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.
Answer:
a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia
c) True. Information is missing to perform the calculation
Explanation:
Let's consider solving this exercise before seeing the final statements.
We use Newton's second law Rotational
τ = I α
T r = I α
T gR = I α
Alf = T R / I (1)
T = α I / R
Now let's use Newton's second law in the mass that descends
W- T = m a
a = (m g -T) / m
The two accelerations need related
a = R α
α = a / R
a = (m g - α I / R) / m
R α = g - α I /m R
α (R + I / mR) = g
α = g / R (1 + I / mR²)
We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant
Let's review the claims
a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia
b) False. Missing data for calculation
c) True. Information is missing to perform the calculation
d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases
When practicing an oral presentation, you can prepare by writing a draft and practice reading aloud what you are going to say before your oral presentation.
Answer:
1.23 m
Explanation:
The vertical distance covered by a free-falling object starting from rest in a time t is
where
g = 9.8 m/s^2 is the acceleration due to gravity
In this problem, we have
t = 0.50 s
So the distance covered is
