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3 years ago

What are earth's three different tidal patterns called?

Physics

1 answer:

mixer [17]3 years ago

8 0

I think those are it but I can't find the other one but I'll keep looking
2.tidal current
3.tidal force

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A ray of light is traveling from water into a diamond. Upon hitting the diamond, a light ray bends toward the normal. With this

Flura [38]

As the light pass through different media, it experiences a refraction. This is a phenomenon where the speed and direction of light changes. If the ray of light moves towards the normal line, the angle is decreased. So, is the speed of that light.

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4 years ago

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A helpful tool for keeping data organized during an experiment is a ???. *

Margarita [4]

Answer:

data table

Explanation:

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4 0

3 years ago

WHICH ONE IS CORRECT

JulsSmile [24]

The correct answer is “B", second option.

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3 years ago

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A parallel-plate capacitor has plates with an area of 451 cm2 and an air-filled gap between the plates that is 2.51 mm thick. Th

Nostrana [21]

To solve this problem we will apply the concepts related to Energy defined in the capacitors, as well as the capacitance and load. From these three definitions we will build the solution to the problem by defending the energy with the initial conditions, the energy under new conditions and finally the change in the work done to move from one point to the other.

Energy in a capacitor can be defined as

$E = \frac{1}{2}CV^2 = \frac{1}{2}\frac{Q^2}{C}$

Here,

V = Potential difference across the capacitor plates

Q = Charge stored on the capacitor plates

At the same time capacitance can be defined as,

$C = \epsilon_0 (\frac{A}{d})$

Here,

$\epsilon_0 =$ Vacuum permittivity constant

A = Area

d = Distance

Replacing with our values we have that,

$C = (8.85*10^{-12})(\frac{0.0451}{2.51*10^{-3}})$

$C = 1.5901*10^{-10}F$

PART A) Energy stored in the capacitor is

$E = \frac{1}{2} CV^2$

$E = \frac{1}{2} (1.5901*10^{-10})(575)^2$

$E = 2.628*10^{-5}J$

PART B) We know first that everything that the load can be defined as the product between voltage and capacitance, therefore

$Q = CV$

$Q = (1.59*10^{-10})(575)$

$Q = 9.1425*10^{-8}C$

Now if $d = 10.04*10^{-3}m$ we have that the capacitance is

$C = \epsilon_0 (\frac{A}{d})$

$C = (8.85*10^{-12})(\frac{0.0451}{10.04*10^{-3}})$

$C = 3.9754*10^{-11}F$

Then the energy stored is

$E = \frac{1}{2} \frac{Q^2}{C}$

$E = \frac{1}{2} (\frac{(9.1425*10^{-8})^2}{3.9754*10^{-11}})$

$E = 1.051*10^{-4} J$

PART C) The amount of work or energy required to carry out this process is the difference between the energies obtained, therefore

$W = 1.051*10^{-4} J -2.628*10^{-5}J$

$W = 7.882*10^{-5} J$

6 0

3 years ago

A particle with charge 7.76×10^(−8)C is moving in a region where there is a uniform 0.700 T magnetic field in the +x-direction.

kodGreya [7K]

Answer:

The z-component of the force is $\= F_z = 0.00141 \ N$

Explanation:

From the question we are told that

The charge on the particle is $q = 7.76 *0^{-8} \ C$

The magnitude of the magnetic field is $B = 0.700\r i \ T$

The velocity of the particle toward the x-direction is $v_x = -1.68*10^{4}\r i \ m/s$

The velocity of the particle toward the y-direction is

$v_y = -2.61*10^{4}\ \r j \ m/s$

The velocity of the particle toward the z-direction is

$v_y = -5.85*10^{4}\ \r k \ m/s$

Generally the force on this particle is mathematically represented as

$\= F = q (\= v X \= B )$

So we have

$\= F = q ( v_x \r i + v_y \r j + v_z \r k ) \ \ X \ ( \= B i)$

$\= F = q (v_y B(-\r k) + v_z B\r j)$

substituting values

$\= F = (7.7 *10^{-8})([ (-2.61*10^{4}) (0.700)](-\r z) + [(5.58*10^{4}) (0.700)]\r y)$

$\= F= 0.00303\ \r j +0.00141\ \r k$

So the z-component of the force is $\= F_z = 0.00141 \ N$

Note : The cross-multiplication template of unit vectors is shown on the first uploaded image ( From Wikibooks ).

7 0

4 years ago