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Most electrical appliances are rated in watts. Does this rating depend on how long the appliance is on? (When off, it is a zero-watt device.) Explain in terms of the definition of power.
Explain, in terms of the definition of power, why energy consumption is sometimes listed in kilowatt-hours rather than joules. What is the relationship between these two energy units?
A spark of static electricity, such as that you might receive from a doorknob on a cold dry day, may carry a few hundred watts of power. Explain why you are not injured by such a spark.
The Crab Nebula (see [link] ) pulsar is the remnant of a supernova that occurred in A.D. 1054. Using data from [link] , calculate the approximate factor by which the power output of this astronomical object has declined since its explosion.
$2\times {\text{10}}^{-\text{10}}$
Suppose a star 1000 times brighter than our Sun (that is, emitting 1000 times the power) suddenly goes supernova. Using data from [link] : (a) By what factor does its power output increase? (b) How many times brighter than our entire Milky Way galaxy is the supernova? (c) Based on your answers, discuss whether it should be possible to observe supernovas in distant galaxies. Note that there are on the order of ${\text{10}}^{\text{11}}$ observable galaxies, the average brightness of which is somewhat less than our own galaxy.
A person in good physical condition can put out 100 W of useful power for several hours at a stretch, perhaps by pedaling a mechanism that drives an electric generator. Neglecting any problems of generator efficiency and practical considerations such as resting time: (a) How many people would it take to run a 4.00-kW electric clothes dryer? (b) How many people would it take to replace a large electric power plant that generates 800 MW?
(a) 40
(b) 8 million
What is the cost of operating a 3.00-W electric clock for a year if the cost of electricity is $0.0900 per $\text{kW}\cdot \mathrm{h}$ ?
A large household air conditioner may consume 15.0 kW of power. What is the cost of operating this air conditioner 3.00 h per day for 30.0 d if the cost of electricity is $0.110 per $\text{kW}\cdot \mathrm{h}$ ?
$149
(a) What is the average power consumption in watts of an appliance that uses $5\text{.}\text{00 kW}\cdot \mathrm{h}$ of energy per day? (b) How many joules of energy does this appliance consume in a year?
(a) What is the average useful power output of a person who does $6\text{.}\text{00}\times {\text{10}}^{6}\phantom{\rule{0.20em}{0ex}}\text{J}$ of useful work in 8.00 h? (b) Working at this rate, how long will it take this person to lift 2000 kg of bricks 1.50 m to a platform? (Work done to lift his body can be omitted because it is not considered useful output here.)
(a) $\text{208 W}$
(b) 141 s
A 500-kg dragster accelerates from rest to a final speed of 110 m/s in 400 m (about a quarter of a mile) and encounters an average frictional force of 1200 N. What is its average power output in watts and horsepower if this takes 7.30 s?
(a) How long will it take an 850-kg car with a useful power output of 40.0 hp (1 hp = 746 W) to reach a speed of 15.0 m/s, neglecting friction? (b) How long will this acceleration take if the car also climbs a 3.00-m-high hill in the process?
(a) 3.20 s
(b) 4.04 s
(a) Find the useful power output of an elevator motor that lifts a 2500-kg load a height of 35.0 m in 12.0 s, if it also increases the speed from rest to 4.00 m/s. Note that the total mass of the counterbalanced system is 10,000 kg—so that only 2500 kg is raised in height, but the full 10,000 kg is accelerated. (b) What does it cost, if electricity is $0.0900 per $\text{kW}\cdot \mathrm{h}$ ?
(a) What is the available energy content, in joules, of a battery that operates a 2.00-W electric clock for 18 months? (b) How long can a battery that can supply $8\text{.}\text{00}\times {\text{10}}^{4}\phantom{\rule{0.20em}{0ex}}\text{J}$ run a pocket calculator that consumes energy at the rate of $1\text{.}\text{00}\times {\text{10}}^{-3}\phantom{\rule{0.20em}{0ex}}\text{W}$ ?
(a) $9\text{.}\text{46}\times {\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{J}$
(b) $2\text{.}\text{54 y}$
(a) How long would it take a $1\text{.}\text{50}\times {\text{10}}^{5}$ -kg airplane with engines that produce 100 MW of power to reach a speed of 250 m/s and an altitude of 12.0 km if air resistance were negligible? (b) If it actually takes 900 s, what is the power? (c) Given this power, what is the average force of air resistance if the airplane takes 1200 s? (Hint: You must find the distance the plane travels in 1200 s assuming constant acceleration.)
Calculate the power output needed for a 950-kg car to climb a $\mathrm{2.00\xba}$ slope at a constant 30.0 m/s while encountering wind resistance and friction totaling 600 N. Explicitly show how you follow the steps in the Problem-Solving Strategies for Energy .
Identify knowns: $m=\text{950 kg}$ , $\text{slope angle}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{2.00\xba}$ , $v=\mathrm{3.00\; m/s}$ , $f=\text{600 N}$
Identify unknowns: power $P$ of the car, force $F$ that car applies to road
Solve for unknown:
$P=\frac{W}{t}=\frac{\text{Fd}}{t}=F\left(\frac{d}{t}\right)=\text{Fv},$
where $F$ is parallel to the incline and must oppose the resistive forces and the force of gravity:
$F=f+w=\text{600 N}+\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta $
Insert this into the expression for power and solve:
$\begin{array}{lll}P& =& \left(f+\text{mg}\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \right)v\\ & =& \left[\text{600 N}+\left(\text{950 kg}\right)\left({\text{9.80 m/s}}^{2}\right)\text{sin 2\xba}\right](\text{30.0 m/s})\\ & =& \text{2.77}\times {\text{10}}^{4}\phantom{\rule{0.25em}{0ex}}\mathrm{W}\end{array}$
About 28 kW (or about 37 hp) is reasonable for a car to climb a gentle incline.
(a) Calculate the power per square meter reaching Earth’s upper atmosphere from the Sun. (Take the power output of the Sun to be $4\text{.}\text{00}\times {\text{10}}^{\text{26}}\phantom{\rule{0.25em}{0ex}}\text{W}\text{.})$ (b) Part of this is absorbed and reflected by the atmosphere, so that a maximum of $1\text{.}{\text{30 kW/m}}^{2}$ reaches Earth’s surface. Calculate the area in ${\text{km}}^{2}$ of solar energy collectors needed to replace an electric power plant that generates 750 MW if the collectors convert an average of 2.00% of the maximum power into electricity. (This small conversion efficiency is due to the devices themselves, and the fact that the sun is directly overhead only briefly.) With the same assumptions, what area would be needed to meet the United States’ energy needs $(1\text{.}\text{05}\times {\text{10}}^{\text{20}}\phantom{\rule{0.25em}{0ex}}\text{J})\mathrm{?}$ Australia’s energy needs $(5\text{.}4\times {\text{10}}^{\text{18}}\phantom{\rule{0.25em}{0ex}}\text{J)?}$ China’s energy needs $(6\text{.}3\times {\text{10}}^{\text{19}}\phantom{\rule{0.25em}{0ex}}\text{J)?}$ (These energy consumption values are from 2006.)
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