdot t (4)

Where v is the final speed of the system, measured in meters per second.

If we know that a = 4.203\,\frac{m}{s^{2}} and t \approx 0.845\,s, then the final speed of the system is:

v = \left(4.203\,\frac{m}{s^{2}} \right)\cdot (0.845\,s)

v = 3.551\,\frac{m}{s}

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

8 0
2 years ago
For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number repr
leonid [27]

<u>Answer:</u> The weight of the object is 29.4 N

<u>Explanation:</u>

To calculate the weight of the object, we use the equation:

W=m\times g

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = 9.8m/s^2

Putting values in above equation, we get:

W=3kg\times 9.8m/s^2\\\\W=29.4N

Hence, the weight of the object is 29.4 N

6 0
3 years ago
A space shuttle is flying north at 6,510 m/s. It
Maksim231197 [3]

Answer:

7808 m/s

Explanation:

Find NE velocity after 60 s  of acceleration in that direction:

  = a t  =  28.4 m/s^2 * 60 s = 1704  m/s

    Vertical component = 1704 sin 45 = 1204.9 m/s

     Horiz component = 1704 cos 45 = 1204.9  m/s

Add the two vertical components

  6510 + 1204.9 = 7714.9 m/s = vertical velocity

Pythagorean theorem to find resultant of vertical and horiz  v's

Vf ^2 = 1204.9^2   + 7714.9^2 0

Vf = 7808. m/s

7 0
2 years ago
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