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katen-ka-za [31]

4 years ago

15

some galaxies have regions that are relatively blue other regions appear redder what does this varuation indicate about the diff

erences between these regions?

1 answer:

Vinvika [58]4 years ago

3 0

Answer:

Blue is hot while the red is cool.

Explanation:

Hot stars appear in blue whereas cooler stars appear to have red hues.

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A car starts from rest and accelerates uniformly over a time of 10.21 seconds for a distance of 210 m. Determine the acceleratio

uysha [10]

Answer:

4.03 m/s/s

Explanation:

d=1/2 a t^2

210 = 1/2 a (10.21)^2

210 = 52.1 a

a = 4.03 m/s/s

6 0

3 years ago

Bambi the young dear was distracted Buy butterfly and jumped into the road in front of the two vehicles as shown in the diagram

bagirrra123 [75]

Speed of car A is given as

$v_a = 70 mph$

now we need to convert it into SI units

1 miles = 1609 m

1 hour = 3600 s

now we have

$v_a = 70 *\frac{1609}{3600} = 31.3 m/s$

now its distance from Bambi is given as

$d_a = 350 m$

time taken by it to hit the Bambi

$t = \frac{d}{v}$

$t = \frac{350}{31.3}$

$t = 11.2 s$

Now other car is moving at speed 50 mph

so its speed in SI unit will be

$v_b = 50* \frac{1609}{3600}$

$v_b = 22.35 m/s$

now its distance from Bambi is given as

$d_b = 590 feet$

as we know that 1 feet = 0.3048 m

$d_b = 590*0.3048 = 179.83 m$

now the time to hit the other car is

$t_2 = \frac{179.83}{22.35}$

$t_b = 8.05 s$

So Car B will hit the Bambi first

7 0

3 years ago

Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

amid [387]

Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$

= $\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]$

∴NOTE: Graph is attached

8 0

4 years ago

If the 50 kg objects slows down to a velocity of 1 m/s how much kinetic energy does it have?

Bogdan [553]

It has 50kg with a velocity of 1 m/s times the speed of the cart divided by 2 and multiplied by kinectic x plus 5

3 0

3 years ago

Read 2 more answers

D section a only still need help on this

SIZIF [17.4K]

Acceleration means speeding up, slowing down, or changing direction. The graph doesn't show anything about direction, so we just have to examine it for speeding up or slowing down ... any change of speed.

The y-axis of this graph IS speed. So the height of a point on the line is speed. If the line is going up or down, then speed is changing.

Sections a, c, and d are all going up or down. Section b is the only one where speed is not changing. So we can't be sure about b, because we don't know if the track may be curving ... the graph can't tell us that. But a, c, and d are DEFINITELY showing acceleration.

5 0

4 years ago