How are stars distributed throughout an elliptical galaxy?
1 answer:
You might be interested in
Answer:
b) 5 J
Explanation:
Work is the energy transferred by an object when acted by a force along a displacement. Work is the product of force and displacement. The SI unit of work is the joules (J)
To calculate the work done by the force, we have to first get the displacement (D) of the object. Hence:
Displacement (D) = Q(3, 8) - P(1, 3) = (3 - 1, 8 - 3) = (2, 5) = 2i + 5j
The work done is the dot product of the force and the displacement. Force = 5i - j. Hence:
Work done = (5i - j)(2i + 5j) = 10 - 5 = 5 J
To solve this problem we will apply the concepts related to volume, as a function of length and area, as of mass and density. Later we will take the same concept of resistance and resistivity, equal to the length per unit area. Once obtained from the known constants it will be possible to obtain the area by matching the two equations:
Mass of copper wire
Density
Resistively of copper
Resistance (R) = 0.390\Omega
Volume is defined as,
(1)
We know that,
(2)
Multiplying equation we have
Therefore the length of the wire is 1.68m
Answer:
Capacitance is 0.572×10⁻¹⁰ Farad
Explanation:
Radius = R₁ = 6.25 cm = 6.25×10⁻² m
Radius = R₂ = 15 cm = 15×10⁻² m
Dielectric constant = k = 4.8
Electric constant = ε₀ = 8.854×10⁻¹² F/m
ε/ε₀=k
ε=kε₀
∴ Capacitance is 0.572×10⁻¹⁰ Farad