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s344n2d4d5 [400]
3 years ago
14

Due to the earth rotation, the sun will rise out of what direction

Physics
1 answer:
ohaa [14]3 years ago
8 0

Answer:  East

Explanation:  Earth rotates or spins toward the east, and that's why the Sun, Moon, planets, and stars all rise in the east and make their way westward across the sky.

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c would be the right answer for this problem

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calculate the period of a wave whose frequency is 5 Hz and whose wavelength is 1 cm give your answer in decimal form
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Period T = 1/f,        where f = frequency = 5 Hz

T =  1/5 = 0.20 s

Period = 0.20 seconds.
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A uniformly charged, straight filament 6 m in length has a total positive charge of 3 µC. An uncharged cardboard cylinder 1 cm i
Lorico [155]

Answer:

Part A : 5.) = 564.704 N*m²/C

Part B:  10.) = 179751.0 V/m

Explanation:

A) Applying Gauss'Law to the straight filament, using a cylindrical gaussian surface with the filament as the central axis of the surface, assuming that the electric field is normal to the surface (which means that no flux exist through the lids of the cylinder) and is constant at any point of the surface (except the lids where is 0), we can find the electric flux, as follows:

E*2*\pi *r*l = \frac{Qenc}{\epsilon 0} (1)

where:

r is the radius of the cylinder = 0.05 m

l is the length of the cylinder = 0.01 m

Qenc, is the net charge on the filament enclosed by the gaussian surface

ε₀ = 8.8542*10⁻¹² C²/N*m²

In order to find the value of Qenc, we need to find first the linear charge density, as follows:

\lambda = \frac{Q}{L} =\frac{+3e-6C}{6m}  = 5e-7 C/m

The net charge enclosed by the gaussian surface will be just the product of the linear change density λ times the length of the gaussian surface:

Qenc = \lambda * l = 5e-7C/m * 0.01 m = 5e-9 C

According Gauss ' Law, the net flux through the gaussian surface must be equal to the charge enclosed by the surface, divided by the permittivity of free space (in vacuum or air), as follows:

Flux = \frac{Qenc}{\epsilon 0} = \frac{5e-9C}{8.8542e-12 C2/N*m2} = 564.704 (N*m2)/C

which is the same as the option 5.

B)  Repeating the equation (1) from above:

E*2*\pi *r*l = \frac{Qenc}{\epsilon 0} (1)

we can solve for E, as follows:

E =  \frac{Qenc}{2*\pi*r*l* \epsilon 0} = \frac{5e-9C}{2*\pi*(0.05m)*(0.01m)*8.8542e-12 C2/N*m2} = 179751.0 V/m

which is the same as the option 10 of part B.

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