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2 years ago

"how do the fundamental laws of physics make manifest that space has 3 dimensions?"

1 answer:

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If you are asking for a proof on having at least 3 dimensions in space, you can find the physical proof anywhere in your daily life activities. Just the fact that solids have volumes is a proof already that we live in a three-dimensional space. We can move forwards, backwards, sidewards and in all other directions possible.

When you go right into detail, the fundamental laws governing these proofs are very technical. They have differential equations to show as proof. It is too detailed to discuss here. The important things is that, these fundamental laws are what explains the science in our basic activities and natural phenomena:

*Gravitation and planetary motion
* Translation, rotation, magnetic field, forces
* Integrals of equations:

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2. Which of the following is an example of work being done on an object? A prism scatters ultraviolet light into visible light.

Liula [17]

A man pushes a couch across the room is the answer!

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2 years ago

Read 2 more answers

A 4.4 nC charge exerts a repulsive force of 36 mN on a second charge which is located

zhenek [66]

The magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C. The principal of the Columb's law is used in the given problem.

<h3>What is Columb's law?</h3>

The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

Charges that are similar repel each other, whereas charges that are diametrically opposed attract each other.

They will repel, moving in opposite directions at the same speed. Because the magnitude and nature of the charge are the same.

The given data in the problem is;

q₁ is the charge 1 = 4.4 nC = 4.4 ×10⁻⁹ C

F is the repulsive force = 36 mN =36 ×10⁶ N

d is the distance = 0.70 m

The Coulomb force is found as;

$\rm F = \frac{Kq_1q_2}{r^2}\\\\\ \rm 36\times 10^6 = \frac{9 \times 10^9 }{(0.7)^2} \times 4.4 \times 10^{-9} \times q_2\\\\\ q_2 = 8.6241 \times 10^{-19 } \ C$

Hence, the magnitude and sign of the second charge will be + 8.6241×10⁻¹⁹ C.

To learn more about Coulomb's law, refer to the link;

brainly.com/question/1616890

#SPJ2

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2 years ago

A man pushes a shopping cart across a level floor. What force resists the effort force? A) gravity B) friction C) the normal for

Kipish [7]

Answer:

B) Friction

Explanation:

Friction is a force that acts when an object is sliding along a surface. Microscopically, this force is due to the fact that the two surfaces are not perfectly smooth, but they have "imperfections" that cause a force that opposes the motion of the object.

For an object sliding on a flat surface, the force of friction has magnitude:

$F_f = \mu_k mg$

where

$\mu_k$ is the coefficient of kinetic friction

m is the mass of the object

g is the acceleration of gravity

The direction of the force of friction is always opposite to the direction of motion of the object.

In reality, friction also acts if the object is at rest and it is pushed by a force; in this case, we talk about static friction, and its magnitude is

$F_f = \mu_s mg$

where $\mu_s$ is called coefficient of static friction, and it is generally larger than the coefficient of kinetic friction.

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3 years ago

Two spheres A and B are projected off the edge of a 1.0 m high table with the same horizontal velocity . sphere A has a mass of

olga2289 [7]

Answer:

c. because A will land first becuase its heavier :)

Explanation:

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2 years ago

A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

ElenaW [278]

Explanation:

The given data is as follows.

Length of beam, (L) = 5.50 m

Weight of the beam, ( $W_{b}$ ) = 332 N

Weight of the Suki, ( $W_{s}$ ) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

$\sum M_{o} = 0$

$-W_{s} \times x + W_{b} \times 1.5$ = 0

x = $\frac{W_{b} \times 1.5}{W_{s}}$

= $\frac{332 N \times 1.5}{505 N}$

= 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

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3 years ago