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kozerog [31]
3 years ago
10

According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f

lash in miles. Assuming that the flash of light arrives in essentially no time at all, estimate the speed of sound in m/s from this rule. Express your answer to one significant figure and include the appropriate units. What would be the rule for kilometers?
Physics
1 answer:
Bond [772]3 years ago
3 0

Answer:

S_{s}=300 m/s

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

Explanation:

In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.

The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.

For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

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An area experiences thunderstorms with high winds and a drop in temperature. Which weather event most likely occurred? A. A stat
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C a fast-moving cold front moved through the area.

Explanation:

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Thus, for the area to experience thunderstorms with high winds and a drop in temperature, <u>a fast-moving cold front moved through the area.</u>

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How many times faster can a rider go if they are pedaling an Ultimate HPV bike with a power of 500 W compared to pedaling a road
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Explanation:

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3 years ago
An object is dropped from a 45 m high building. At the same time, another object is thrown
Dafna11 [192]

Answer:

-92.33 (meaning the objects will not meet above the ground).

Explanation:

We can use the kinematic equation <em>displacement = initial velocity*time + 1/2*acceleration*time^2.</em>

We can plug in the known values of the 2 objects into the equation, where t is the time and x is the displacement:

x = 0*t + 1/2*(-9.8)*t^2+45

x = 8.5*t + 1/2*(-9.8)*t^2

We need to first solve for t to solve for x. Since both equations are equal to x, we can set them equal to each other and solve for t:

0*t + 1/2*(-9.8)*t^2+45 = 8.5*t + 1/2*(-9.8)*t^2

-4.9*t^2 +45 = 8.5*t + -4.9*t^2

45 = 8.5*t

t = 45/8.5 ≈5.294

Now, we can plug t as 5.294 into any of the equations above to solve for x:

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3 years ago
If a spring is stretched 4m from its starting length when 20n of force is applied, then how much work (in joules) is done by the
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250 J

STEP-BY-STEP EXPLANATION:

F = 20N is required to stretch the spring by 4 meters

We know that the force is equal to:

F=k\cdot x

We solve for k (spring constant):

k=\frac{F}{x}=\frac{20}{4}=5\text{ N/m}

The work done in stretching the spring is given by the following equation (in this case the stretch is 10 meters:

\begin{gathered} W=\frac{1}{2}k\cdot x^2 \\ \text{ Replacing} \\ W=\frac{1}{2}\cdot5\cdot10^2 \\ W=250\text{ J} \end{gathered}

The work required is 250 joules.

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1 year ago
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