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Levart [38]
3 years ago
8

Sally travels by car from one city to another. She drives for 26.0 min at 83.0 km/h, 52.0 min at 41.0 km/h, and 45.0 min at 60.0

km/h, and she spends 15.0 min eating lunch and buying gas.
What's the average speed (km/h)?
What's the total distance traveled(km)?
Physics
1 answer:
Anna007 [38]3 years ago
6 0
The average speed is determined by the following formula:

average speed = [sum of (speed * time for which that speed was traveled)] / total time

average speed = [(83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60] / [(26 + 52 + 45 + 15) / 60]
*note: The division by 60 is to convert minutes to hours. We see that the 60 cancels from the top and bottom of the division

average speed = 50.65 km/hr

The total distance traveled is equivalent to the numerator of the fraction we used in the first part. This is:
Distance = (83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60

Distance = 116.5 kilometers
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A hill that has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction. At
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The angle of incline of the hill above the horizontal is 8.81°.

Since the hill has a 15.5% grade is one that rises 15.5 m vertically for every 100.0 m of distance in the horizontal direction.

<h3>Tangent of the angle of the incline of the hill,</h3>

The tangent of the angle of the incline of the hill, Ф is

tanФ = vertical rise/horizontal distance = grade of hill

Now, the vertical rise = 15.5 m and the horizontal distance = 100.0 m

So, substituting the values of the variables into the equation, we have

tanФ = vertical rise/horizontal run

tanФ = 15.5 m/100.0 m

tanФ = 0.155

<h3>Angle of incline of the hill</h3>

Taking inverse tan of both sides, we have

Ф = tan⁻¹(0.155)

Ф = 8.81°

So, the angle of incline of the hill above the horizontal is 8.81°.

Learn more about angle of incline of a hill here:

brainly.com/question/10056962

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A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the
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Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

U=\dfrac{1}{2}kx^2...(I)

Using gravitational potential energy

U' =mgh....(II)

Using energy of conservation

E_{i}=E_{f}

U_{i}+U'_{i}=U_{f}+U'_{f}

\dfrac{1}{2}kx^2+0=0+mgh

h=\dfrac{kx^2}{2mg}

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

h=\dfrac{5365\times(0.097)^2}{2\times0.221\times9.8}

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Hence, The maximum height above the point of release is 11.653 m.

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