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
Answer:
Mass of the other object is 38.45 kg.
Explanation:
Given:
The gravitational force between two objects is,
N
Mass of one object is, 
Distance between the objects is, 
Gravitational constant is, 
Let the mass of the other object be 
Gravitational force is given as:

Plug in the given values and solve for
. This gives,

Therefore, the mass of the other object is 38.45 kg.
Answer:
Explanation:
a. Landing height is
H=1.3m
Velocity of lander relative to the earth is, i.e this is the initial velocity of the spacecraft
u=1.3m/s
Velocity of lander at impact, i.e final velocity is needed
v=?
The acceleration due to gravity is 0.4 times that of the one on earth,
Then, g on earth is approximately 9.81m/s²
Then, g on Mars is
g=0.4×9.81=3.924m/s²
Then using equation of motion for a free fall body
v²=u²+2gH
v²=1.3²+2×3.924×1.3
v²=1.69+10.2024
v²=11.8924
v=√11.8924
v=3.45m/s
The impact velocity of the spacecraft is 3.45m/s
b. For a lunar module, the safe velocity landing is 3m/s
v=3m/s.
Given that the initial velocity is 1.2m/s²
We already know acceleration due to gravity on Mars is g=3.924m/s²
The we need to know the maximum height to have a safe velocity of 3m/s
Then using equation of motion
v²=u²+2gH
3²=1.2²+2×3.924H
9=1.44+7.848H
9-1.44=7.848H
7.56=7.848H
H=7.56/7.848
H=0.963m
The the maximum safe landing height to obtain a final landing velocity of 3m/s is 0.963m
Answer:
-24.76 kJ/mol
Explanation:
given,
mass of solid magnesium burned = 0.1375 g
the temperature increases by(ΔT) 1.126°C
heat capacity of of bomb calorimeter (C_{cal})= 3024 J/°C
heat absorbed by the calorimeter




heat released by the reaction


energy density will be equal to heat released by the reaction divided by the mass of magnesium
Energy density = 
Energy density = -24.76 kJ/mol
heat given off by burning magnesium is equal to -24.76 kJ/mol