What happens if you reverse the galaxy motion and go back in time?

What happens to a theory as new evidence is found?

algol13

New evidence may support the theory -> then nothing

New evidence conflicts with the theory => rework theory / create a new one

6 0

3 years ago

A 700 kg racecar slowed from 30 m/s to 15 m/s. What is the change in momentum

Lerok [7]

Momentum = (mass) x (speed)

Original momentum = (700 kg) x (30 m/s) = 21,000 kg-m/s

Final momentum = (700 kg) x (15 m/s) = 10,500 kg-m/s

Change in momentum = - 10,500 kg-m/s .

8 0

3 years ago

heat and pressure deep beneath Earth's surface can change any rock into A. chemical rock B. organic rock C. metamorphic rock D.

Y_Kistochka [10]

It should be a metamorphic rock as those rocks are formed due fue to heat and pressure

4 0

3 years ago

Read 2 more answers

What are the major features of the human eye and to what are they analogous in a camera?

adelina 88 [10]

Answer:

The human major features with the analogous of camera are mentioned.

Explanation:

There are the following major features of human eye which are analogue with the camera:

Camera will have shutters which controls light entering into it, human eye consists of Diaphragm which also functions same.

Both gives Real & inverted images.

In camera film records image, in eye image is focused on retina and it is converted into electrical impulses.

Pupil and iris acts as aperture of camera.

8 0

3 years ago

Consider a wire of a circular cross-section with a radius of R = 3.17 mm. The magnitude of the current density is modeled as J =

Sloan [31]

Answer:

The current is $I = 8.9 *10^{-5} \ A$

Explanation:

From the question we are told that

The radius is $r = 3.17 \ mm = 3.17 *10^{-3} \ m$

The current density is $J = c\cdot r^2 = 9.00*10^{6} \ A/m^4 \cdot r^2$

The distance we are considering is $r = 0.5 R = 0.001585$

Generally current density is mathematically represented as

$J = \frac{I}{A }$

Where A is the cross-sectional area represented as

$A = \pi r^2$

=> $J = \frac{I}{\pi r^2 }$

=> $I = J * (\pi r^2 )$

Now the change in current per unit length is mathematically evaluated as

$dI = 2 J * \pi r dr$

Now to obtain the current (in A) through the inner section of the wire from the center to r = 0.5R we integrate dI from the 0 (center) to point 0.5R as follows

$I = 2\pi \int\limits^{0.5 R}_{0} {( 9.0*10^6A/m^4) * r^2 * r} \, dr$

$I = 2\pi * 9.0*10^{6} \int\limits^{0.001585}_{0} {r^3} \, dr$

$I = 2\pi *(9.0*10^{6}) [\frac{r^4}{4} ] | \left 0.001585} \atop 0}} \right.$

$I = 2\pi *(9.0*10^{6}) [ \frac{0.001585^4}{4} ]$

substituting values

$I = 2 * 3.142 * 9.00 *10^6 * [ \frac{0.001585^4}{4} ]$

$I = 8.9 *10^{-5} \ A$

5 0

3 years ago