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2 years ago

How far away is the closest star?

Physics

2 answers:

sukhopar [10]2 years ago

8 0

Answer: Proxima Centauri is the closet star about 40,208,000,000,000 km away.

Explanation:

balu736 [363]2 years ago

4 0

Answer:

The closest star is about 25,300,000,000,000 miles (39,900,000,000,000 kilometers) away

Explanation:

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answer: in pic

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3 years ago

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6. As the sky diver's fall the______energy is converted into _______.

notka56 [123]

Answer: B

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3 years ago

Calculate the magnitude of the focal length and power of a spherical mirror that has a radius of 5.00 cm.

yarga [219]

Answer:

2.5 cm

40 D

Explanation:

When the radius of curvature of a lens is divided by 2 we get the focal length of the lens.

Focal length is given by

$f=\dfrac{R}{2}\\\Rightarrow f=\dfrac{5}{2}\\\Rightarrow f=2.5\ cm$

The focal length of the lens is 2.5 cm

When we divide 1 by the focal length in the unit of meters we get the power of a lens

Power of a lens is given by

$P=\dfrac{1}{f}\\\Rightarrow P=\dfrac{1}{0.025}\\\Rightarrow P=40\ D$

The power of the lens is 40 D

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4 years ago

On a frictionless horizontal surface, a 1.50 kg mass traveling at 3.50 m/s collides with and sticks to a 3.00 kg mass that is in

My name is Ann [436]

Answer:

√ $\frac{3}{5} m$

Explanation:

Hope it helps!.

4 0

3 years ago

What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago