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IgorLugansk [536]
3 years ago
14

Which of the following are thought to be rapidly rotating neutron stars with intense magnetic fields? Choose one answer.a. pulsa

rsb. white dwarfsc. black holesd. flare starse. accretion disks
Physics
1 answer:
frutty [35]3 years ago
8 0
<h2>Answer: Pulsars</h2>

A <u>pulsar</u> is a neutron star that emits very intense electromagnetic radiation at short and periodic intervals ( rotating really fast) due to its intense magnetic field that induces this emission.  

Nevertheless, it is important to note that all pulsars are neutron stars, but not all neutron stars are pulsars.

Let's clarify:

A neutron star, 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.  

However, the way to know that a pulsar is a neutron star is because of its high rotating speed.  

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. In single particle problem whole mass is concentrated at a single point so it has a single displacement, single velocity and single acceleration. while, in rigid body mass is distributed
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a train traveling at 30m/s comes to a complete stop in 20 sec.how far did the train go during the deceleration
elixir [45]

Answer:

600m

Explanation:

v=30m/s

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A 10-kg package drops from chute into a 25-kg cart with a velocity of 3 m/s. The cart is initially at rest and can roll freely w
amid [387]

Answer:

(a) the final velocity of the cart is 0.857 m/s

(b) the impulse experienced by the package is 21.43 kg.m/s

(c) the fraction of the initial energy lost is 0.71

Explanation:

Given;

mass of the package, m₁ = 10 kg

mass of the cart, m₂ = 25 kg

initial velocity of the package, u₁ = 3 m/s

initial velocity of the cart, u₂ = 0

let the final velocity of the cart = v

(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;

m₁u₁  + m₂u₂ = v(m₁  +  m₂)

10 x 3   + 25 x 0   = v(10  +  25)

30  = 35v

v = 30 / 35

v = 0.857 m/s

(b) the impulse experienced by the package;

The impulse = change in momentum of the package

J = ΔP = m₁v - m₁u₁

J = m₁(v - u₁)

J = 10(0.857 - 3)

J = -21.43 kg.m/s

the magnitude of the impulse experienced by the package = 21.43 kg.m/s

(c)

the initial kinetic energy of the package is calculated as;

K.E_i = \frac{1}{2} mu_1^2\\\\K.E_i = \frac{1}{2} \times 10 \times (3)^2\\\\K.E_i = 45 \ J\\\\

the final kinetic energy of the package;

K.E_f = \frac{1}{2} (m_1 + m_2)v^2\\\\K.E_f = \frac{1}{2} \times (10 + 25) \times 0.857^2\\\\K.E_f = 12.85 \ J

the fraction of the initial energy lost;

= \frac{\Delta K.E}{K.E_i} = \frac{45 -12.85}{45} = 0.71

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Answer:

Explanation:

70.0(5.00)(9.81) = 3,433.5 = 3430 N

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