Answer:
Voltage across the capacitor is 30 V and rate of energy across the capacitor is 0.06 W
Explanation:
As we know that the current in the circuit at given instant of time is
i = 2.0 mA
R = 10 k ohm
now we know by ohm's law



so voltage across the capacitor + voltage across resistor = V


Now we know that

here rate of change in energy of the capacitor is given as



Answer:
500km
Explanation:
Given parameters:
Speed = 200km/hr
Time taken = 2.5hrs
Unknown:
Distance = ?
Solution:
To solve this problem, we use the speed, time and distance equation.
Therefore;
Distance = Speed x time
So;
Distance = 200 x 2.5 = 500km
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f =
we substitute the values
v_f =
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f =
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a =
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s
Answer:
Linear motion is a one-dimensional motion along a straight line, and can therefore be described mathematically using only one spatial dimension.
Explanation: