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faust18 [17]
3 years ago
10

How long does it take the Earth to orbit around the sun

Physics
2 answers:
schepotkina [342]3 years ago
7 0
Hey there!

<span>How long does it take the Earth to orbit around the sun?

It takes 365.25 days for the Earth to make 1 orbit around the sun. That means it takes more than 1 whole year.

Hope this helps
Have a great day (:
</span>
NeX [460]3 years ago
5 0
365 days 5hours 58min 56s  (Sun year), or 365 days 6hours 19min 10s (Stellar year)
Or 1 year
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A substance that is present in a solution in a smaller amount and is dissolved by the solvent is called the solute.
Hunter-Best [27]
<span>a solution is a liquid mixture  </span>
4 0
4 years ago
A body goes from P to Q with a velocity of 10 m/s and comes back from Q to P with a velocity of 20 m/s. What is the average velo
Pachacha [2.7K]
Let's say the distance is D. Then the time going is D/10 sec. The time returning is D/20 s. The total time is 3D/20 s, and the total distance is 2D. The average speed for the round trip is (total distance)/(total time). That's (2D) ÷ (3D/20). That's (40D/3D) which is 13-1/3 m/s. (I thought it was going to depend on the distance, but it doesn't.)
5 0
3 years ago
A large crate is at rest on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.500. A f
Nookie1986 [14]

Answer:

93.4 kg

Explanation:

Draw a free body diagram.  There are three four forces:

Weight force mg pulling down,

Normal force N pushing up,

Friction force Nμ pushing left,

Applied force F pulling up and to the right, 30.0° above the horizontal.

Sum of forces in the y direction:

∑F = ma

N + F sin 30.0° − mg = 0

N = mg − ½ F

Sum of forces in the x direction:

∑F = ma

F cos 30.0° − Nμ = 0

½√3 F = Nμ

Substitute:

½√3 F = (mg − ½ F) μ

½√3 F / μ = mg − ½ F

½√3 F / μ + ½ F = mg

½F (√3 / μ + 1) = mg

m = F (√3 / μ + 1) / (2g)

Plug in values:

m = 410 N (√3 / 0.500 + 1) / (2 × 9.8 m/s²)

m = 93.4 kg

4 0
3 years ago
Read 2 more answers
The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper
meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, v_{d} , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

v_{d} = \frac{I}{qnA}             ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = \frac{I}{A} = \frac{E}{p}

Equation (i) can then be written as follows;

v_{d} = \frac{J}{qn} = \frac{E}{qnp}

v_{d} = \frac{E}{qnp}      ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

v_{d} = \frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}

v_{d} = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

4 0
3 years ago
Can Someone Help ? !5 Points!!!
Ugo [173]
A)float
D)<span>density is greater than the weight of the object

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4 years ago
Read 2 more answers
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