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Greeley [361]
3 years ago
10

A student examines a 20-meter long rectangular stream channel and takes the following measurements: width of stream = 4 meters,

depth of stream = 2 meters, velocity of stream = 5 meters per second. What is the discharge of the stream at this location.
Physics
1 answer:
Scrat [10]3 years ago
3 0

Answer:

The discharge of the stream at this location is 40 cubic meters per second.

Explanation:

The discharge is the volume flow rate of the water in the stream. For this purpose we can use the following formula:

Discharge = Volume Flow Rate = (Cross-Sectional Area)(Velocity of Stream)

Volume Flow Rate = (Width of Stream)(Depth of Stream)(Velocity of Stream)

Volume Flow Rate = (4 meters)(2 meters)(5 meters per second)

<u>Volume Flow Rate = 40 cubic meters per second</u>

Therefore, the discharge of the stream at this location is found to be <u>40 cubic meters per second</u>

This result shows that 40 cubic meters volume of water passes or discharges through this point in a time of one second. Hence, this is called the volume flow rate or the discharge of the stream.

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a horizontal force of 50N acts on a body of mass 20KG initially at rest and displaces it by striking it along the entire 20m pat
Aneli [31]

Answer:

1000 J

Explanation:

Work = force × distance

W = (50 N) (20 m)

W = 1000 J

3 0
3 years ago
The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.2
sergejj [24]

Answer:

The car strikes the tree with a final speed of 4.165 m/s

The acceleration need to be of -5.19 m/seg2 to avoid collision by 0.5m

Explanation:

First we need to calculate the initial speed V_{0}

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62.5m=V_{0} *4.15s+\frac{1}{2} *-5.25\frac{m}{s^{2} } *(4.15^{2} )\\V_{0}=25.953\frac{m}{s}

Once we have the initial speed, we can isolate the final speed from following equation:

V_{f} =V_{0} +a*t  V_{f}= 4.165 \frac{m}{s}  

Then we can calculate the aceleration where the car stops 0.5 m before striking the tree.

To do that, we replace 62 m in the first formula, as follows:

x=V_{0} *t+\frac{1}{2} *a*(t^{2} )\\62m=25.953\frac{m}{s}*4.15s+\frac{1}{2} *-a\frac{m}{s^{2} } *(4.15^{2} )\\a=-5.19\frac{m}{s^{2} }

3 0
3 years ago
Read 2 more answers
The space station is 4.41 x 10^5 kg and orbits the earth 6.78 x 10^6 m from the center of earth. The mass of earth is 5.97 x 10^
allochka39001 [22]

Answer:

3 820 885 N

Explanation:

Gravitational equation

   F = G  m1 m2 / r^2    

         G = gravitational constant = 6.6713 x 10^-11 m^3/kg-s^2

F = 6.6713 x 10^-11   *   4.41 x 10^5  * 5.97 x 10^24  / ( 6.78x 10^6)^2

 = 3820885 .3 N

6 0
2 years ago
You are standing in a building whose height (40m) you throw a ball downward at a angle of -30 at a speed of (10m/s) acceleration
jeyben [28]

Answer: 3.41 s

Explanation:

Assuming the question is to find the time t the ball is in air, we can use the following equation:

y=y_{o}+V_{o}sin \theta t-\frac{1}{2}gt^{2}

Where:

y=0m is the final height of the ball

y_{o}=40 m is the initial height of the ball

V_{o}=10 m/s is the initial velocity of the ball

t is the time the ball is in air

g=9.8 m/s^{2} is the acceleration due to gravity  

\theta=30\°

Then:

0 m=40 m+(10 m/s)(sin(30\°))t-\frac{1}{2}9.8 m/s^{2}t^{2}

0 m=40 m+5m/s t-4.9 m/s^{2}t^{2}

Multiplying both sides of the equation by -1 and rearranging:

4.9 m/s^{2}t^{2}-5m/s t-40 m=0

At this point we have a quadratic equation of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}

Where:

a=4.9

b=-5

c=-40

Substituting the known values:

t=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(4.9)(-40)}}{2(4.9)}

Solving the equation and choosing the positive result we have:

t=3.41 s  This is the time the ball is in air

5 0
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alekssr [168]

Answer:

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