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Flauer [41]
2 years ago
9

A black hole is a region of spacetime where gravity is so strong that nothing—no particles or even electromagnetic radiation suc

h as light—can escape from it.
So were do you go if you get sucked into a black hole? What happens to Your body? Do you travel to another solar system?
Physics
1 answer:
Lady_Fox [76]2 years ago
4 0

Answer:

I think you would die instantly-

Explanation:

wouldnt the air force be to much and your insides will be like jelly?? and wouldnt you snap in to your something? or am i wrong??

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The force required to stretch a Hooke’s-law
Setler [38]

I think this is correct, but I am not entirely certain.

Find the force constant of the spring:

F = - KX

(0 - 62.4) = -K(0.172m)

-362.791 = -K

362.791 N/m = K


Find the work done in stretching the spring:

W = (1/2)KX

W = (1/2)(362.791)(0.172m)

W = 31.2 J


5 0
2 years ago
Which statement provides the best description of a computational model?
Varvara68 [4.7K]

Answer:

The answer is "Option C".

Explanation:

Computation modeling is used as the software for math, physics, or software engineering in the simulation and study of complicated processes. The software framework incorporates many parameters, that characterize the model under study. It is a computer program, in which the mathematical formula is generated by computers and research complex systems.

4 0
2 years ago
Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs
Temka [501]
The motion of the ball on the vertical axis is an accelerated motion, with acceleration 
a=g=-9.81 m/s^2
The following relationship holds for an uniformly accelerated motion:
2aS=v_f^2 - v_i^2
where S is the distance covered, vf the final velocity and vi the initial velocity.

If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
v_f =0
So we can rewrite the equation as
2(-9.81 m/s^2) h=-v_i^2
from which we can isolate h
h= \frac{v_i^2}{19.62} (1)

Now let's assume that v_i is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: 2v_i. So the maximum height of the second ball is
h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62} (2)

Which is 4 times the height we found in (1). Therefore, the maximum height of ball 2 is 4 times the maximum height of ball 1.
8 0
2 years ago
A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the
Bingel [31]

Answer:

E_T= 28J

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2

Where:

\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since \Delta x=0, so:

E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2

Where in this case:

m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m

Therefore:

E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J

8 0
3 years ago
A ruler of length 0.30m is pivoted at its centre. Equal and opposite forces of magnitude 2.0N are applied to the ends of the rul
kow [346]

Answer:

0.3858 Nm

Explanation:

The torque of the couple is the dot product of the force vector and the couple vector from 1 end of the ruler to the center. This equals to the product of their magnitude times the cosine() of the angle made by their direction:

T = \vec{F} \cdot \vec{s} = Fscos(50^0) = 2 * 0.3 * 0.643 = 0.3858 Nm

6 0
3 years ago
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