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3 years ago

What is a situation when you might travel at a high velocity, but with low acceleration?

2 answers:

7 0

Rocket in space, once it gets into space and we're to just orbit the earth you're traveling at a high velocity because velocity is change in direction or speed, and you're constantly changing in direction relative to earth

Alexandra [31]3 years ago

5 0

Cruising at 35,000 feet in the air, straight toward the east at 500 miles per hour. Since your speed is not changing and your moving in a straight line, your acceleration is 0.

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How does an airbag help protect a passenger in a car following a car accident?

Roman55 [17]

Answer: A

Explanation:

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3 years ago

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An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th

kvv77 [185]

Answer:

The plane would need to travel at least $8,\!580\; {\rm ft}$ ( $8.58 \times 10^{3}\; {\rm ft}$ .)

The $10,\!000\; {\rm ft}$ runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to $\rm ft\cdot s^{-1}$ :

$\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}$ .

Initial velocity of the plane: $u = 0\; {\rm ft \cdot s^{-1}}$ .

Take-off velocity of the plane $v =264\; {\rm ft\cdot s^{-1}}$ .

Let $x$ denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation $x = ((u + v) / 2) \, t$ .

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration $t$ times average velocity $(u + v) / 2$ .

The distance that the plane need to cover would be:

$\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}$ .

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3 years ago

An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine

katrin [286]

Answer:

If efficiency is .22 then W = .22 * Q where Q is the heat input

Heat Input Q = 2510 / .22 = 11,400 J

Heat rejected = 11.400 - 2510 = 8900 J of heat wasted

Also, 8900 J / (4.19 J / cal) = 2120 cal

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3 years ago

What kind of image is formed by a convex lens when the object is less than a focal length away from the object?

maw [93]

Bigger image and can get clearer

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3 years ago

A spaceship hovering over the surface of Venus drops an object from a height of 17 m. How much longer does it take to reach the

Paraphin [41]

1.96s and 1.86s. The time it takes to a spaceship hovering the surface of Venus to drop an object from a height of 17m is 1.96s, and the time it takes to the same spaceship hovering the surface of the Earth to drop and object from the same height is 1.86s.

In order to solve this problem, we are going to use the motion equation to calculate the time of flight of an object on Venus surface and the Earth. There is an equation of motion that relates the height as follow:

$h=v_{0} t+\frac{gt^{2}}{2}$

The initial velocity of the object before the dropping is 0, so we can reduce the equation to:

$h=\frac{gt^{2}}{2}$

We know the height h of the spaceship hovering, and the gravity of Venus is $g=8.87\frac{m}{s^{2}}$ . Substituting this values in the equation $h=\frac{gt^{2}}{2}$ :

$17m=\frac{8.87\frac{m}{s^{2} } t^{2}}{2}$

To calculate the time it takes to an object to reach the surface of Venus dropped by a spaceship hovering from a height of 17m, we have to clear t from the equation above, resulting:

$t=\sqrt{\frac{2(17m)}{8.87\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{8.87\frac{m}{s^{2} } } }=1.96s$

Similarly, to calculate the time it takes to an object to reach the surface of the Earth dropped by a spaceship hovering from a height of 17m, and the gravity of the Earth $g=9.81\frac{m}{s^{2}}$ .

$t=\sqrt{\frac{2(17m)}{9.81\frac{m}{s^{2} } }} =\sqrt{\frac{34m}{9.81\frac{m}{s^{2} } } }=1.86s$

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3 years ago

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