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shtirl [24]
4 years ago
12

Question 8 (1 point)

Physics
1 answer:
bixtya [17]4 years ago
3 0

Answer:

2.46 m/s

Explanation:

speed = distance / time

9350/3800 = 2.46

speed = 2.46 m/s

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A person who weighs 150 pounds on Earth would weigh ____ pounds on the moon.
enot [183]

Answer:

25lb

Explanation:

You haven't changed (you are made up of the same atoms), but the force exerted on you is different. Physicists like to say that your mass hasn't changed, only your weight.

6 0
3 years ago
Two particles with charges +6e and -6e are initially very far apart (effectively an infinite distance apart). They are then fixe
JulijaS [17]

Answer:

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}\ J.

Explanation:

Given charges are:

\rm q_1 = +6e.\\q_2 = -6e.

The electric potential energy of a charge due to the electric field of another charge is given by

\rm EPE=\dfrac{kq_1q_2}{r}.

where,

  • k = Coulomb's constant, having value = \rm 9\times 10^9\ Nm^2/C^2.
  • r = distance between the charges.

When the charges are infinite distance apart, \rm r = \infty,

\rm EPE_{initial} = \dfrac{kq_1q_2}{r}=0\ J.

When the charges are \rm 5.61\times 10^{-12}\ m apart, \rm r=5.61\times 10^{-12}\ m,

\rm EPE_{final}=\dfrac{kq_1q_2}{r}\\=\dfrac{(9\times 10^9)\times (+6e)\times (-6e)}{5.61\times 10^{-12}}\\=-5.775\ e^2\times 10^{22}.

Here, e is the charge on one electron, such that, \rm e = -1.6\times 10^{-19}\ C.

Therefore,

\rm EPE_{final}=-5.775\times (-1.6\times 10^{-19})^2\times 10^{22} = -1.478\times 10^{-15}\ J.

Thus,

\rm EPE_{final}-EPE_{initial}=-1.478\times 10^{-15}-0=-1.478\times 10^{-15}\ J.

4 0
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Which of the following can be accurately determined using Newtonian physics?
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The correct option is B.
Newtonian physics refers to the study of the motion of bodies and the forces acting on them according to the Newton laws of motion.This aspect of physics is also known as classical mechanics. Newton laws of motions deal with the speed of object in motion and those that are at rest. Using Newton laws of motion, the momentum of a car can be easily determined. 
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