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Alona [7]
3 years ago
11

Which of the following correctly shows the chain of energy transfers that create surface currents on the ocean?

Physics
1 answer:
Colt1911 [192]3 years ago
8 0
D.
Solar energy is converted to wind energy which then drive surface currents.
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Which set of terms is in the proper order, from left to right, for labeling the
Rufina [12.5K]
A. Base , salt , water,acid
3 0
2 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
A 24 liter tank contains ideal helium gas at 27 degrees C anda pressure of 22 atm. How many moles of gas are in the tank?
frozen [14]

Answer:

d)21.5 moles

Explanation:

Given that

L = 24 L

T = 27 °C = 300 K

P = 22 atm

We know that ideal gas equation

P V = n R T

P=Pressure ,V= Volume ,n=Moles ,R=Universal gas constant ,T=Temperature

Now by putting the values

22 x 24 = n x 0.08206 x 300

n= 21.447 moles

n= 21.5 moles

Therefore the number of moles will be 21.5 moles.

The answer is "d".

5 0
3 years ago
If the height of the ramp was 1.2 m above the floor, how long would it take for the marble to hit the ground after it left the r
Evgesh-ka [11]
It would take at less 10 minte i guess this the right awnser

8 0
3 years ago
How much work is done if I use 5N of force to move an object 2 meters
sweet-ann [11.9K]
It’s 10 joules. W=FD, W=5•2=10
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3 years ago
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