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3 years ago

Can someone explain why light from distant galaxies is projected. Plz

1 answer:

8 0

I'm not exactly sure what you mean by 'projected'. The light from other galaxies
is the light generated by all the stars in that galaxy. It comes out of a distant star,
spends millions of years traveling straight through space, and millions of years
later, a tiny bit of that same light zooms straight into your eye. Practically nothing
happened to it on the way between the star in the distant galaxy and the back of
your eye. Pretty amazing !

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What is the force of a punch if it accelerates at 0.2m/s2 and has a mass of 2kg?

OleMash [197]

Answer:

0.4 N

Explanation:

We know ,

F = ma
F = 0.2 * 2 N
F = 0.4N

6 0

3 years ago

As your skateboard coasts uphill, your speed changes from 3 m/s to 1 m/s in

vlada-n [284]

Answer:

$a=-0.33\ m/s^2$

Explanation:

<u>Accelerated Motion</u>

The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by

$\displaystyle a=\frac{\Delta v}{t}$

Or, equivalently

$\displaystyle a=\frac{v_f-v_o}{t}$

Where vf and vo are the final and initial speeds respectively. The problem gives us these values: v0 = 3 m/s, vf = 1 m/s, t = 3 seconds. Computing a

$\displaystyle a=\frac{1-3}{3}=-0.33\ m/s^2$

The negative sing of a indicates there is deceleration or decreasing speed

7 0

3 years ago

Which of these is an example of physical weathering? Rock breaking up due to carbonic acid Rock breaking up due to mechanical fo

neonofarm [45]

Answer:

Biological weathering

Explanation:

7 0

3 years ago

What forms when two air masses meet and create weather?

Minchanka [31]

When two different air masses meet, a boundary is formed. the boundary between two air masses is called a front. weather at a front is usually cloudy and stormy. there at four different fronts: cold, warm, stationary, and occluded

4 0

3 years ago

Read 2 more answers

You drop a steel ball bearing, with a radius of 2.40 mm, into a beaker of honey. Note that honey has a viscosity of 6.00 Pa/s an

Stells [14]

Answer:

The “terminal speed” of the ball bearing is 5.609 m/s

Explanation:

Radius of the steel ball R = 2.40 mm

Viscosity of honey η = 6.0 Pa/s

$\text { Viscosity has Density } \sigma=1360 \mathrm{kg} / \mathrm{m}^{3}$

$\text { Steel has a density } \rho=7800 \mathrm{kg} / \mathrm{m}^{3}$

$\left.\mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2} \text { (g is referred to as the acceleration of gravity. Its value is } 9.8 \mathrm{m} / \mathrm{s}^{2} \text { on Earth }\right)$

While calculating the terminal speed in liquids where density is high the stokes law is used for viscous force and buoyant force is taken into consideration for effective weight of the object. So the expression for terminal speed (Vt)

$V_{t}=\frac{2 \mathrm{R}^{2}(\rho-\sigma) \mathrm{g}}{9 \eta}$

Substitute the given values to find "terminal speed"

$\mathrm{V}_{\mathrm{t}}=\frac{2 \times 0.0024^{2}(7800-1360) 9.8}{9 \times 6}$

$\mathrm{V}_{\mathrm{t}}=\frac{0.0048 \times 6440 \times 9.8}{54}$

$\mathrm{V}_{\mathrm{t}}=\frac{302.9376}{54}$

$\mathrm{V}_{\mathrm{t}}=5.609 \mathrm{m} / \mathrm{s}$

The “terminal speed” of the ball bearing is 5.609 m/s

7 0

3 years ago