make a sketch showing the direction of the magnetic field from the solenoid. on your sketch, label the induced north magnetic po
1 answer:
A coil of wire with electric current running through it is a solenoid. A solenoid is depicted in the Figure below.
A magnetic field with north and south poles, similar to a bar magnet, is created as current flows through the coil. The magnetic field inside the solenoid has a parallel, linear pattern. A solenoid's magnetic field moves from its north pole to its south pole in one direction only. The north and south poles of a solenoid can be determined using the clock face rule. A solenoid is enclosed by a magnetic field and has north and south magnetic poles similar to a bar magnet.
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The time between the successive crusts will give the time period of the wave.
So, the time period, T of the given wave is 0.2 s.
The frequency of a wave is the reciprocal of its time period. That is,
Frequency, ν =
=
= 5 Hz
Thus, the frequency of the given travelling periodic wave is 5 Hz.
Answer:
B
Explanation:
kinetic energy (KE) is the energy possessed by moving bodies. It can be expressed as:
KE = m
Where: m is the mass of the object, and v its speed.
For example, a stone of mass 2kg was thrown and moves with a speed of 3 m/s. Determine the kinetic energy of the stone.
Thus,
KE = x 2 x
= 9
KE = 9.0 Joules
Assume that the speed of the stone was 4 m/s, then its KE would be:
KE = x 2 x
= 16
KE = 16.0 Joules
Therefore, it can be observed that as speed increases, the kinetic energy increases. Thus option B is appropriate.
Answer:
The location of the shear center o is 0.033 or 33 m
Explanation:
Solution
Recall that,
The moment of inertia of the section is = I = 0.05 * 0.4 ^3 /12 + 0.005 * 0.2 ^3/12
= 30 * 10 ^ ⁻⁶ m⁴
Now,
The first moment of inertia is
Q =ῩA = [ (0.1 -x) + x/2] (0.005 * x)
= 0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²
Thus,
The shear flow is,
q = VQ/I
so,
P = (0.5x * 10 ^⁻³ - 2.5 x * 10⁻³ x²)/ 30 * 10 ^⁻⁶
P = (16.67 x - 83. 33 x²)
The shear force resisted by the shorter web becomes
Vw,₂ = 2∫ = ₀.₁ and ₀ = P (16.67 x - 83. 33 x²) dx = 0.11x
Then,
We take the moment at a point A
∑Mₐ = 0
- ( p * e)- (Vw₂ * 0.3 ) = 0
e = 0.11 p * 0.3/p
which gives us 0.033 m
= 33 m
Therefore the location of the shear center o is 0.033 or 33 m
Note: Kindly find an attached diagram to the question given above as part of the explanation solved with it.
