Q3. (a) 0m/s, as they are asking for initial velocit.
(b)(i) The paper has a large surface area or weighs less than the coin,thus,falls smoothly.
(ii)The coin has more mass than the paper.
(c) They fall at the same acceleration and hit the bottom at the same time.
Their mass doesn't matter in the vacuum.
Answer:
a = -7.29 m / s²
Explanation:
For this exercise we must use Newton's second law,
F -W = m a
Force is electrical force
F = k q₁ q₂ / r²
k q₁ q₂ / r² -mg = m a
indicate that the charge of the two spheres is equal
q₁ = q₂ = q
a = (k q² / r² - m g) / m
a = k q² / m r² - g
Let's reduce the magnitudes to the SI system
m = 0.19 g (1kg / 1000 g) = 1.9 10⁻⁴ kg
q1 = q2 = q = -23.0 nC (1C / 10⁹ nC) = -23.0 10⁻⁹ C
r = 10.0 cm (1m / 100cm) = 0.1000 m
let's calculate
a = 9 10⁹ (23.0 10⁻⁹)² / (0.1000² 1.9 10⁻⁴) - 9.8
a = -7.29 m / s²
The negative sign indicates that the direction of this acceleration is downward
Answer:
3430000 J
Explanation:
The formula for potential energy is PE=mgh.
M being the mass, g being the force of gravity, and h being the height.
First thing you want to do is convert 250 kg to g (grams).
From there you get 25000g and you have to multiply that by 14m and 9.8m/s^2 (the force of gravity is constant, at least on earth).
Answer:
v = -v₀ / 2
Explanation:
For this exercise let's use kinematics relations.
Let's use the initial conditions to find the acceleration of the electron
v² = v₀² - 2a y
when the initial velocity is vo it reaches just the negative plate so v = 0
a = v₀² / 2y
now they tell us that the initial velocity is half
v’² = v₀’² - 2 a y’
v₀ ’= v₀ / 2
at the point where turn v = 0
0 = v₀² /4 - 2 a y '
v₀² /4 = 2 (v₀² / 2y) y’
y = 4 y'
y ’= y / 4
We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.
v² = v₀² -2a y’
v² = 0 - 2 (v₀² / 2y) y / 4
v² = -v₀² / 4
v = -v₀ / 2
We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.
