Displacement equals (Velocity times Time) plus half times the (acceleration times time squared). =. (48 * 4) + 1/2 * (12 *12^2) = 288meters
Answer:
L2 = 1.1994 m
the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m
Explanation:
Given;
Initial length L1 = 1.2m
Initial temperature T1 = 27°C
Final temperature T2 = 0.0°C
Linear expansion coefficient of brass x = 1.9 × 10^-5 /°C
The change i length ∆L;
∆L = L2 - L1
L2 = L1 + ∆L ...........1
∆L = xL1(∆T)
∆L = xL1(T2 - T1) ......2
Substituting the given values into equation 2;
∆L = 1.9 × 10^-5 /°C × 1.2m × (0 - 27)
∆L = 1.9 × 10^-5 /°C × 1.2m × (- 27)
∆L = -6.156 × 10^-4 m
From equation 1;
L2 = L1 + ∆L
Substituting the values;
L2 = 1.2 m + (- 6.156 × 10^-4 m)
L2 = 1.2 m - 6.156 × 10^-4 m
L2 = 1.1993844 m
L2 = 1.1994 m
the length of the pendulum rod when the temperature drops to 0.0°C is 1.1994 m
We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.
it is the point at infinity where it is at a distance from the curve equal to the radius of curvature lying on the normal vector. Sorry no diagram
