58775.51 J kinetic energy is lost,
Energy lost = 0.5*mv²
= 0.5*8000/9.8 * 12*12
= 58775.51 J
To calculate the energy loss of a moving object, subtract the initial kinetic energy from the final kinetic energy, where the initial and final kinetic energies are calculated using the formula 1/2*M*V^2.
When energy is transformed from one form to another or moved from one area to any other, or from one device to another there's energy loss. This means whilst energy is converted to a different form, some of the entering energy is become an especially disordered form of energy, like heat.
Learn more about kinetic energy here: brainly.com/question/8101588
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<h2>Answer: Because they rotate really fast.
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First, it is necessary to clarify what is a pulsar (pulsating star) and what is a neutron star:
<u>A pulsar</u> is a neutron star that emits very intense electromagnetic radiation at short and periodic intervals due to its intense magnetic field that induces this emission.
<u>A neutron star</u>, is the name given to the remains of a supernova. In itself it is the result of the gravitational collapse of a massive supergiant star after exhausting the fuel in its core.
Neutron stars have a small size for their very high density and they rotate at a huge speed.
<h2>So, the way to know that a pulsar is a neutron star is because of its <u>high rotating speed.
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Nevertheless, it is important to note that all pulsars are neutron stars, but not all neutron stars are pulsars.
Answer:
We know that ΔK = Kf - Ki = 1/2 m Vf^2 - 1/2 m Vi^2 = 1/2m(Vf^2-Vi^2) = 1/2 m ΔV^2.
The mass remains the same, just calculate the difference of squared velocities and multiply it by half of the mass.
Answer:
(a) E=λ/(2\pi e0 r)
(b) E = 0
Explanation:
(a) We can use the Gaussian's Law to calculate the electric field at any distance r from the axis. By using a cylindrical Gaussian surface we have:
where λ is the total charge per unit length inside the Gaussian surface. In this case we have that the Electric field vector is perpendicular to the r vector. Hence:
(b) outside of the outer cylinder there is no net charge inside the Gaussian surface, because charge of the inner radius cancel out with the inner surface of the cylindrical conductor.
Hence, we have that E is zero.
hope this helps!!
The answer would be false