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3 years ago

Whose famous conjecture is paraphrased as "space, time, and matter"?

Physics

2 answers:

GaryK [48]3 years ago

7 0

The person who came up with this conjecture was Albert Einstein

leonid [27]3 years ago

3 0

Answer:

John Archibald Wheeler

Explanation:

the famous conjecture "space, time, and matter" is given by John Archibald Wheeler .

Albert Einstein stated that the fourth dimension is space time.

But John Archibald Wheeler was the physicist who later collaborator of Albert Einstein and work on the mission to achieve unified field theory.

John Archibald Wheeler stated that space time tells matter how to move and matter curves space time.

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The moon's illumination changes in a periodic way that can be modeled by a trigonometric function. On the night of a full moon,

kupik [55]

Answer:

$L(t) = 0.125 (cos (\frac{\pi}{14.765}(t+7)) + 0.125$

Explanation:

The expression for the trigonometric function is :

L(t) = A (cos (B(t - C)))+ D ----- equation (1)

where ;

$A = \frac{max-min}{2}$

$A = \frac{0.25-0}{2}$

A = 0.125

D = $\frac{0+.025}{2}$

D = 0.125

Period of the lunar cycle = 29.53

Then;

$\frac{2 \pi}{B} = 29.53$

$29.53 \ \ B = 2 \pi$

$B = \frac{2 \pi}{29.53}$

$B = \frac{\pi}{29.53}$

Also; we known that December 25 is 7 days before January 1.

Then L(-7) = 0.025

Plugging all the values into trigonometric function ; we have:

$0.125 ( cos ( \frac{\pi}{14.765}((-7)-C)))+0.125 = 0.25 \\ \\ \\ ( cos ( \frac{\pi}{14.765}((-7)-C))) = \frac{0.25-0.125}{0.125}$

$( cos ( \frac{\pi}{14.765}((-7)-C))) = 1$

$( \frac{\pi}{14.765}((-7)-C))= cos^{-1} (1)$

$}((-7)-C))=0$

$C= -7$

$L(t) = 0.125 (cos (\frac{\pi}{14.765}(t-(-7))) + 0.125$

$L(t) = 0.125 (cos (\frac{\pi}{14.765}(t+7)) + 0.125$

7 0

3 years ago

Consider the magnetic field (B) of a wire with a constant current (I). A Magnetic Field sensor is placed at a radius (r). Will t

iren [92.7K]

Answer:

No, the magnitude of the magnetic field won't change.

Explanation:

The magnetic field produced by a wire with a constant current is circular and its flow is given by the right-hand rule. Since this field is circular with center on the wire the magnitude of the magnetic field around the wire will be given by B = [(\mi_0)*I]/(2\pi*r) where (\mi_0) is a constant, I is the current that goes through the conductor and r is the distance from the wire. If the field sensor will move around the wire with a fixed radius the distance from the wire won't change so the magnitude of the field won't change.

8 0

3 years ago

True or false kinetic energy is the energy of a moving tennis ball

Kay [80]

answer = true

reasoning

kinetic energy is the energy of movement

potential energy is built up kinetic energy

hope this helps!

5 0

3 years ago

What is inside of a blackhole ?

horsena [70]

We don't know. A black hole is a star that has collapsed into its own gravity. The gravity in fact, is so strong that even light cannot get through it. That's why it looks black to us.

7 0

3 years ago

a circus performer launches himself from a springboard with an initial velocity of 21 m/s at an angle of 75 toward a platform ha

kherson [118]

Answer:

The circus performer falls back down to the ground

Explanation:

The question parameters are;

The initial velocity of the circus performer = 21 m/s

The angle in which the performer launches himself = 75° towards the platform

The height of the platform above the ground = 20 m

The horizontal distance of the platform from the springboard = 15 m

The vertical motion of the circus performer is given by the following projectile motion relation;

y = y₀ + v₀·sinθ₀t-1/2·g·t²

Where;

y = Height reached by the circus performer

y₀ = Initial height of the the circus performer (the springboard) = 0 m

v₀ = Initial velocity of the the circus performer = 21 m/s

θ₀ = The angle with which the circus performer launches himself = 75°

t = The time of ,light of the circus performer

g = The acceleration due to gravity

Therefore, when the height is 20 m, we have;

20 = 21*sin(75)*t - 1/2*9.81*t²

Which gives;

21*sin(75)*t - 1/2*9.81*t² - 20 = 0

Factorizing using a graphing calculator, gives;

t = 1.623 or t = 2.513

Therefore, the circus performer passes the 20 m mark twice, in his motion, where the first time is when he is on his way up while the second time is when he is on his way down

The horizontal motion of the circus performer is given by the following projectile motion relation;

x = x₀ + v₀*cos(θ₀)* t

Where;

x₀ = The initial position of the circus performer in relation to the final position = 0

Plugging in the value of t when y = 20, we get;

x = 21×cos(75)×1.623 = 8.82 m, which is less than the 15 m platform distance from the spring board

Checking the other time value, we have;

x = 21×cos(75)×2.513 = 13.66 m which is also less than the 15 m platform distance from the spring board

Therefore, the circus performer misses the platform and falls back down to the ground.

8 0

3 years ago