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3 years ago

What is the change in air pressure over a given distance called?.

1 answer:

5 0

The change in pressure measured across a given distance called a Pressure Gradient. The pressure gradient creates a net force that is directed from higher to lower pressure and is called the Pressure Gradient Force. ... As air increases in velocity, it is deflected by the Coriolis Force.

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About ________ percent of wais scores fall between 70 and 130.

stiv31 [10]

It is about 95%. The Wechsler Adult Intelligence Scale-IV is an IQ test that is given by therapists and measures worldwide scholarly working. It incorporates both verbal and nonverbal parts. The normal score for all tests and subtests is 100; therefore, a score of more than 100 is better than expected and beneath 100 is underneath normal.

5 0

3 years ago

A forklift pushes a box with a force of 500 N across the floor for a distance of 5.0 m, then turns around and pushes with the sa

yanalaym [24]

Answer:

5,000J

Explanation:

Work = Force x Distance

Distance back and forth is canceled out, so either the answer is + or -

5.0m + 5.0m = 10.0m

500N x 10.0m = 5,000J

4 0

3 years ago

Natalie accelerate her skateboard along a straight path from 0 m/s to 4.0 m/s in 2.5 s. find her average acceleration

Vikentia [17]

Acceleration=(speed end - speed start)/ time
Data:
speed end=4 m/s
speed start=0 m/s
time=2.5 s

acceleration=(4 m/s - 0 m/s)/2.5 s=1.6 m/s²

Answer: the acceleration would be 1.6 m/s²

8 0

3 years ago

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

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

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

For downgrade of 4%, Substituting values

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

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

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

3 0

3 years ago

Calculate the gravitational potential energy of a 1200 kg car at the top of a hill that is 42 m high.

GalinKa [24]

It IS PE = (1200 kg)(9.8 m/s²)(42 m) = 493,920 J

6 0

3 years ago