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2 years ago

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At room temperature, sound travels at a speed of about 344 m/s in air. You see a distant flash of lightning and hear the thunder

arrive 7.8 seconds later.
How many miles away was the lightning strike? (Assume the light takes essentially no time to reach you.)

1 answer:

beks73 [17]2 years ago

4 0

Answer:

$d=1.67mi$

Explanation:

Assuming the light takes essentially no time to reach you, the distance at which the lightning occurred can be calculated by multiplying the speed of sound by the time it takes to hear the thunder:

$d=v_st\\d=344\frac{m}{s}(7.8s)\\d=2683.2m*\frac{1mi}{1609.34m}=1.67mi$

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Explanation:

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3 years ago

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Suppose you are climbing a hill whose shape is given by the equation z = 2000 − 0.005x2 − 0.01y2, where x, y, and z are measured

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Answer:

(a) Ascend at 0.8 vertical meter/meter

(b) Descend at -0.2·√2 vertical meter/meter

(c) In the (-0.6, -0.8) direction. The path begins at 45° to the horizontal

Explanation:

The given equation of the shape of the hill is z = 2000 - 0.005·x² - 0.01·y²

The current location = (60, 40, 1966)

The direction of the positive x-axis = east

The direction of the positive y-axis = north

(a) Walking due south = Reducing the y-value 40

From the equation, the elevation varies inversely with the motion towards the north

Therefore, walking south increases the elevation, and we ascend

The rate is given by the partial derivative at in the -j direction, which is 0.02

The rate is therefore 40 × 0.02 = 0.8

(b)The unit vector in the northwest direction u = 1/√2·(-1, 1)

∴ The rate = (-0.01(60), -0.02(40))·u = (-0.6, -0.8)·1/√2·(-1, 1) = -0.2·√2

Therefore we descend

(c) The slope is largest in the grad of the function at the point (60, 40) which is given as follows;

d(2000 - 0.005·x² - 0.01·y² )/dx, d(2000 - 0.005·x² - 0.01·y² )/dy = (-0.6, -0.8)

Therefore, the direction is tan⁻¹(-0.8/-0.6) ≈ S 36.87° W

The slope =(√((-0.4)² + (-0.8)²) = 1

Therefore, the angle is 45° to the horizontal.

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2 years ago

Gravitational potential energy is a form of potential energy

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True because it has "falling" ability

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3 years ago

A mass of 0.5 kg hangs motionless from a vertical spring whose length is 1.10 m and whose unstretched length is 0.50 m. Next the

ser-zykov [4K]

Answer:

The maximum length during the motion is $L_{max} = 1.45m$

Explanation:

From the question we are told that

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The unstretched length is $L_{un} = 1.30m$

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The new length of the spring $L_{new} = 1.30 m$

The spring constant k is mathematically represented as

$k = -\frac{F}{y}$

Where F is the force applied $= m * g = 0.5 * 9.8=4.9N$

y is the difference in weight which is $=1.10-0.50=0.6m$

The negative sign is because the displacement of the spring (i.e its extension occurs against the force F)

Now substituting values accordingly

$k = \frac{4.9}{0.6}$

$= 8.17 N/m$

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where D is this the is the displacement

Since Energy is conserved the total elastic potential energy would be

$E_T = initial \ elastic\ potential \ energy + kinetic \ energy$

$E_T = \frac{1}{2} k D_{max}^2 = \frac{1}{2} k D^2 + \frac{1}{2} mv^2$

Substituting value accordingly

$\frac{1}{2} *8.17 *D_{max}^2 =\frac{1}{2} * 8.17*(1.30 - 0.50)^2 + \frac{1}{2} * 0.5 *1.30^2$

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$D_{max} = 0.950m$

So to obtain total length we would add the unstretched length

So we have

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2 years ago

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