Answer:
1.507×10⁻⁴ T
Explanation:
a = 5 cm = 0.05 m
b = 7 cm = 0.07 m
j = 1 A/cm²
Distance from magnetic field = d = 10 cm = 0.1 m
μ₀ = Vacuum permeability = 4π×10⁻⁷ H/m
Magnetic of hollow cylinder
∴ Magnitude of the magnetic field at a distance d = 10 cm from the axis of the cylinder is 1.507×10⁻⁴ T
Answer:
P.E = 980 J
Explanation:
Given,
The mass of the rock, m = 1 Kg
The height of the rock from the ground, h = 100 m
The acceleration due to gravity, a = 9.8 m/s²
The gravitational potential energy of the stone is given by the formula
P.E = mgh joules
Substituting the given values in the above equation,
P.E = 1 x 100 x 9.8
= 980 J
Hence, the gravitational potential energy of the stone is, P.E = 980 J
Earth is the third planet closest to the sun.
We will use the Cosine Law:
m D² = A² + B² - 2 AB cos D
∠D = 65° + 58° = 123°
m D² = 220² + 140² - 2 * 220 * 140 * cos 123°
m D² = 48,000 + 19,600 + 33,854.72
m D = 319.15 km
After that we will use the Sine Law:
340 / sin 123° = 220 / sin (theta)
sin (theta) = 0.578
∠(theta) = sin^(-1)0.578 = 35.3°
Answer:
The magnitude of the vector D is 319.15 km and points 35.3° south to east.
Answer:
173.45 K
Explanation:
This an Adiabatic process because no energy is lost by thermal conduction on expansion. We will be using this Adiabatic condition and ideal gas equation to solve the question.
From idea gas equation;
PV = nRT
-----(1)
where
P is pressure of the gas
V is volume of the gas
n is number of moles
R is gas constant 8.31441 J K-1 mol-1.
T is temperature in Kelvin
For Adiabatic Condition;
----(2)
Substituting equation(1) into equation(2)
Eliminating the constants and simplify with exponent
Making T₂ the subject of the formula;
The temperature when the initial volume has quadrupled ⇒ V₂ = 4V₁
⇒
Since air is diatomic, we assume k = 1.4
∴
T₂ = 173.45 K