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3 years ago

Which star makes up Orion's knee to your right, or the end of his tunic?

Physics

1 answer:

lord [1]3 years ago

7 0

Answer:

Betelgeuse and Rigel

Explanation:

100% correct

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How much power is required to lift a 60 kg bed to the top of a 3 meter flight of stairs in 10 seconds?

Zigmanuir [339]

Answer:

176.58Watts

Explanation:

Power= work done /time

Where mass(m)=60kg

Height (h) =3m

Time(s)=10s

Force of gravity = 9.81m/s^2

Power=mgh/t

Power= (60kg) * (9.81m/s^2) * (3m)/10s

Power= 176.58Watts

8 0

3 years ago

Why does your hand feel cool if you spill some alcohol on it?

son4ous [18]

<span>It takes heat to make something evaporate, so it takes heat from your arm. Alcohol easily evaporates at room temperature, so it feels cool. This is also why you feel cool when getting out of the pool. The water on your skin evaporates.

</span>

6 0

3 years ago

Read 2 more answers

Can someone pls help ASAP:(

diamong [38]

I’m not sure but I think it’s
△ m=5 and △= -3 and so

Answer: 5/△-3 m/s

So sorry if it’s wrong

6 0

3 years ago

Question 4

bagirrra123 [75]

That's <em>false</em>. It's just the opposite. As you become more fit, your heart becomes able to accomplish more with each beat, so your resting heart rate DEcreases.

4 0

3 years ago

Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi

nadya68 [22]

Answer:

$E = 1.85*10^{12}\frac{N}{C}$

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction, and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

$E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}$

Then the electric field at the point of interest is estimated as:

$E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}$

6 0

4 years ago