The atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.
<h3>What is an atomic bomb?</h3>
An atomic bomb is type of nuclear weapon that releases a vast amount of energy upon explosion in the form of fission reaction.
During an explosion of atomic bomb, a cloud of mushroom fire is formed which vaporises anything found within it. The extent of destruction depends on:
Therefore, the atomic bomb explosion, depends on the energy released, W, the elapsed time, t, and the air density which altogether lead to the vaporisation of the contents within the mushroom cloud created.
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Answer:
R = 4Ω
Explanation:
If we have two resistors with resistances R1 and R2 in series the total resistance is R = R1 + R2
If the resistances are in parallel, the total resistance is given by:
1/R = 1/R1 + 1/R2.
First, we have a resistor with R1 = 1.5Ω
This resistor is connected in series with a parallel part (let's find the resistance of this parallel part), in one branch we have two resistors in series with resistances:
R2 = 8Ω and R3 = 4Ω
Because these are in series, the resistance of that branch is:
R = 8Ω + 4Ω = 12Ω
In the other branch, we have a single resistor of R4 = 4Ω
The resistance of the parallel part is:
1/R = 1/12Ω + 1/4Ω = 1/12Ω + 3/12Ω = 4/12Ω = 1/3Ω
1/R = 1/3Ω
R = 3Ω
Then we have a resistor (the first one, R1 = 1.5Ω) in series with a resistor of 3Ω.
Then the total resistance is:
R = 1Ω + 3Ω = 4Ω
Answer:
0.0016 T
Explanation:
Parameters given:
Diameter of wire = 5 mm = 0.005 m
Radius of wire, R = 0.0025 m
Number of turns, N = 200
Current through the wire, I = 0.10A
The magnitude of the magnetic field is given as:
B = (u₀NI) / (2πR)
Where u = magnetic permeability of free space.
B = (1.257 * 10⁻⁶ * 200 * 0.1) / (2 * π * 0.0025)
B = 0.0016 T
The magnitude of the Magnetic field is 0.0016 T.
Answer:
Explanation:
To solve this problem we can use the Gauss' Theorem
Hence, we have:
where QN is the total net charge inside the Gaussian surface, r is the point where we are going to compute E and ε0 is the dielectric permitivity. For each value of r we have to take into account what is the net charge inside the Gaussian surface.
a) r=4.80m (r>R2)
QN=+2.50 μC+2.70 μC = 5.2 μC
b) r=0.70m (R1<r<R2)
QN=+2.50 μC
c) r=0.210 (r<R1)
Inside the spherical shell of radius R1 the net charge is zero. Hence
E=0N/C
- For the calculation of the potential we have
Thus, we compute the potential by using the net charge of the Gaussian surface
d) r=0.210 (r<R1)
Inside the spherical shell the net charge is zero, thus
E=0N/C
e) r=1.40m (R1<r<=R2)
In this case we take the net charge from the first spherical shell
QN=+2.50 μC
f) r=0.70m
QN=+2.50 μC
V=3.164*10^{4}Nm/C
g) r=0.52
QN=0
V=0
h) r=0.2
QN=0
V=0
HOPE THIS HELPS!!