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3 years ago

What are two processes that transfer water into the atmosphere

1 answer:

4 0

Water enters the atmosphere through evaporation, transpiration, excretion and sublimation: Transpiration is the loss of water from plant

Hope this helped you! :D

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What is 178.57 rounded to the nearest whole number

postnew [5]

<h2><em>THANK</em><em> </em><em>YOU</em><em> </em></h2>

4 0

3 years ago

Read 2 more answers

A pail in a water well is hoisted by means of a frictionless winch, which consists of a spool and a hand crank. When Jill turns

Katena32 [7]

Answer:

166 W

Explanation:

Power is the rate at which work is done.

$\text{Power} = \dfrac{\text{Work done}}{\text{time}}$

The work done by Jill is the product of the weight of the pail and the height it moves.

The weight is the product of the mass and acceleration of gravity, <em>g</em>. Taking <em>g</em> as 9.81 m/s², the weight is

<em>W</em> = (6.90 kg)(9.81 m/s²) = 67.689 N

Work done = (67.689 N)(27.0 m) = 1827.603 J

Power = (1827.603 J) ÷ (11.0 s) = 166 W

4 0

3 years ago

Calculate the acceleration of a car if it's velocity increases from 15 m/s to 75m/s in 5 seconds.

Andreas93 [3]

initial velocity=u=15m/s
Final velocity=v=75m/s
Time=t=5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{75-15}{5}$

$\\ \sf\longmapsto Acceleration=\dfrac{60}{5}$

$\\ \sf\longmapsto Acceleration=12m/s^2$

7 0

3 years ago

Which organ system do you think is the most interesting? Why?

klasskru [66]

The reproductive system because without that we would not exist , but the same thing could be said about any other systems but the reproductive system is the reason why were here today .

3 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

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

<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.

3 0

3 years ago