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DIA [1.3K]
3 years ago
12

LW Hydraulic engineers in the United States often use, as a unit of volume of water, the acre-foot, defined as the volume of wat

er that will cover 1 acre of land to a depth of 1 ft. A severe thunderstorm dumped 2.0 in. of rain in 30 min on a town of area 26 km2. What volume of water, in acre-feet, fell on the town?
Physics
1 answer:
ruslelena [56]3 years ago
4 0

Answer:

volume of water is 1066.50 acre feet

Explanation:

given data

dia = 2 in = 0.166 feet

area = 26 km² = 26 × 247.105 acre = 6424.73 acre

to find out

volume of water

solution

we know volume of water is express as

volume = area × dia     ....................1

put here value in equation 1

volume = 6424.73 acre × 0.166 feet

volume = 1066.50

so volume of water is 1066.50 acre feet

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A bowling ball has a mass of 5 kg what happens to its momentum when its speed increases from 1m/s to 2m/s?
Mama L [17]
Here, Initial momentum = mu = 5*1 = 5 Kg m/s
Final momentum = mv = 5*2 = 10 Kg m/s

So, Momentum has been increased from 5 Kg m/s to 10 Kg m/s. Hence, Your Final answer is option B

Hope this helps!
8 0
3 years ago
A man standing in front of a mountain beats a drum at regular intervals. The rate of drumming is generally increased and he find
Likurg_2 [28]
I think it is d I hope this help you if not let me know if it is not right
3 0
3 years ago
What is the acceleration of a 349 kg object that moved with a force of 750 N?
zavuch27 [327]

Answer:

<h3>The answer is 2.15 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

where

f is the force

m is the mass

From the question we have

a =  \frac{750}{349}  \\  = 2.14899713...

We have the final answer as

<h3>2.15 m/s²</h3>

Hope this helps you

4 0
2 years ago
Determine the magnitude of the effective value of g⃗ at a latitude of 60 ∘ on the earth. assume the earth is a rotating sphere.
dezoksy [38]
In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr) 
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>
5 0
3 years ago
Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff.
givi [52]

Answer:

A. speed = 7.14 Km/s

B. distance = 1820.7 Km

Explanation:

Given that: a = 14.0 m/s^{2}, t = 8.50 minutes.

But,

t = 8.50 = 8.50 x 60

  = 510 seconds

A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;

v = u + at

where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

u = 0

So that,

v = 14 x 510

 = 7140 m/s

The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.

B. the distance traveled can be determined by applying second equation of motion.

s = ut + \frac{1}{2}at^{2}

where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.

u = 0

s = \frac{1}{2}at^{2}

  = \frac{1}{2} x 14 x (510)^{2}

 = 7 x 260100

 = 1820700 m

The distance that the shuttle has traveled during the given time is  1820.7 Km.

5 0
2 years ago
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