Answer:
a) 0.3 m
b) r = 0.45 m
Explanation:
given,
q₁ = 0.44 n C and q₂ = 11.0 n C
assume the distance be r from q₁ where the electric field is zero.
distance of point from q₂ be equal to 1.8 -r
now,
E₁ = E₂
1.8 = 6 r
r = 0.3 m
<h3>b) zero when one charge is negative.</h3>
let us assume q₁ be negative so, distance from q₁ be r
from charge q₂ the distance of the point be 1.8 +r
now,
E₁ = E₂
1.8 =4 r
r = 0.45 m
Question: A. The state highway patrol radar guns use a frequency of 9.15 GHz. If you're approaching a speed trap driving 30.1 m/s, what frequency shift will your FuzzFoiler 2000 radar detector see?
B. The radar gun measures the frequency of the radar pulse echoing off your car. By what percentage is the measured frequency different from the original frequency? (Enter a positive number for a frequency increase, negative for a decrease. Just enter a number, without a percent sign.)?
Answer:
The frequency change percentage is 9.94%
Explanation:
The frequency shift can be calculated as follows.
= 
=9.95 GHz
So the frequency change seen by the detector is 9.95 - 9.05
% difference
= 9.94%
Satellites are placed into orbit at least 150 km above Earth's surface to be above the atmosphere.
Answer:
u" + 40u' + 49u = 2 sin(t/6)
upp + 40up + 49u = 2 sin(t/6)
Explanation:
Step 1: Data given
mass = 5 kg
L = 20 cm = 0.2 m
F = 10 sin(t/6)N
Fd(t) = - 6 N
u(0) = 0.03 m/s
u(0) = 0
u'(0) = 3 cm/s
Step 2:
ω =kL
k = ω/L = m*g /L = (5*9.8)/0.2 = 245 kg/s²
Since Fd(t) = -γu'(t) we know:
γ =- Fd(t) / u'(t) = 6N/ 0.03 m/s = 200 Ns/m
The initial value problem which describes the motion of the mass is given by
5u" + 200u' + 245u = 10 sin(t/6) u(0) = 0 ; u'(0) = 0.03
This is equivalent to:
u" + 40u' + 49u = 2 sin(t/6) u(0) = 0 ; u'(0) = 0.03
upp + 40up + 49u = 2 sin(t/6)
With u in m and t in s
Answer:
The answers are in the explanation section below
Explanation:
1) The generalization that can be made from the exploration is that as we move away from the positive electrode, the potential energy gets lower. If we move away from the negative electrode, then the potential energy becomes higher.
2) The positive test charge will have the least potential energy when it gets to the negative electrode point.
3) To move one electron 1m in a direction along one of the equal potential lines, there is no energy needed. Zero work will be required for a charge to move on the equipotential line.
4) If lightning strikes a tree 20m away, it would be better to face the tree or have our back facing the tree. This is because the equipotential line will be present at the point where our body stands, this will protect from electric shock.
The pattern to be sketched is attached.
