A) No, the equations presented above are the product of the derivation of position and velocity when the acceleration is constant.
The equations change to polynomial function of the second degree for the description of the acceleration when described as a function of time.
B) Yes, when the acceleration is zero it is concluded that the velocity is constant, therefore they could be used to describe the position as a function of the change in velocity.
<h2>Activities illustrates the Nature of a Dialectical Process</h2>
Dialectical method is at center a conversation among two or more characters carrying diverse localities of opinion regarding a topic but hoping to build the fact by reasoned disputes. Dialectic relates a discussion but cleft of biased components such as passionate plea and the recent derogatory knowledge of eloquence.
It can be compared with the dialectic method where one faction of the communication informs the another. Logic is alternatively identified as lesser reasoning, as objected to superior thought or commentary.
A car A house A phone they all can be renewable
Answer:
Explanation:
Tension T in the rope will create torque in solid cylinder ( axle ). If α be angular acceleration
T R = 1/2 M R²α ( M is mass and R is radius of cylinder )
= 1/2 M R² x a / R ( a is linear acceleration )
T = Ma / 2
For downward motion of the bucket
mg - T = m a ( m is mass and a is linear acceleration of bucket downwards )
mg - Ma / 2 = ma
a = mg / ( M /2 + m )
Substituting the values
a = 14.7 x 9.8 / ( 5.8+ 14.7 )
= 7 m / s²
A )
T = Ma / 2
= 5.8 x 7
= 40.6 N
B ) v² = u² + 2 a h
= 2 x 7 x 10.3
v = 12 m /s
C )
v = u + a t
12 = 0 + 7 t
t = 1.7 s
Answer:
V = -RC (dV/dt)
Solving the differential equation,
V(t) = V₀ e⁻ᵏᵗ
where k = RC
Explanation:
V(t) = I(t) × R
The Current through the capacitor is given as the time rate of change of charge on the capacitor.
I(t) = -dQ/dt
But, the charge on a capacitor is given as
Q = CV
(dQ/dt) = (d/dt) (CV)
Since C is constant,
(dQ/dt) = (CdV/dt)
V(t) = I(t) × R
V(t) = -(CdV/dt) × R
V = -RC (dV/dt)
(dV/dt) = -(RC/V)
(dV/V) = -RC dt
∫ (dV/V) = ∫ -RC dt
Let k = RC
∫ (dV/V) = ∫ -k dt
Integrating the the left hand side from V₀ (the initial voltage of the capacitor) to V (the voltage of the resistor at any time) and the right hand side from 0 to t.
In V - In V₀ = -kt
In(V/V₀) = - kt
(V/V₀) = e⁻ᵏᵗ
V = V₀ e⁻ᵏᵗ
V(t) = V₀ e⁻ᵏᵗ
Hope this Helps!!!
