<span>There are two possible arrangements of q1,q2,and q3 in this problem. They are:
q3, 2 cm gap, q1, 2 cm gap, q2
or
q1, 2/3 cm gap, q3, 4/3 cm gap, q2
We really don't care about the absolute magnitude of q, so the fact that it's 1.00 nano Coulombs is totally irrelevant to this problem. The only thing important is the relative charge and distances between the particles.
The force exerted between two particles is expressed as
F = q1*q2/r^2.
q1,q2 = charges on the particles.
r = distance between the particles.
Depending upon the relative charge (positive or negative) the force may be either attraction, or repulsion. But since the signs of all the charges mentioned are the same, I'll assume that the force will be repulsive.
For the distance between q1 and q3 I'll use the value "r". And since q1 and q2 are 2 cm apart, for the distance between q3 and q2, I'll use the value (2-r). So we have the following equations.
Force between q1 and q3
F = q1*q3/r^2
Force between q2 and q3
F = q2*q3/(2-r)^2
Set the 2 equations equal to each other
q1*q3/r^2 = q2*q3/(2-r)^2
Substitute the known values and solve for r.
q1*q3/r^2 = q2*q3/(2-r)^2
q*q/r^2 = 4q*q/(2-r)^2
q^2/r^2 = 4q^2/(4 - 4r + r^2)
1/r^2 = 4/(4 - 4r + r^2)
(4 - 4r + r^2)/(r^2(4 - 4r + r^2)) = 4/(4 - 4r + r^2)
(4 - 4r + r^2)/(r^2(4 - 4r + r^2)) = 4r^2/(r^2(4 - 4r + r^2))
0 = (4r^2-(4 - 4r + r^2))/(r^2(4 - 4r + r^2))
0 = (4r^2 - 4 + 4r - r^2)/(r^2(4 - 4r + r^2))
0 = (3r^2 - 4 + 4r)/(r^2(r-2)(r-2))
Now let's look at the numerator and denominator of the expression and see where we can get a value of 0. The denominator is allowed to have any value EXCEPT 0 and that will occur at r = 2, or r=0. And nothing else in the denominator will help the expression become 0. But if the numerator is 0, then the expression is 0. So let's see at what values the numerator is 0. Using the quadratic formula with A=3, B = 4 and C=-4, we get zeros at r = -2 and r = 2/3. Both of those values make sense. If r = -2, that means that the charges are arranges q3, q1, q2 with q1 being 2 cm from q1 and 4 cm from q2. And for r = 2/3, that also makes sense with the charges being arranged q1, q3, q2 and q3 is 2/3cm from q1 and 4/3cm from q2. In both cases, q3 is twice as from from q2 as it is from q1.</span>
??? What do you need help with/
A deep zone.........................
Answer:
60 days.
Explanation:
Let the original mass (N₀) = 1 g
Amount remaining (N) = 6.25% of its original mass
= 6.25% × 1
= 6.25/100 × 1
= 0.0625 g
Half life (t½) = 15 days
Time (t) =?
Next, we shall determine the rate of decay. This can be obtained as follow:
Decay constant (K) = 0.693/ half life
K = 0.693 / t½
Half life (t½) = 15 days
Decay constant (K) =?
K = 0.693 / t½
K = 0.693 / 15
K = 0.0462 / day
Finally, we shall determine the time taken for the sample of the isotope to decay to 6.25% of its original mass.
This can be obtained as follow:
Original amount (N₀) = 1 g
Amount remaining (N) = 0.0625 g
Decay constant (K) = 0.0462 / day
Time (t) =?
Log (N₀/N) = Kt/2.3
Log (1/0.0625) = 0.0462 × t / 2.3
Log 16 = 0.0462 × t / 2.3
1.2041 = 0.0462 × t /2.3
Cross multiply
0.0462 × t = 1.2041 × 2.3
Divide both side by 0.0462
t = (1.2041 × 2.3)/0.0462
t = 59.9 ≈ 60 days
Therefore, the time taken for the sample of the isotope to decay to 6.25% of its original mass is 60 days
Answer: Δβ (dB) = -13.1dB
Explanation:
The intensity of sound is inversely proportional to the square of the distance between them.
I ∝ 1/r²
I₁/I₂= r₂²/r₁² .....1
When the listener increases his distance from the source by a factor of 4.49.
Then,
r₂/r₁= 4.49
From equation 1
I₁/I₂ = (4.49)²
I₁/I₂ = 20.16
I₂/I₁ = 1/20.16
The change in sound intensity in dB can be given as
Δβ (dB) = 10 log(I₂/l₁) = 10log(1/20.6) = -13.1dB