Answer:
Wind the long piece of thin wire around the uniform glass rod multiple times, find the length of the total diameters using the metre ruler, and divide by the number of times you wound it around the rod.
Explanation:
Since the diameter of one long piece of thin wire is too thin to be measured by a metre ruler, you can wind it multiple times and push it side by side to get a length you can measure.
For example, if you wound it around 20 times and the total length of 20 diameters of the wire side-by-side is 2.0 cm, one winding, which is the diameter would be 2.0cm ÷ 20 = 0.10cm or 1mm.
Answer
given,
Side of copper plate, L = 55 cm
Electric field, E = 82 kN/C
a) Charge density,σ = ?
using expression of charge density
σ = E x ε₀
ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²
now,
σ = 82 x 10³ x 8.85 x 10⁻¹²
σ = 725.7 x 10⁻⁹ C/m²
σ = 725.7 nC/m²
change density on the plates are 725.7 nC/m² and -725.7 nC/m²
b) Total change on each faces
Q = σ A
Q = 725.7 x 10⁻⁹ x 0.55²
Q = 219.52 nC
Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC
Answer:
Volume = 1,015 acre-feet (Approx)
Explanation:
Given:
Rain = 1.7 in
Time = 30 min
Area = 29 km²
Find:
Volume in acre-feet
Computation:
1 km = 1,000 m
1 m = 3.28 feet
1 km² = 247.105 acre
d = 1.7 in = 1.7 / 12 = 0.14167 ft
Area = 29 × 247.105 = 7,166.045 acre
Volume = 7,166.045 acre × 0.14167 ft
Volume = 1,015 acre-feet (Approx)
If all the energy she put into bending the bow is completely
transmitted to the arrow, then the arrow has the 100 joules
of kinetic energy when it leaves the bow.
Kinetic energy = (1/2) (mass) (speed)²
100 J = (1/2) (0.5 kg) (speed²)
Divide each side by 0.25 kg: 100 J / 0.25 kg = speed²
[ joule ] = [ newton-meter ] = kg-m²/sec²
100 kg-m²/sec² / 0.25 kg = speed²
400 m²/sec² = speed²
Take the square root of each side: speed = √400 m/s
20 m/s
(about 44.7 mph)
Answer:

Explanation:
The electrostatic potential energy is given by the following formula

Now, we will apply this formula to both cases:

So, the change in the potential energy is
