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3 years ago

A 7.00-kg bowling ball from a 2.00-m shelf. Just before hitting the floor, what will be it’s kinetic energy?

Physics

1 answer:

san4es73 [151]3 years ago

4 0

I'm not sure of the numbers sense I dont have a calculator right now. The formula is 1/2*M*V^2.

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Normal force is exerted _____to the surface of an object.

Elanso [62]

The normal force is always (underline, bold) is always perpendicular to the surface an object is sitting on. If the object is on an inclined plane, then the normal will not be vertical but it will be perpendicular to the angle of the incline.

The diagram below (left) shows a normal force (GH) that is not vertical, but it is perpendicular to the surface. The object on the right is the more usual normal a mass on a table top.

The vertical line on the right is the normal and it points up.

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3 years ago

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Explain the why a constant speed has a slope of 0 on a graph of speed v.time.

Studentka2010 [4]

Because if I do 0*time you get the answer as 0

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3 years ago

3. How does Earth’s rotation affect our view of stars

Elis [28]

Since we ride along with the Earth while it's doing whatever it does,
the Earth's rotation causes our eyes to constantly point in a different
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If we try to keep watching one star, we have to keep changing the
direction of our eyes to keep looking at the same star.

We can't feel the Earth rotating, so our brains say that the star ... and
the sun and the moon too ... is actually moving across the sky.

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3 years ago

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The design speed of a multilane highway is 60 mi/hr. What is the minimum stopping sight distance that should be provided on the

kicyunya [14]

Answer:

Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.

Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.

Explanation:

Part a

When Road is Level

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is this case is 0 as the road is level

Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0)}\\SSD=220.5 +342.86 ft\\SSD=563.36 ft$

So the minimum stopping sight distance is 563.36 ft.

Part b

When Road has a maximum grade of 4%

The stopping sight distance is given as

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}$

Here

SSD is the stopping sight distance which is to be calculated.
u is the speed which is given as 60 mi/hr
t is the perception-reaction time given as 2.5 sec.
a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade

For upgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35+0.04)}\\SSD=220.5 +307.69 ft\\SSD=528.19 ft$

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>

For downgrade of 4%, Substituting values

$SSD=1.47 ut +\frac{u^2}{30 (\frac{a}{g} \pm G)}\\SSD=1.47 \times 60 \times 2.5 +\frac{60^2}{30 \times (0.35-0.04)}\\SSD=220.5 +387.09 ft\\SSD=607.59ft$

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>

As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.

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3 years ago

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Afina-wow [57]

Answer:

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Explanation:

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2 years ago

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