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ollegr [7]
4 years ago
9

A student is investigating two types of waves. Which question can the student ask to determine whether a wave is electromagnetic

?
A Can the wave travel through a vacuum?
B
Can the wave travel through a liquid?
C
Can the wave travel through a solid?
D
Can the wave travel through a gas?
Physics
1 answer:
Romashka-Z-Leto [24]4 years ago
6 0

Answer: A

Can the wave travel through a vacuum

Explanation:

Electromagnetic waves can propagate through space, and vacuum carrying electromagnetic radiant energy without the help of any medium for its propagation. But mechanical wave can't propagate through a vacuum without the help of a medium.

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A small but measurable current of 1.2 10-10A exists in a copper wire whose diameter is 2.0 mm. Assume the current is uniform. (a
Rom4ik [11]

Q: A small but measurable current of 1.2×10⁻¹⁰A exists in a copper wire whose diameter is 2.0 mm. Assume the current is uniform. (a) Calculate the current density.

Answer:

3.82×10⁻⁵ A/m

Explanation:

Current density: This can be defined as the amount of charge passing through a conductor, per unit area, per unit time. The S.I unit of current density is A/m²

From the question above, the expression for current density is given as,

τ = I/A............... Equation 1

Where τ = current density of the copper wire, I = current flowing  through the copper wire, A = cross sectional area of the copper wire

But,

A = πd²/4................. Equation 2

Substitute equation 2 into equation 1

τ = 4I/(πd²)............ Equation 3

Given: I = 1.2×10⁻¹⁰ A, d = 2 mm = 2×10⁻³ m, π = 3.14

Substitute into equation 3

τ = 4(1.2×10⁻¹⁰)/[3.14×(2×10⁻³)²]

τ = (4.8×10⁻¹⁰)/(1.256×10⁻⁵)

τ = 3.82×10⁻⁵ A/m

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-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.

-- We know that the y-component of velocity is the derivative of the
y-component of position.

-- We're given the y-component of position as a function of time.

So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.

Now, the position function may look big and ugly in the picture.  But with the
exception of  't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation.  The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.

From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶

First derivative . . . y' (t) = (a₀ - g) t  -  6 (a₀ / 30t₀⁴ ) t⁵  =  (a₀ - g) t  -  (a₀ / 5t₀⁴ ) t⁵

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and there's your acceleration . . . /\ .
That's the one you're supposed to graph.

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t₀  is how long the model rocket engine burns

Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.

The big name in model rocketry is Estes.  Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.


6 0
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