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3 years ago

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

1 answer:

8 0

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:

You might be interested in

If a spaceship has a momentum of 30,000 kg-m/s to the right and a mass of

m_a_m_a [10]

Answer:

75m/s

Explanation:

...................

8 0

3 years ago

You are designing a 108 cm3 right circular cylindrical can whose manufacture will take waste into account. There is no waste in

FinnZ [79.3K]

Explanation:

It is given that,

The volume of a right circular cylindrical, $V=108\ cm^3$

We know that the volume of the cylinder is given by :

$V=\pi r^2 h$

$108=\pi r^2 h$

$h=\dfrac{108}{\pi r^2}$ ............(1)

The upper area is given by :

$A=32r^2+2\pi rh$

$A=32r^2+2\pi r\times \dfrac{108}{\pi r^2}$

$A=32r^2+\dfrac{216}{r}$

For maximum area, differentiate above equation wrt r such that, we get :

$\dfrac{dA}{dr}=64r-\dfrac{216}{r^2}$

$64r-\dfrac{216}{r^2}=0$

$r^3=\dfrac{216}{64}$

r = 1.83 m

Dividing equation (1) with r such that,

$\dfrac{h}{r}=\dfrac{108}{\pi r}$

$\dfrac{h}{r}=\dfrac{108}{\pi 1.83}$

$\dfrac{h}{r}=59 \pi$

Hence, this is the required solution.

8 0

3 years ago

Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. If the coconut from the shorter

valentinak56 [21]

Wee can use here kinematics

as we know that

$y = v*t + \frac{1}{2} at^2$

for shorter tree we know that

$y = 0 + \frac{1}{2}*9.8 * 2^2$

$y = 19.6 meter$

now since we know that other tree is twice high

So height of other tree is y = 39.2 m

now again by above equation

$y = v*t + \frac{1}{2} at^2$

$39.2 = 0 + \frac{1}{2}*9.8 * t^2$

$t = 2.83 s$

so the time taken is 2.83 s

4 0

3 years ago

A baseball player friend of yours wants to determine his pitching speed. you have him stand on a ledge and throw the ball horizo

zhenek [66]

Answer:

The pitching speed of your friend is 33.20 m/s

Explanation:

Lets explain how to solve the problem

Your friend throw the ball horizontally that means the vertical initial

component of velocity is zero ( $u_{y}=0$ ).

The ball is thrown from a height 4 meters above the ground.

The height $h=u_{y}t+\frac{1}{2}gt^{2}$

Remember: the height is negative value because its below the point of

thrown (initial position)

h = -4 m , $u_{y}=0$ and g = -9.8 m/s²(downward)

Substitute these values in the rule above

⇒ $4=0-\frac{1}{2}(9.8)t^{2}$

⇒ -4 = -4.9t² (multiply both sides by -1)

⇒ 4 = 4.9t² (divide both sides by 4.9)

⇒ 0.81633 = t² (take √ for both sides)

⇒ t = 0.9035

Then the time of the ball to land on the ground is 0.9035 seconds

The range of the ball on the ground is 30 m

The range $R=u_{x}*t$ , where $u_{x}$ is the horizontal

component of the initial velocity

R = 30 meters and t = 0.9035

⇒ $30=u_{x}(0.9035)$ (divide both sides by 0.9035)

⇒ $u_{x}=33.20$ m/s

The pitching speed of your friend is 33.20 m/s 

4 0

3 years ago

PLEASE HELP!!! i’ll mark brainliest if you’re correct

dedylja [7]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

The net force acting on the block is ~

$\qquad \sf \dashrightarrow \: 10 + 5$

$\qquad \sf \dashrightarrow \: 15 \:N$

So, the Answer in the boxes will be ~

$\boxed{ \sf15} \: \boxed{ \sf N}$

7 0

2 years ago