Is it raining anywhere else because it is about to start pouring where I live
2 answers:
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Answer:
p = 8N/mm2
Explanation:
given data ;
diameter of cylinder = 150 mm
thickness of cylinder = 6 mm
maximum shear stress = 25 MPa
we know that
hoop stress is given as =
axial stress is given as =
maximum shear stress = (hoop stress - axial stress)/2
putting both stress value to get required pressure
t = 6 mm
d = 150 mm
therefore we have pressure
p = 8N/mm2
Answer:
The maximum induced emf in the rotating coil = 29.66V
The induced emf in the rotating coil when (t = 1.00 s) = 26.66V
The maximum rate of change of the magnetic flux through the rotating coil = 0.674Wb/s
Explanation:
Lets state the parameters we are being given right from the question:
Number of rectangular coil, (N) = 44
Length of Coil, l =17cm in meters we have; (l) = 17 × 10⁻² m
Width of Coil, w =8.10cm in meters we have; (w) = 8.10 × 10⁻² m
Magnitude of Uniform Magnetic Field (B) = 767mT= 765 × 10⁻³ T
Angular Speed of Coil, (ω) = 64 rad/s
(a)
To calculate the induced emf in the rotating cell,we can use the formula:
emf = NBAωsin(ωt)
For maximum induced emf, the value of sin(ωt) will be 1
= NBAω ; if (A = l × w) , we have:
= NB(l × w)ω
subsitituting the parameters into the above equation; we have:
= 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64
= 29.66V
(b)
At t = 1s, the induced emf is calculated as:
emf = NBAωsin(ωt)
substituting the parameters into the equation, we have:
emf = 44 × 765 × 10⁻³ ( 17 × 10⁻² × 8.10 × 10⁻² ) × 64 × sin (64 × 1)
=26.66V
(c)
To calculate the maximum rate of change of the magnetic flux through the rotating coil; we need to reflect on the equation for the maximum induced emf in terms of magnetic flux.
i.e =
since = 29.66 and N = 44; we have:
29.66 =
=
= 0.674 Wb/s
Answer:
(1) passed through the foil
Explanation:
Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.
- When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.
- While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive.
- And some were acutely deflected due to the field effect of the positive charge of the proton inside the nucleus. All these conclusions were made based upon the spot of glow on the fluorescent screen.
