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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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If the force is 4 newtons between two charged spheres separated by 3 centimeters, calculate the force between the same spheres s

il63 [147K]

Answer:

1 Newton

Explanation:

F=9*10^9*q0q1/r^2]]

F=9*10^9*(q0q1)/ r^2

r=3cm

F=4N

F=9*10^9*(q0q1)/3^2

4=9*10^9*(q0q1)/9

4=10^9 q0q1

q0q1=4/10^9

q0q1=4*10^-9

To calculate the force between the forces at a distance of 6 cm

F=9*10^9*(q0q1)/ r^2

=9*10^9*(4*10^-9)/6^2

=9*10^9*(4*10^-9)/36

=10^9*4*10^-9/4

=10^9*10^-9

=1 Newton

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3 years ago

WILL GIVE 40 PTS AND BRAINLIEST IF ANSWERED ASAP AND CORRECTLY!!! When a glass falls and breaks on the floor, potential energy i

bekas [8.4K]

Answer:

Explanation:

1.C

2.B

3.D

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6 0

4 years ago

Read 2 more answers

Can u answer 4 and 5 for me

Luda [366]

I don’t know what the answers are

8 0

4 years ago

When a wire loop is connected to a battery, what is produced in the loop?

Vinil7 [7]

Answer:

magnetic field

Explanation:

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3 years ago

A hollow cylindrical (inner radius = 1.0 mm, outer radius = 3.0 mm) conductor carries a current of 80 A parallel to its axis. Th

Pavlova-9 [17]

Answer:

The magnetic field is $B = 3 mT$

Explanation:

From the question we are told that

The inner radius is $r_i = 1.00 mm =1*10^{-3} \ m$

The outer radius is $r_2 = 3.00 \ mm = 3.0 *10^{-3} \ m$

The distance from the axis of the conductor is $d =2.0 \ mm = 2.0 *10^{-3} \ m$

The current carried by the conductor is $I = 80 A$

According to Ampere's circuital law , the magnetic field at a point that is $r_3$ from the axis of the conductor

$B = \frac{\mu_oI}{2 \pi d } [\frac{d - r_1}{r_2 -r_1} ]$

Where $\mu_o$ is the permeability of free space with a value of $\mu_o = 4 \pi *10^{-7} N/A^2$

substituting values

$B = \frac{(4 \pi *10^{-7})(80)}{2 * 3.142 * 2 *10^{=3} } [\frac{(2^2 - 1 ^2 )*10^{-3}}{(3^2 - 1^2) *10^{-3}} ]$

$B = 3 mT$

5 0

4 years ago