Answer:
![1.08 * 10^{14} J](https://tex.z-dn.net/?f=1.08%20%2A%2010%5E%7B14%7D%20J)
Explanation:
Energy and mass are related by the famous equation developed by Albert Einstein:
![E = mc^2](https://tex.z-dn.net/?f=E%20%3D%20mc%5E2)
where m = mass and c = speed of light
This equation explains that an object with very small mass can produce a large amount of energy in reactions such as a nuclear reaction.
Hence, the energy produced by the explosion of a Plutonium bomb containing 3.6 grams of matter is:
![E = 3.6 * 10^{-3} * (3 * 10^8)^2](https://tex.z-dn.net/?f=E%20%3D%203.6%20%2A%2010%5E%7B-3%7D%20%2A%20%283%20%2A%2010%5E8%29%5E2)
E = ![1.08 * 10^{14} J](https://tex.z-dn.net/?f=1.08%20%2A%2010%5E%7B14%7D%20J)
Answer:
a )1.851
b ) 32.68°
Explanation:
The relation between critical angle and index of refraction is as follows
Sin C = 1 / μ
C is critical angle and μ is refractive index.
Given C = 47.3
Sin 47.3 = 1 / μ
0.735 = 1 / μ
μ = 1 / .735
= 1.36
This is with respect to ethanol .
1.36 = refractive index of material / refractive index of ethanol
refractive index of material = 1.36 x refractive index of ethanol
= 1.36 x 1.361 = 1.851
If critical angle for this substance be A
Sin A = 1 / 1.851 = .54
A = 32.68° .
Answer:
Explanation:
Given that,
Current in wire are 1.3A and 3.15A
Distance between wire is d= 2.25cm
d = 2.25/100 = 0.025m
Force per unit length F/l?
Let us consider the field produced by wire 1 and the force it exerts on wire 2 (call the force F2).
The field due to I1 at a distance r is given to be
B1 = μo• I1 / 2πr
This field is uniform along wire 2 and perpendicular to it, and so the force F2 it exerts on wire 2 is given by
F=ILBsinθ
with sinθ=1:
F2=I2 • L •B1
By Newton’s third law, the forces on the wires are equal in magnitude, and so we just write F for the magnitude of F2. (Note that F1=−F2.) Since the wires are very long, it is convenient to think in terms of F/l, the force per unit length. Substituting the expression for B1 into the last equation and rearranging terms gives
F/l = μo• I1• I2 / 2πr
Where μo is constant
μo = 4π×10^7 Tm/A
Then,
F/l = μo• I1• I2 / 2πr
F/l = 4π ×10^-7 × 1.3×3.15/(2π×0.025)
F/l = 3.276×10^-5 N/m
the magnitude of the force per unit length that one wire exerts on the other is 3.276×10^-5 N/m
Answer:
Force exerted = 48.89 N
Explanation:
Force = Mass x Acceleration
Mass = 44 kg
Acceleration is rate of change of velocity.
Acceleration, ![a=\frac{2-7}{4.5}=-1.11m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B2-7%7D%7B4.5%7D%3D-1.11m%2Fs%5E2)
Force = Mass x Acceleration = 44 x -1.11 = -48.89 N
Force exerted = 48.89 N