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3 years ago

If the sun were to collapse into a black hole, what would the radius be?

Physics

2 answers:

Diano4ka-milaya [45]3 years ago

7 0

Answer:

It would be 1/3rd size of a neutron star

Explanation:

the Schwarzschild radius would be about 3 kilometers; thus, the entire black hole would be about one-third the size of a neutron star of that same mass.

Marianna [84]3 years ago

4 0

Answer:

If the Sun, with its mass of 1 MSun, were to become a black hole the Schwarzschild radius would be about 3 kilometers; thus, the entire black hole would be about one-third the size of a neutron star of that same mass

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An object, with mass 72 kg and speed 28 m/s relative to an observer, explodes into two pieces, one 2 times as massive as the oth

IRINA_888 [86]

Answer:

14112 J

Explanation:

When the 72 Kg mass explodes into two, one mass is twice the other so 72/3=24 Kg

M1= 24 kg, M2= 72-24=48 kg

From law of conservation of linear momentum, the sum of initial and final momentum are equal. p=mv where p is momentum, m is mass and v is velocity. Fir this case, since the less massive piece stops, its final velocity is zero.

72*28=48v2

V2=72*28/48=42 m/s

Difference between initial and final kinetic energy will be

$\triangle KE= 0.5(Mv^{2}-m2v2^{2})\\\triangle KE= 0.5(72*28^{2}-48*42^{2})=-14112 J$

Therefore, from observers reference, kinetic energy of 14112 J is added

5 0

4 years ago

Please help I don't have a ruler lol

Ivahew [28]

Answer:

hi you can just buy new ruler lol joke

1. energy and air

8 0

3 years ago

Suppose the voltage source for a series RL-circuit were given as V0sin(ωt) instead of V0cos(ωt). Write an expression for the cur

Gala2k [10]

Answer:

Explanation:

This is an RL circuit, therefore:

Impedance; z = $\mathbf{\sqrt{R^2+L^2}}$

$\mathbf{z = \sqrt{R^2+(Lw)^2}}$

Current amplitude

$\mathbf{I_o = \dfrac{V_o}{z}} \\ \\ \mathbf{I_o = \dfrac{V_o}{\sqrt{R^2+L^2\omega ^2}}}$

Given that:

$V_o = 1.9 \ V \\ \\ \omega= 51 \ rad/s\\\\ R = 21 \Omega \\ \\ L = 0.52 H$

∴

$I_o= \dfrac{1.9}{\sqrt{21^2+(0.52\times 51)^2}}$

$\mathbf{I_o= 0.0562} \\ \\ \mathbf{I_o = 56.2 \ mA}$

Phase constant :

$tan \ \phi = \dfrac{L \omega}{R } \\ \\ tan \ \phi = \dfrac{0.52 \times 51}{21} \\ \\ tan \phi = 1.263$

$\text{Phase constant : }\phi = tan^{-1} (1.263) \\ \\ \phi = 51.6^0\\ \\\text{To radians} \phi = 51.6 \times \dfrac{\pi}{180} \\ \\ \phi = 0.287 \pi \\ \\ \mathbf{\phi = 0.9 \ rad}$

3 0

3 years ago

Mechanical energy that has been ‘lost' to friction isn't really lost. It just is no longer in its mechanical form. True or False

Vadim26 [7]

Answer:

True.

Explanation:

Energy can be defined as the ability (capacity) to do work. The two (2) main types of energy are;

a. Gravitational potential energy (GPE): it is an energy possessed by an object or body due to its position above the earth.

b. Kinetic energy (KE): it is an energy possessed by an object or body due to its motion.

Furthermore, the mechanical energy of a physical object or body is the sum of the potential energy and kinetic energy possessed by the object or body.

Mathematically, it is given by the formula;

Mechanical energy = G.P.E + K.E

Mechanical energy that has been ‘lost' to friction isn't really lost. It just is no longer in its mechanical form. This is ultimately in accordance with the law of conservation of energy, which states that energy cannot be destroyed but can only be converted or transformed from one form to another.

Hence, Mechanical energy that has been ‘lost' to friction isn't really lost but converted into heat energy.

4 0

3 years ago

Two climbers are on a mountain. Simon, of mass m, is sitting on a snow covered slope that makes an angle θ with the horizontal.

elena-14-01-66 [18.8K]

Answer:

Explanation:

It is required that the weight of Joe must prevent Simon from being pulled down . That means he is not slipping down but tends to be towed down . So in equilibrium , force of friction will act in upward direction on Simon.

Let in equilibrium , tension in rope be T

For balancing Joe

T = M g

For balancing Simon

friction + T = mgsinθ

μmgcosθ+T = mgsinθ

μmgcosθ+Mg = mgsinθ

M = (msinθ - μmcosθ)

M = m(sinθ - μcosθ)

5 0

3 years ago