To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

Here,
q = Charge
t = Time
R = Resistance
C = Capacitance
When the charge reach its half value it has passed 10ms, then the equation is,




We know that RC is equal to the time constant, then

Therefore the time constant for the process is about 14ms
Looks like you need to review through the lesson and take notes as it tells you in the lesson what each of these are.
Unlike a longitudinal wave, a transverse wave moves about, perpendicular to the direction of propagation. The particles in a transverse wave do not travel along the direction of propagation, but only oscillate up and down on its equilibrium position. With this, the displacement can be determined by measuring (in the case of electronic waves, using an oscilloscope or spectrum analyzer) and setting the desired units to measure the wave in.
Answer:
8 time increase in K.E.
Explanation:
Consider Mass of truck = m kg and speed = v m/s then
K.E. = 1/2 ×mv²
If mass and speed both are doubled i.e let m₀ = 2m and v₀ = 2v then
(K.E.)₀ = 1/2 ×2m(2v)²
(K.E.)₀ = 8 (1/2 × mv²) = 8 × K.E.
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