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Vilka [71]
3 years ago
12

. A pulsar is a type of __________ that rotates very rapidly and emits electromagnetic radiation.. . . protostar. . red giant. .

neutron star.
Physics
2 answers:
damaskus [11]3 years ago
7 0
A pulsar is a type of neutron star that rotates <span>very rapidly and emits electromagnetic radiation. The correct option among all the options that are given in the question is the last option or the third option. I hope that this is the answer that has actually come to your desired help.</span>
spin [16.1K]3 years ago
7 0

Answer:

neutron star

Explanation:

A neutron star is formed when a star which had a mass of between 10 and 29 solar masses collapses. They have a radius of about 10 kilometers and a mass of about 1.4 solar masses. When star loses its original mass the rotation speed increases drastically.

This rotation increases the electromagnetic field generated by the star which exits from the magnetic poles of the star. When there is misalignment of the stars rotational and magnetic axis the electromagnetic radiation is released once every rotation. This causes the EM radiation to be emitted in pulses.

You might be interested in
A truck pulls a trailer at a constant velocity for 100 m while exerting a force of 480 N. Calculate the work done.
Arisa [49]
The answer is c because:

Work done =force x distance 
5 0
3 years ago
Three children are riding on the edge of a merry-go-round that is 182 kg, has a 1.60 m radius, and is spinning at 15.3 rpm. The
Korolek [52]

Answer:

The new angular velocity of the merry-go-round is 18.388 revolutions per minute.

Explanation:

The merry-go-round can be represented by a solid disk, whereas the three children can be considered as particles. Since there is no external force acting on the system, we can apply the principle of angular momentum conservation:

\left(\frac{1}{2}\cdot M + m_{1}+m_{2} + m_{3} \right)\cdot R^{2}\cdot \dot n_{o} = \left(\frac{1}{2}\cdot M + m_{1} + m_{3})\cdot R^{2}\cdot \dot n_{f} (1)

Where:

M - Mass of the merry-go-round, in kilograms.

m_{1}, m_{2}, m_{3} - Masses of the three children, in kilograms.

R - Radius of the merry-go-round/Distance of the children with respect to the center of the merry-go-round, in meters.

\dot n_{o}, \dot n_{f} - Initial and final angular speed, in revolutions per minute.

If we know that M = 182\,kg, m_{1} = 17.4\,kg, m_{2} = 28.5\,kg, m_{3} = 32.8\,kg, R = 1.60\,m and \dot n_{o} = 15.3\,\frac{rev}{min}, then the final angular speed of the system is:

\dot n_{f} = \dot n_{o}\cdot \left(\frac{\frac{1}{2}\cdot M + m_{1} + m_{2} + m_{3} }{\frac{1}{2}\cdot M + m_{1} + m_{3} } \right)

\dot n_{f} = \left(15.3\,\frac{rev}{min} \right)\cdot \left[\frac{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg +28.5\,kg + 32.8\,kg }{\frac{1}{2}\cdot (182\,kg) + 17.4\,kg + 32.8\,kg } \right]

\dot n_{f} = 18.388\,\frac{rev}{min}

The new angular velocity of the merry-go-round is 18.388 revolutions per minute.

7 0
3 years ago
Question in picture.
Gelneren [198K]
<h2>Hello!</h2>

The answer is: A. 19.3 joules

<h2>Why?</h2>

Since it's an elastic collision, the kinetic energy after and before the collision will be the same.

Kinetic energy can be calculated using the following equation:

KE=\frac{1}{2}mv^{2}

Where:

KE=KineticEnergy\\m=mass\\v=velocity

So,

First object, (going to the right):

m=7.20kg\\v=2\frac{m}{s}

KE_{1}=\frac{1}{2}*7.20Kg*(2\frac{m}{s})^{2}=14.4Joules

Second object:, (going to the left):

m=5.75kg\\v=-1.30\frac{m}{s}

KE_{2}=\frac{1}{2}*5.75kg*(-1.30\frac{m}{s})^{2}=4.86Joules

Remember,

1Joule=1Kg.\frac{m^{2}}{s^{2} }

Hence,

The total kinetic energy after the collision will be:

T=KE_{1}+KE_{2}=14.4Joules+4.86joules=19.26joules=19.3joules

The total kinetic energy after the collision is 19.3 joules (rounded to the nearest tenth)

Have a nice day!

6 0
3 years ago
A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
bagirrra123 [75]

Answer:

a) 520m

b) 10.30 s

c) 100,95 m/s

Explanation:

a) According the given information, the rocket suddenly stops when it reach the height of 520m, because the engines fail, and then it begins the free fall.

This means the maximum height this rocket reached before falling  was 520 m.

b) As we are dealing with constant acceleration (due gravity) g=9.8 \frac{m}{s^{2}} we can use the following formula:

y=y_{o}+V_{o} t-\frac{gt^{2}}{2}   (1)

Where:

y_{o}=520 m  is the initial height of the rocket (at the exact moment in which it stops due engines fail)

y=0  is the final height of the rocket (when it finally hits the launch pad)

V_{o}=0 is the initial velocity of the rocket (at the exact moment in which it stops the velocity is zero and then it begins to fall)

g=9.8m/s^{2}  is the acceleration due gravity

t is the time it takes to the rocket to hit the launch pad

Clearing t:

0=520 m+0-\frac{9.8m/s^{2} t^{2}}{2}   (2)

t^{2}=\frac{-520 m}{-4.9 m/s^{2}}   (3)

t=\sqrt{106.12 s^{2}   (4)

t=10.30 s   (5)  This is the time

c) Now we need to find the final velocity V_{f} for this rocket, and the following equation will be perfect to find it:

V_{f}=V_{o}-gt  (6)

V_{f}=0-(9.8 m/s^{2})(10.30 s)  (7)

V_{f}=-100.95 m/s  (8) This is the final velocity of the rocket. Note the negative sign indicates its direction is downwards (to the launch pad)

7 0
3 years ago
In a random sample of 200 students from Northern State University, 150 said that they own an Therefore, probably at least 50% of
kirza4 [7]

Answer:

No fallacy

Explanation:

The statement that at least  50% of the students at Northern State own an iPod is best categorized as no fallacy.

This statement is very true and consistent with the facts presented in the passage.

A fallacy is a mistaken belief usually based on biased arguments.

Total number of students sampled = 200

Number of iPod owners = 150

Percentage of iPod owners = \frac{number of iPod owners}{Number of sample} x 100

Percentage of iPod owners = \frac{150}{200} x 100 = 75%

We see that it is consistent to say that at least, 50% of the students in the sample are iPod owners.

6 0
3 years ago
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