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.
Answer:
A dependent variable is a variable that is tested in an experiment. An independent variable is that can be modified. Depending on what you are testing, the dependent variable will change accordingly to the dependent variable.
- I'm reading this back and it doesn't make much sense, if you want me to reword this I can
Answer:
the electroscope separate by the presence of charge carriers
Explanation:
Metal bodies are characterized by having free (mobile) electrons. In the electroscope the plates are in balance; when the external metal ball is touched, a charge is introduced into the device, when the body that touched the ball is separated, an excess charge remains. This charge, being a metal, is distributed over the entire surface, giving a uniform density and an electric force of repulsion is created between the two charged sheets, which tends to separate the sheets. This force is counteracted by the tension component as the sheets are separated at a given angle, the separation reaches the point where
Fe - Tx = 0
Fe = Tx
In summary, the electroscope separate its leaves by the presence of charge carriers
Answer:
The magnitude of the horizontal displacement of the rock is 7.39 m/s.
Explanation:
Given that,
Initial speed = 11.5 m/s
Angle = 50.0
Height = 30.0 m
We need to calculate the horizontal displacement of the rock
Using formula of horizontal component
Put the value into the formula
Hence, The magnitude of the horizontal displacement of the rock is 7.39 m/s.