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3 years ago

? If Proxima Centauri is 4.2 light-years away from the Sun, how many AU is that?

Physics

1 answer:

Likurg_2 [28]3 years ago

5 0

Answer:

here is explanation

Explanation:

Proxima Centauri The closest star to our own Is still 40,208,000,000,000 km away.

(OR ABOUT 268,770 AU)

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Once a disk forms around a star, the process of planetary formation can begin. Rank the evolutionary stages for the formation of

PolarNik [594]

Answer: See explanation

Explanation:

The evolutionary stages for the formation of planets from earliest to latest will be:

1. Dust keeps matter inside the disk cool enough for planet formation to start

2. Dust grains form condensation nuclei on which surrounding atoms condense to form small clumps of matter.

3. Small clumps of matter stick together via the process of accretion to form planetesimals a few hundred kilometers in diameter.

4. Planetesimals begin to accrete, forming protoplanets.

5. A collection of a few planet-sized protoplanets remain in a fairly cleared out disk around the star

6 0

3 years ago

What, roughly, is the percent uncertainty in the volume of a spherical beach ball whose radius is 5.66 0.09 m?

iren2701 [21]

Answer:

4.77 %

Explanation:

We know that the volume V for a sphere of radius r is

$V(r) = \frac{4}{3} \ \pi \ r^3$

If we got an uncertainty $\Delta r$ the formula for the uncertainty of V is:

$\Delta V(r) = \sqrt{ (\frac{dV}{dr} \Delta r)^2 }$

We can calculate this uncertainty, first we obtain the derivative:

$\frac{dV}{dr} = 3 * \frac{4}{3} \ \pi \ r^2$

$\frac{dV}{dr} = 4 \ \pi \ r^2$

And using it in the formula:

$\Delta V(r) = \sqrt{ (4 \ \pi \ r^2\Delta r)^2 }$

$\Delta V(r) = \sqrt{ 4^2 \ \pi^2 \ r^4 \Delta r^2 }$

$\Delta V(r) = 4 \ \pi \ r^2 \Delta r$

The relative uncertainty is:

$\frac{\Delta V(r)}{V(r)}$

$\frac{ 4 \ \pi \ r^2 \Delta r }{ \frac{4}{3} \ \pi \ r^3}$

$\frac{ 3 \Delta r }{ r}$

Using the values for the problem:

$\frac{ 3 * 0.09 m }{ 5.66 m} = 0.0477$

This is, a percent uncertainty of 4.77 %

4 0

3 years ago

Why are very high temperatures and pressures required for fusion to occur?

Ber [7]

Explanation:

to overcome the repulsion between the protons in the nuclei that join

7 0

3 years ago

Read 2 more answers

The drawing shows a hydraulic chamber with a spring (spring constant = 1600 N/m) attached to the input piston and a rock of mass

Triss [41]

Answer:

$\Delta x=245\ mm$

Explanation:

Given:

spring constant of the spring attached to the input piston, $k=1600\ N.m^{-1}$

mass subjected to the output plunger, $m=40\ kg$

Now, the force due to the mass:

$F=m.g$

$F=40\times 9.8$

$F=392\ N$

Compression in Spring:

$\Delta x=\frac{F}{k}$

$\Delta x=\frac{392}{1600}$

$\Delta x=0.245\ m$

$\Delta x=245\ mm$

8 0

4 years ago

A current-carrying wire oriented north-south and laid over a compass deflects the compass 8° to the east. What is the magnitude

zhenek [66]

$2.8 \times 10^{-6}\ T$ is the magnitude of the magnetic field made by the current

Explanation:

Given data:

$\theta=8^{\circ}$

Magnetic field of earth, $B_{\text {earth}}=2 \times 10^{-5}\ \text {tesla}$

We need to find the magnetic field of wire, $B_{\text {wire }}$

The compass needle moves toward a direction of magnetic field. The current in wire makes a magnetic field in available space where the compass is on the ground. The vector sum of the Earth's magnetic field and the wire's magnetic field represents the net magnetic field, as shown in the attached drawing, expressing the angle:

$\tan \theta=\frac{B_{\text {wire}}}{B_{\text {earth}}}$

By substituting the given values, we get

$B_{\text {wire}}=\tan \theta \times B_{\text {earth}}=\tan 8^{\circ} \times 2 \times 10^{-5}=0.1405 \times 2 \times 10^{-5}$

$B_{\text {wire}}=0.28 \times 10^{-5}=2.8 \times 10^{-6}\ T$

4 0

3 years ago