Answer:
The options are not shown, so let's derive the relationship.
For an object that is at a height H above the ground, and is not moving, the potential energy will be:
U = m*g*H
where m is the mass of the object, and g is the gravitational acceleration.
Now, the kinetic energy of an object can be written as:
K = (1/2)*m*v^2
where v is the velocity.
Now, when we drop the object, the potential energy begins to transform into kinetic energy, and by the conservation of the energy, by the moment that H is equal to zero (So the potential energy is zero) all the initial potential energy must now be converted into kinetic energy.
Uinitial = Kfinal.
m*g*H = (1/2)*m*v^2
v^2 = 2*g*H
v = √(2*g*H)
So we expressed the final velocity (the velocity at which the object impacts the ground) in terms of the height, H.
Answer:
A. -2.16 * 10^(-5) N
B. 9 * 10^(-7) N
Explanation:
Parameters given:
Distance between their centres, r = 0.3 m
Charge in first sphere, Q1 = 12 * 10^(-9) C
Charge in second sphere, Q2 = -18 * 10^(-9) C
A. Electrostatic force exerted on one sphere by the other is:
F = (k * Q1 * Q2) / r²
F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²
F = -2.16 * 10^(-5) N
B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:
Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))
= - 6 * 10^(-9) C
Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C
Hence the electrostatic force between them is:
F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²
F = 9 * 10^(-7) N
A. True
Hope this helps :)
Mars.
Water exists as small amounts of ice on Mars and as water vapor. It is suspected that Mars used to have flowing water on it, but that there is none left now.
Lets se
And
So
If spring constant is doubled mass must be doubled
