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kondor19780726 [428]
2 years ago
13

SOLVE Complete the final column in the table of nuclear decay by choosing the

Physics
1 answer:
Alecsey [184]2 years ago
8 0

Answer:

Select all systems that are equivalent to this system: {6d+4.5e5d+0.5e=16.5=4

{

6d+4.5e =16.5 5d+0.5e =4

.

A {6d+4.5e45d+4.5e=16.5=4

{6d+4.5e45d+4.5e=16.5=4

D {6d+4.5e6d+0.6e=16.5=4.8

{6d+4.5e6d+0.6e=16.5=4.8

B {30d+22.5e5d+0.5e=82.5=4

{30d+22.5e5d+0.5e=82.5=4

E {12d+9e10d+0.5e=33=8

{12d+9e10d+0.5e=33=8

C {30d+22.5e30d+3e=82.5=24

{30d+22.5e30d+3e=82.5=24

F {6d+4.5e10d+e=16.5=8

Explanation:

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bonufazy [111]

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3 years ago
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Taya2010 [7]

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7 0
3 years ago
Two very large parallel sheets are 5.00 cm apart. sheet a carries a uniform surface charge density of -9.70 μc/m2 , and sheet b,
MAVERICK [17]
Question is missing. Found on internet the complete text of the problem:

"<span>Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of −9.70µC/m2, and sheet B, which is to the right or A, carries a uniform charge density of −11.5 µC/m2. Assume the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point (a) 4.00 cm to the right of sheet A; (b) 4.00 cm to the left of sheet A; (c) 4.00 cm to the right of sheet B."

Solution:

(a) The electric field produced by a uniformly charged sheet at any distance is given by
</span>E= \frac{\sigma}{2 \epsilon _0}
where \sigma is the charge density and \epsilon _0 = 8.85\cdot 10^{-12} F/m is the vacuum permittivity.

First of all, let's compute the fields generated by the two sheets separately. The two densities of charge are \sigma_A = -9.70 \mu C/m^2=-9.70\cdot 10^{-6}C/m^2 and \sigma_B = -11.5 \mu C/m^2 = -11.5\cdot 10^{-6} C/m^2.

Sheet a gives an electric field of
E_A= \frac{\sigma_A}{2\epsilon _0}= \frac{-9.7\cdot 10^{-6} C/m^2}{2\cdot 8.85\cdot 10^{-12} F/m} = -5.48\cdot 10^5 V/m
where the negative sign means the field points towards sheet A, in any point of the space.

The electric field produced by sheet B is given by:
E_B = \frac{\sigma_A}{2\epsilon _0}= \frac{-11.5\cdot 10^{-6} C/m^2} {2\cdot 8.85\cdot 10^{-12} F/m} =-6.50\cdot 10^{5} V/m
and again, the negative sign means that the field at any point of the space points towards sheet B.

The point at which we have to compute the total field is at 4.00 cm right of sheet A. Since the two sheets are 5.00cm far apart, it means that this point is between the two sheets. Therefore, in this point the two fields point into opposite directions. Therefore, the total field is
E=E_1-E_2= -5.48\cdot 10^5 V/m - (-6.50\cdot 10^{5} V/m)=1.02\cdot 10^5 V/m
And the direction is towards sheet B, since it has a field with stronger intensity.

(b) Field at 4.00 cm to the left of sheet A: in this point of the space, the two fields point towards same direction (on the right, towards both sheet A and sheet B). So, the total field is simply the sum of the two fields:
E=E_1+E_2=-11.98\cdot 10^5 V/m
towards right.

(c) Field at 4.00 cm to the right of sheet B. As before, the two fields in this point have same direction (both towards left, pointing towards both sheet A and sheet B). And so, the total field is simply the sum of the two fields:
E=E_1+E_2=11.98 \cdot 10^5 V/m
towards left.
7 0
4 years ago
3 positive test charge lie on the x axis,q1=25microcoloum,is at the origin ,q2=10microcoloumb is at x=2m and q3=20microcloumb is
alexira [117]

Answer:

the link is right here

ok

6 0
3 years ago
what of instead of other forms of energy we use sound energy to power every thing since sound is there every where
Bad White [126]

So is light, but we don't use light either to power anything, even though we do know how to capture them and extract their energy.

The reason we don't use sound or light as sources of energy is because the amount of energy in the sound and light that surrounds us is pitifully small ... in the milliwatts or microwatts, whereas the things we build and operate with oil or electricity need watts or kilowatts to function.

1 watt of sound will drive you out of the room and wake the neighbors, and 1 watt of light ... if it all enters your eye ... can permanently damage your sight.

4 0
3 years ago
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