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3 years ago

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Physics

1 answer:

tatuchka [14]3 years ago

5 0

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A metal ball with a charge of -8 C is brought in contact with a similar metal ball that had a charge of +2 C. How much charge wi

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I think the answer is -6

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2 years ago

Suppose oil spills from a ruptured tanker and spreads in a circular pattern. If the radius of the oil spill increases at a const

vesna_86 [32]

Answer:

84π m²/s or 263.76 m²/s

Explanation:

see the attached file

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3 years ago

A block of mass m witha given inital speed slides down a rough incline with constant speed. If a smaller block that has a mass o

arsen [322]

Answer:

c. Slowly slide down the incline and then stop.

Explanation:

The first block slides down a rough incline since friction forces are strong enough to countereffect the effects of gravity forces and friction force is directly proportional to mass. Hence, a greater mass means higher friction forces. Net acceleration is zero.

The smaller block has a greater mass, it slides down the incline and decelerates before stopping. Net acceleration is negative.

The correct answer is c.

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3 years ago

A tank holds a 2.38-m thick layer of oil that floats on a 1.24-m thick layer of brine. Both liquids are clear and do not intermi

kozerog [31]

Answer:

1.22m

Explanation:

Since

sinθ = refraction-of-the-oil/refraction-of-the-brine =1.27/1.81 = 0.702

θ = $sin^-^1$ (0.702)

Hence

Critical angle = θ = 44.58°

tan(θ) = d/1.24

tan(44.58°) = d/1.24

Hence, 0.98 = d/1.24

The distance d = 0.98 x 1.24 = 1.22m

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4 years ago

A straight, nonconducting plastic wire 8.00 cm long carries a charge density of 100 nC/m distributed uniformly along its length.

Sphinxa [80]

Answer:

(A) 16641 N/C (B) 72000 N/C

Explanation:

We have given the length of the wire = 8 cm =0.08 m

Charge density = 100 nC/m

So charge = 100 ×0.08 = 8 nC $=8\times 10^{-9}C$

(A) Electric field E that produce 6 cm directly above its midpoint will be

$E=\frac{KQ}{X\sqrt{X^2+a^2}}=\frac{9\times 10^9\times 8\times 10^{-9}}{0.06\sqrt{0.06^2+(\frac{0.08}{2})^2}}=16641N/C$

Here X is the distance where we have to find the electric field

(B) Now electric field due to flat ring will be

$E=\frac{KQX}{(X^2+a^2)^\frac{3}{2}}$

Here X is the distance where we have to find the electric field

So $E=\frac{9\times 10^9\times 8\times 10^{-9}}{(0.06^2+0.08^2)^\frac{3}{2}}=72000N/C$

7 0

3 years ago