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:
F=m(11.8m/s²)
For example, if m=10,000kg, F=118,000N.
Explanation:
There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:
Solving for the force F, we obtain that:
Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:
From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:
In this case, the force from the rocket's thrusters is equal to 118,000N.
Answer:
what language is that
Explanation:
i don't understand the languge u used please can you change it
Answer:
9.36*10^11 m
Explanation
Orbital velocity v=√{(G*M)/R},
G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,
M = mass of the star
R =distance from the planet to the star.
v=ωR, with ω as the angular velocity and R the radius
ωR=√{(G*M)/R},
ω=2π/T,
T = orbital period of the planet
To get R we write the formula by making R the subject of the equation
(2π/T)*R=√{(G*M)/R}
{(2π/T)*R}²=[√{(G*M)/R}]²,
(4π²/T²)*R²=(G*M)/R,
(4π²/T²)*R³=G*M,
R³=(G*M*T²)/4π²,
R=∛{(G*M*T²)/4π²},
Substitute values
R=9.36*10^11 m
