In our solar system, most asteroids are found _____.A.In the kuiper beltb.Orbiting close to the sunc.Between mars and jupiterd.B
2 answers:
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Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA = / ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.
(A) We can solve the problem by using Ohm's law, which states:
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know
and 
, therefore we can rearrange Ohm's law to find the current through the device:
(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
where
is the resistivity of the material of the wire
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
where
V is the potential difference across the electrical device
I is the current through the device
R is its resistance
For the heater coil in the problem, we know
(B) The resistance of a conductive wire depends on three factors. In fact, it is given by:
where
L is the length of the wire
A is the cross-sectional area of the wire
Basically, we see that the longer the wire, the larger its resistance; and the larger the section of the wire, the smaller its resistance.
F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
