Answer:
Explanation:
Given that,
Current in loops are
i1 = 12A
i2 = 20A
The loops are 3.4cm apart
The magnetic field at the center is found to be zero, so when want to find the radius of bigger loop
Magnetic Field is given as
B= μoi/2πr
Where,
μo is a constant = 4π×10^-7 Tm/A
r is the distance between the two wires
i is the current in the wires
B is the magnetic field
NOTE
Field due to large loop should be equal to the smaller loop.
B1 = B2
μo•i1 / 2π•r1 = μo•i2 / 2π•r2
Then, μo, 2π cancels out, so we have
i1 / r1 = i2 / r2
Make r2 subject of formula
i1•r2 = i2•r1
r2 = i2•r1 / i2
r2 = 20×3.4/12
r2 = 5.67cm
The radius of the bigger loop is 5.67cm.
Atmosphere
Atmospheric gas from prehistoric eras is found trapped in glaciers in the form of bubbles. These gas bubbles are the basis of studying ice cores as they provide us with accurate estimates of the conditions of past climates. The bubbles allow us to determine the composition of atmospheric air, such as the carbon dioxide and methane concentrations, as well as allow us to determine air temperatures in the past.
Answer: Take your pick
Explanation:
if they are all in parallel 1 /(1/100 + 1/300 + 1/50) = 30 Ω
if 50 is in parallel with 2 in series 1 / (1/(100 + 300) + 1/50) = 44.444...Ω
if 100 is in parallel with 2 in series 1 / (1/(50 + 300) + 1/100) = 77.777...Ω
if 300 is in parallel with 2 in series 1 / (1/(100 + 50) + 1/300) = 100 Ω
If 50 is in series with 2 in parallel 50 + 1/(1/100 + 1/300) = 125 Ω
If 100 is in series with 2 in parallel 100 + 1/(1/50 + 1/300) = 142.857...Ω
If 300 is in series with 2 in parallel 300 + 1/(1/50 + 1/100) = 333.333...Ω
If they are all in series 100 + 300 + 50 = 450 Ω
Any ride that oscillates back and forth or moves only in a complete circle utilizes periodic motion.
<h2>
The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = 3 s
Substituting
s = ut + 0.5 at²
s = 0 x 3 + 0.5 x 9.81 x 3²
s = 44.145 m
The seagull's approximate height above the ground at the time the clam was dropped is 4 m
