Answer:
A ) Find the direct runoff depth; d =V A 22,698 _ 2.25x106 =0.01m =1cm The above result shows that the given hydrograph qualifies as a unit hydrograph. .
Explanation:
Brainlyist
Answer:
Δt = 5.85 s
Explanation:
For this exercise let's use Faraday's Law
emf =
- d fi / dt
= B. A
\phi = B A cos θ
The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.
suppose a linear change of the magnetic field
emf = - A 
Dt = - A 
the final field before a fault is zero
let's calculate
Δt = - 0.046 (0- 1.4) / 0.011
Δt = 5.85 s
The range of potential energies of the wire-field system for different orientations of the circle are -
θ U
0° 375 π x 
90° 0
180° - 375 π x 
We have current carrying wire in a form of a circle placed in a uniform magnetic field.
We have to the range of potential energies of the wire-field system for different orientations of the circle.
<h3>What is the formula to calculate the Magnetic Potential Energy?</h3>
The formula to calculate the magnetic potential energy is -
U = M.B = MB cos 
where -
M is the Dipole Moment.
B is the Magnetic Field Intensity.
According to the question, we have -
U = M.B = MB cos 
We can write M = IA (I is current and A is cross sectional Area)
U = IAB cos 
U = Iπ
B cos 
For
= 0° →
U(Max) = MB cos(0) = MB = Iπ
B = 5 × π ×
× 3 ×
=
375 π x
.
For
= 90° →
U = MB cos (90) = 0
For
= 180° →
U(Min) = MB cos(0) = - MB = - Iπ
B = - 5 × π ×
× 3 ×
=
- 375 π x
.
Hence, the range of potential energies of the wire-field system for different orientations of the circle are -
θ U
0° 375 π x 
90° 0
180° - 375 π x 
To solve more questions on Magnetic potential energy, visit the link below-
brainly.com/question/13708277
#SPJ4
<span>The fluid in a graduated cylinder should be read at the BOTTOM of the meniscus.</span>








<h3>☯ <u>By using formula of Lens</u> </h3>











<h3>☯ <u>Now, Finding the magnification </u></h3>





<h3>☯ <u>Hence</u>,

</h3>

