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erik [133]
4 years ago
11

Continents affect the directions of currents true or false

Physics
1 answer:
sladkih [1.3K]4 years ago
7 0
True is the correct answer
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You are designing a new piece of fishing equipment that will allow you to catch a fish from the surface. You know that water ref
OleMash [197]

Answer:

Law of refraction

Explanation:

An experiment to analyze the refraction of light in water can easily be performed with a laser pointer and protractor.

We throw the fishing rod line into the water, place the protractor at the point where the line touches the water and use the direction of the line for the direction of the laser pointer (on), the laser is visible by the reflection on the particles in the air.

The vertical line is called Normal and all angles must be measured with respect to this reference in optics.

Having these angles and the refractive index of water we can use the law of refraction

         n₁ sin θ₁  = n₂ sin θ₂

         θ₂ = sin^{-1} ( \frac{n_1}{n_2} \ sin \ \theta_1 )

we can repeat several times to analyze several different input points (different angles) and to decrease the errors in the measurements.

the refractive index of air is n1 = 1 and n2= 1.33  (water)

4 0
3 years ago
Thank you in advance
Musya8 [376]

m1 =   \frac{- v2m2 - m2v1}{v2 - v1}

8 0
3 years ago
If there is 8 g of a substance before a physical change , how much will there be afterwards?
Dimas [21]
A substance undergoing a physical change will still weigh the same even after the change. This is in accordance to the law of conservation of mass which states that mass is neither created nor destroyed. so an 8 g substance remains of the same weight even after undergoing a physical change.
8 0
4 years ago
Read 2 more answers
PLEASE HELP WITH THIS!
jok3333 [9.3K]

Answer: A

Explanation:

Parallel circuit. The source current is three times the current of a single bulb

7 0
3 years ago
A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc
marta [7]
This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so: 

v_{d} =  \frac{J}{n\left | q \right |}

<span>The current I is any motion of charge from one region to another, so this is given by:

</span>I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)

The magnitude of the current density is:

J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})

Being:

A=2mm^{2} = 2x10^{-6}m^{2}
<span>
Finally, for the drift velocity magnitude vd, we find:

</span>v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
6 0
3 years ago
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