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HACTEHA [7]
3 years ago
13

What is the result of convection currents in the atmosphere?

Physics
1 answer:
Degger [83]3 years ago
4 0
Wind is the result of convection currents. :)
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A total resistance of 3.03 Ω is to be produced by connecting an unknown resistance to a 12.18 Ω resistance. (a) What must be the
insens350 [35]

Answer:

(a) 4.0334Ω

(b)parallel

Explanation:

for resistors connected in parallel;

\frac{1}{R_{eq} } =\frac{1}{R1}+\frac{1}{R2}

Req =3.03Ω , R1 =12.18Ω

\frac{1}{3.03 } =\frac{1}{12.18}+\frac{1}{R2}

\frac{1}{R2}=\frac{1}{3.03 }-\frac{1}{12.18}

\frac{1}{R2}=0.2479

R2=1/0.2479

R2=4.0334Ω

(b)parallel connection is suitable for the desired total resistance. series connection can not be used to achieve a lower resistance as the equation for series connection is.

Req = R1+R2

3 0
3 years ago
If a cup of coffee is at 90°C and a person with a body temperature of 36'C touches it,how will heat flow between them
Arlecino [84]

Answer:

the heat always transfers from high temperature to low temperature body without aid of any external energy to this law the heat transfers from cup of coffee to the person body until both bodies reaches to the equilibrium state    

Explanation:

5 0
3 years ago
Is there air resistance in space?........HELP!!!!!
Leokris [45]
There's no air in space, so there's no air resistance there.
5 0
3 years ago
Read 2 more answers
If opposite poles repel each other, why does the north end of a compass point to the north pole?
lina2011 [118]
Ratios are fractions
6 0
3 years ago
A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius
ahrayia [7]

Answer:

The specific question is not stated, however the general idea is given in the attached picture. The electric field in each region can be found by Gauss’ Law.

at r < R:

Since the solid sphere is conducting, the total charge Q is distributed over the surface, and the electric field inside the sphere is zero.

E = 0.

at R < r < 2R:

The electric field can be found by Gauss’ Law as in the attachment. The green pencil shows this exact region.

at 2R < r:

The electric field can again be found by Gauss’ Law, the blue pencil shows the calculations for this region.

Explanation:

Gauss’ Law is straightforward when applied to spheres. The area of the sphere is A = 4\pi r^2, and the enclosed charge is given in the question as Q for the inner sphere, and 2Q for the whole system.

3 0
3 years ago
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