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xxTIMURxx [149]
3 years ago
8

The absolute magnitude of a star________.

Physics
1 answer:
serg [7]3 years ago
8 0
<span>The absolute magnitude of a star is how bright it would appear to us
if it were located ten parsecs (about 32.6 light years) from us.  So it's
a way of treating all stars equally ... on a "level playing field" ... and it
describes each star's actual brightness. </span>
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What is one way in which radioactive decay is used to benefit society?
snow_lady [41]
<span>They are used to measure and map effluent and pollution discharges from factories and sewerage plants, and the movement of sand around harbours, rivers and bays. Radioactive materials used for such purposes have short half-lives and decay to background levels within days.</span>
7 0
3 years ago
Read 2 more answers
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
Mature salmon swim upstream, returning to spawn at their birthplace. During the arduous trip they leap vertically upward over wa
Katarina [22]

Answer: minimum speed of launch must be 7.45m/s

Explanation:

Given the following:

Height or distance (s) = 2.83m

The final velocity(Vf) at maximum height = 0

Upward motion, acceleration due to gravity(g) us negative = -9.8m/s^2

From the 3rd equation of motion:

V^2 = u^2 - 2gs

Where V = final velocity

u = initial velocity

Therefore, u = Vi

u = √Vf^2 - 2gs

u = √0^2 - 2(-9.8)(2.83)

u = √0 + 55.468

u = √55.468

u = 7.4476 m/s

u = 7.45m/s

7 0
3 years ago
A VW Beetle goes from 0 to 60.0 mi/h with an acceleration of +2.35 m/s^2. (a) How much time does it take for the beetle to reach
Firlakuza [10]
Im going to tell you what to do but not the result. So pay close attention: the first thing you need to do is convert miles/h to m/s. Then for the part a) <span>divide the final velocity by the initial velocity. That will give you the amount of it will take to accelerate to the final velocity.Now for the part b you </span>use the formula v=vo+at. I hope this can help you
8 0
3 years ago
According to Newton's 3rd Law, an object's momentum depends on it's velocity and mass. 4 dog-sled teams competed to see who coul
podryga [215]

Answer:

C.) Sled Team C 28 kg moving at 12m/s

I'm pretty sure.

7 0
3 years ago
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