Ummm I’m not sure let me do the work
In order to compute the final velocity of the trains, we may apply the principle of conservation of momentum which is:
initial momentum = final momentum
m₁v₁ = m₂v₂
The final mass of the trains will be:
10,000 + 10,000 = 20,000 kg
Substituting the values into the equation:
10,000 * 3 = 20,000 * v
v = 1.5 m/s
The final velocity of the trains will be 1.5 m/s
Answer:
2677.3 N
Explanation:
v₀ = initial speed of the hand = 4.75 m/s
v = final speed of the hand = 0 m/s
m = Total mass of hand and forearm = 1.55 kg
t = time interval for hand to come to rest = 2.75 ms = 0.00275 s
F = Force applied on the leg
Using Impulse-change in momentum equation
F t = m (v - v₀)
F (0.00275) = (1.55) (0 - 4.75)
F = - 2677.3 N
magnitude of force = 2677.3 N
Answer:
The potential difference between the ends of a wire is 60 volts.
Explanation:
It is given that,
Resistance, R = 5 ohms
Charge, q = 720 C
Time, t = 1 min = 60 s
We know that the charge flowing per unit charge is called current in the circuit. It is given by :
I = 12 A
Let V is the potential difference between the ends of a wire. It can be calculated using Ohm's law as :
V = IR
V = 60 Volts
So, the potential difference between the ends of a wire is 60 volts. Hence, this is the required solution.
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.