16, 5 , 3 = 16+5+3= 24 + 3
So at the end put 24 + 3 cm
And put 16 for the lengths
For the value 5 and for the diving thingy 3
Given: distance 1 d₁ = 40 m; distance 2 d₂ = 3.8 m g = -9.8 m/s²
Initial Velocity Vi = 0 Final Velocity of stone 2 is unknown = ?
Total distance dₓ = d₁ - d₂ = 40 m - 3.8 m = 36.2 m
Formula: a = Vf² - Vi²/2d derive for Final Velocity Vf
acceleration is now due to gravity, therefore a = g
Vf = √2gd Vf = √2(9.8 m/s²)(36.2 m)
Vf = 26.64 m/s
Reason: The second stone will still start from rest.
Answer:
Explanation:
Initial separation of plate = d
final separation = 2d
The capacitance of the capacitor will reduce from C to C/2 because
capacitance = ε A / d
d is distance between plates.
As the batteries are disconnected , charge on the capacitor becomes fixed .
Initial charge on the capacitor
= Capacitance x potential difference
Q = C ΔV
Final charge will remain unchanged
Final charge = C ΔV
Final capacitance = C/2
Final potential difference = charge / capacitance
= C ΔV / C/2
= 2 ΔV
Potential difference is doubled after the pates are further separated.
Kinetic energy because the horse is in motion
Answer:
a) X = 17.64 m
b) X = 17.64 + 4∆t^2 + 16.8∆t
c) Velocity = lim(∆t→0)〖∆X/∆t〗 = 16.8 m/s
Explanation:
a) The position at t = 2.10s is:
X = 4t^2
X = 4(2.10)^2
X = 17.64 m
b) The position at t = 2.10 + ∆t s will be:
X = 4(2.10 + ∆t)^2
X = 17.64 + 4∆t^2 + 16.8∆t m
c) ∆X is the difference between position at t = 2.10s and t = 2.10 + ∆t so,
∆X= 4∆t^2 + 16.8∆t
Divide by ∆t on both sides:
∆X/∆t = 4∆t + 16.8
Taking the limit as ∆t approaches to zero we get:
Velocity =lim(∆t→0)〖∆X/∆t〗 = 4(0) + 16.8
Velocity = 16.8 m/s
