PE = mgh,
Mass, m = 5.24kg, g ≈ 9.8m/s², h = 1.63m
= 5.24 * 9.8 *1.63 = 83.70376 ≈ 84
PE ≈ 84 J
Answer: find the attached figure for a and b
Explanation:
A) The second figure depict electric field lines and equipotential lines for two equal but opposite charges. The equipotential lines can be drawn by making them perpendicular to the electric field lines. The potential is greatest (most positive) near the positive charge and least (most negative) near the negative charge.
B) The figure attached depicts an isolated point charge Q with its electric field lines in blue and equipotential lines in green. The potential is the same along each equipotential line, meaning that no work is required to move a charge anywhere along one of those lines. Work is needed to move a charge from one equipotential line to another. Equipotential lines are perpendicular to electric field lines in every case.
Please find the attached file for the figure
the person's phenotype would be red hair. phenotype is what you see on the outside, genotype is what the dna says. <3
Answer:
<h3>
2.3125m/s²</h3>
Explanation:
Using the equation of motion v² = u²+2aS
v is the final velocity = 120km/hr
120km/hr = 120 * 1000/1 * 3600 = 33.3m/s
u is the initial velocity = 0m/s
a is the acceleration
S is the distance covered = 240m
On substituting the given parameters
33.3² = 0²+2a(240)
33.3² = 480a
1110 = 480a
a = 1110/480
a = 2.3125m/s²
Hence the minimum constant acceleration that the aircraft require to be airborne after a takeoff run of 240 m is 2.3125m/s²