(7.5)-(5)
----------- = 2 m/s^2
(1.25-0)
Average acceleration is calculated using the equation V(f)-V(i) / t(f)-t(i). (Final velocity minus initial velocity divided by final time minus initial time)
Answer:
The electrical loads in parallel circuits each have the same voltage drop, with equals the total applied voltage of the circuit.
Explanation:
I did some research and the voltage drop across any branch of a parallel circuit is the same as the applied voltage.
ΔU = Q + W
- heat in, Q +
- heat out, Q -
- does work , W -
- work in, W +
93-58 = 63 + W
35 = 63 + W
W = - 28 J (does work/being done by system)
Define
u = 16 m/s, the vertical launch velocity
g = acceleration due to gravity, measured positive downward
s = vertical distance traveled
t = 21.2 s, total time of travel.
The vertical motion obeys the equation
s = ut - (1/2)gt²
When the rock is at ground level, s = 0.
Therefore
(16 m/s)(21.2 s) - 0.5*(g m/s²)*(21.2 s)² = 0
339.2 - 224.72g = 0
g = 1.5094 m/s²
Answer:
The acceleration due to gravity is 1.509 m/s² measured positive downward.
