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andrezito [222]
2 years ago
9

an interplantetary speedcarft moving at 20000m/s.how far will it travell in one day?(give your answer in km)

Physics
1 answer:
Scilla [17]2 years ago
8 0

Explanation:

Step I: 1 day means 24 hours * 60 minutes * 60 seconds.

So 1 day have 86400 seconds.

Step I:

Now,

To calculate the travel distance 20,000m/s* 84600 is 1728000000m

Step Ill:

Now convert the meter in kilometer

Because 1 km = 1000 m

So, = 1728000000/1000 = 172800OKm

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The nearest star to our planet other than the sun , is 4.4 light year away.One light-year is the distance light travels in a yea
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Answer:

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The main reason for using light years, however, is because the distances we deal with in space are immense. If we stick to miles or kilometers we quickly run into unwieldy numbers just measuring the distance to the nearest star: a dim red dwarf called Proxima Centauri that sits a mere 24,000,000,000,000 miles away!

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3 years ago
A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has coordinates and . What is the
Natali5045456 [20]

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

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$x=1 \ m$

$x=2 \ m$

where x and y are in meters. Point $P$ has coordinates $P_x=1 \ m$ and $P_y=-2 \ m$. What is the moment of inertia $I_P$ of the plate about the point $P$ ?

Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1 $

$x=2 $

and $\rho = 2 \ kg/m^2$ , $P_x=1 \ $ , $P_y=-2 \ $.

So,

$dI = dmr^2$

$dI = \rho \ dA  \ r^2$  ,           $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \  dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

  $= 3050.19 \ kg \ m^2$

So the moment of inertia is  $3050.19 \ kg \ m^2$.

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emmainna [20.7K]

Answer:

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Credit: NASA

I hope this helps, if it doesn't then just message me and ill be more than happy to help :)

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