The salesman is telling you the average magnitude of the car's acceleration.

| Acceleration | = (change in speed) / (time for the change)

| Acceleration | = (60 mi/hr) / (6 sec)

| Acceleration | = 10 miles/hr-sec

That would be 36,000 miles per hour squared,
or 0.0028 mile per second squared.

5 0

3 years ago

What are the characteristics of an index fossil?

vichka [17]

I think as a mold. when the Flood came in Genesis, i believe that when the fish were washed away, the kinda made a mold in a rock.

4 0

2 years ago

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An airplane takes off from Dallas Texas to fly to new york city traveling ne for 2,760 km the Same plane returns that day to Dal

Kipish [7]

<h3>Answer</h3>

1104 km/hour

<h3>Explanation</h3>

Distance between Dallas Texas to New York = 2760 km

Time the plane took from Dallas to New York = 2 hours

Time the plane took from New York back to Dallas = 2.5 hours

Formula to use

<h3>distance = speed x time </h3>

Speed the plane took from Dallas to New York

2760 = 2 x speed

speed = 2760 / 2

= 1380 km/hour

Speed the plane took from New York to Dallas (ROUND TRIP)

2760 = 2.5 x speed

speed = 2760 / 2.5

= 1104 km/hour

3 0

2 years ago

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A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

6 0

2 years ago

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