"4. Two long, parallel wires each carry the same current I in the same direction (Fig. OQ 30.4) . Is the total magnetic field at
1 answer:
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Answer:
The equation of motion is
Explanation:
Lets calculate
The weight attached to the spring is 24 pounds
Acceleration due to gravity is
Assume x , is spring stretched length is ,4 inches
Converting the length inches into feet
The weight (W=mg) is balanced by restoring force ks at equilibrium position
mg=kx
⇒
The spring constant ,
= 72
If the mass is displaced from its equilibrium position by an amount x, then the differential equation is
Auxiliary equation is,
=
Thus , the solution is
The mass is released from the rest x'(0) = 0
=0
Therefore ,
Since , the mass is released from the rest from 4 inches
inches
feet
feet
Therefore , the equation of motion is
Answer:
5308.34 N/C
Explanation:
Given:
Surface density of each plate (σ) = 47.0 nC/m² =
Separation between the plates (d) = 2.20 cm
We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:
Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,
Now, plug in for 'σ' and
for
and solve for the electric field. This gives,
Therefore, the electric field between the plates has a magnitude of 5308.34 N/C
Answer:
Da=(1/4)Db
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
When s = Da, t = t
When s = Db, t = 2t
Dividing the two equations
Hence, Da=(1/4)Db