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2 years ago

7

How does the steepness of the line on a speed graph relate to the objects speed

1 answer:

Likurg_2 [28]2 years ago

3 0

That depends on what is being graphed.

-- On a speed/time graph, the higher the speed, the higher the point on the line.
A constant speed will be a horizontal piece of line.

-- On a distance/time graph, the higher the speed, the steeper the line is sloped.

-- On a temperature/time graph, the line on the graph is not related
to the object's speed at all.

-- etc.

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In which situation will the lowest resistance occur?

insens350 [35]

Answer:

thick wire and cold temperatures

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2 years ago

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PLEASE HELP The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the w

d1i1m1o1n [39]

Answer:

$F = 0.3\ Hz$

Explanation:

Given

See attachment for the graph

Required

Determine the frequency

Frequency (F) is calculated as:

$F = \frac{1}{T}$

Where

T = Time to complete a period

From the attachment, the wave complete a cycle or period in 3 seconds..

So:

$F = \frac{1}{3s}$

$F = 0.333\ Hz$

$F = 0.3\ Hz$ --- Approximated

7 0

2 years ago

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The intensity of an earthquake wave passing through the earth is measured to be 2.5×106 j/(m2⋅s) at a distance of 43 km from the

vampirchik [111]

r₁ = distance of the point from the source = 43 km = 43000 m

I₁ = intensity of earthquake wave at distance "r₁" = 2.5 x 10⁶ W/m²

r₂ = distance of the point from the source = 1.5 km = 1500 m

I₂ = intensity of earthquake wave at distance "r₂" = ?

we know that , for a constant power , the intensity of wave is inversely proportional to the distance from the source .

I α 1/r² where I = intensity of wave , r = distance from source

hence we can write

I₁/I₂ = r₂²/r₁²

inserting the values

(2.5 x 10⁶) /I₂ = (1500/43000)²

I₂ = 2.1 x 10⁹ W/m²

4 0

2 years ago

You are given a parallel plate capacitor that has plates of area 27 cm^2 which are separated by 0.0100 mm of nylon (dielectric c

deff fn [24]

Answer:

Applied voltage should be 13.5396 kV

Explanation:

The charge stored by a capacitor when subjected to a potential difference 'v' across it's plates is given by

$Q=\frac{K\epsilon _{o} A}{d}V$

Solving for V we get

$V=\frac{Qd}{K\epsilon _{o}A}$

thus to store a charge of 0.11mC we solve for V as follows

Applying values we get

$V=\frac{0.11\times 10^{-3}\times 0.01\times 10^{-3}}{27\times 10^{-4}\times 3.4\times 8.85\times 10^{-12}}$

$\therefore V=13.5396kV$

7 0

3 years ago

What measures the amount of energy in a wave ?

Sliva [168]

Frequency measure the energy in a wave.

5 0

2 years ago