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elena55 [62]
3 years ago
10

Gaussian surfaces A and B enclose the same positive charge+Q. The area of Gaussian surface A is three times larger than that of

Gaussian surface B. The flux of electric field through Gaussian surface A is A) nine times larger than the flux of electric field through Gaussian surface B. B) three times smaller than the flux of electric field through Gaussian surface B. C) unrelated to the flux of electric field through Gaussian surface B. D) equal to the flux of electric field through Gaussian surface B. E) three times larger than the flux of electric field through Gaussian surface B
Physics
1 answer:
GalinKa [24]3 years ago
4 0

Answer:

D) equal to the flux of electric field through the Gaussian surface B.

Explanation:

Flux through S(A) = Flux through S (B ) =  Charge inside/ ∈₀

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3 years ago
A stone is thrown towards a wall with an initial velocity of v0=19m/s and an angle = 71 with the horizontal, as illustrated in t
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(b) 16.5 m

(c) 21.7 m

(d) 22.7 m

Explanation:

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v₀ₓ = 19 cos 71° m/s

v₀ᵧ = 19 sin 71° m/s

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

(a) Find Δy when t = 3.5 s.

Δy = v₀ᵧ t + ½ aᵧ t²

Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²

Δy = 2.85 m

(b) Find Δy when vᵧ = 0 m/s.

vᵧ² = v₀ᵧ² + 2 aᵧ Δy

(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy

Δy = 16.5 m

(c) Find Δx when t = 3.5 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²

Δx = 21.7 m

(d) Find Δx when Δy = 0 m.

First, find t when Δy = 0 m.

Δy = v₀ᵧ t + ½ aᵧ t²

(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²

0 = t (18.0 − 4.9 t)

t = 3.67

Next, find Δx when t = 3.67 s.

Δx = v₀ₓ t + ½ aₓ t²

Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²

Δx = 22.7 m

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3 years ago
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