When a candle is burning the candle is releasing thermal and radiant energy
S = u + at u = 0 278 = 3.7t t = 278/3.7 = 75.135.. v = ut + 0.5at^2 u = 0 v = 0.5 * 3.7 * 75.135^2 = 10,443 m/sec
<u>Hydroelectric power,</u> also called hydropower is the electricity produced from generators driven by turbines that convert the potential energy of falling or fast-flowing water into mechanical energy.
Answer: 9.9%
Explanation: efficiency = (work output /work input) × 100
Note that, 1 kilocalorie = 4184 joules, hence 22kcal = 22× 4184 = 92048 joules.
Work output = 9200 j and work input = 92048 j
Efficiency = (9200/92048) × 100 = 0.099 × 100 = 9.9%
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
