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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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4 years ago
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1. When you shine light with a wavelength of 400 nm at 50% intensity, what is observed?
riadik2000 [5.3K]

Answer:

I think it might be A

Explanation:

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6 0
3 years ago
A body of mass 8 kg moves in a (counterclockwise) circular path of radius 10 meters, making one revolution every 10 seconds. You
Sav [38]

Answer:

Centripetal force is equal to 31.55 N

Explanation:

We have given mass of the body m = 8 kg

Radius of the circular path r = 10 m

It is given that it makes 1 revolution in 10 seconds

Distance traveled in 10 seconds is equal to d=2\pi r=2\times 3.14\times 10=62.8m

Velocity is equal to velocity=\frac{distance}{time}=\frac{62.8}{10}=6.28m/sec

We have to find the centripetal force

Centripetal force is equal to F=\frac{mv^2}{r}=\frac{8\times 6.28^2}{10}=31.55N

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8 0
3 years ago
A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi
Mamont248 [21]

Answer:

a=368.97\ m/s^2

Explanation:

Given that,

Initial angular velocity, \omega=0

Acceleration of the wheel, \alpha =7\ rad/s^2

Rotation, \theta=14\ rotation=14\times 2\pi =87.96\ rad

Let t is the time. Using second equation of kinematics can be calculated using time.

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s

Let \omega_f is the final angular velocity and a is the radial component of acceleration.

\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s

Radial component of acceleration,

a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2

So, the required acceleration on the edge of the wheel is 368.97\ m/s^2.

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