Answer:
<em> -4.7 x 10^-3 J/K-s</em>
Explanation:
The Power generated by metabolizing food = 80 W
The watt W is equivalent to the Joules per sec J/s
therefor power = 80 J/s
20% of this energy is not used for heating, amount available for heating is
==> H = 80% of 80 = 0.8 x 80 = 64 J/s
The inner body temperature = 37 °C = 273 + 37 = 310 K
The entropy of this inner body ΔS = ΔH/T
ΔS = 64/310 = 0.2065 J/K-s
The skin temperature is cooler than the inner body by 7 °C
Temperature of the skin = 37 - 7 = 30 °C = 273 + 30 = 303 K
The entropy of the skin = ΔS = ΔH/T
ΔS = 64/303 = 0.2112 J/K-s
change in entropy of the person's body = (entropy of hot region: inner body) - (entropy of cooler region: skin)
==> 0.2065 - 0.2112 =<em> -4.7 x 10^-3 J/K-s</em>
Answer:
150156.25 Ω
Explanation:
Resistance: This can be defined as the opposition to the flow of electric current in a circuit. The S.I unit of resistance is Ohm's (Ω)
The expression for resistance is given as
P = V²/R................ equation 1
Where P = power, V = Voltage, R = Resistance.
Making R the subject of the equation,
R = V²/P.................. Equation 2
Given: V = 115 V , P = 0.16 W.
Substitute into equation 2
R = 155²/0.16
R = 150156.25 Ω
Hence,
The resistance = 150156.25 Ω
In order to decrease the friction on the slide,
we could try some of these:
-- Install a drippy pipe across the top that keeps continuously
dripping olive oil on the top end of the slide. The oil oozes
down the slide and keeps the whole slide greased.
-- Hire a man to spread a coat of butter on the whole slide,
every 30 minutes.
-- Spray the whole slide with soapy sudsy water, every 30 minutes.
-- Drill a million holes in the slide,and pump high-pressure air
through the holes. Make the slide like an air hockey table.
-- Keep the slide very cold, and keep spraying it with a fine mist
of water. The water freezes, and a thin coating of ice stays on
the slide.
-- Ask a local auto mechanic to please, every time he changes
the oil in somebody's car, to keep all the old oil, and once a week
to bring his old oil to the park, to spread on the slide. If it keeps
the inside of a hot car engine slippery, it should do a great job
keeping a simple park slide slippery.
-- Keep a thousand pairs of teflon pants near the bottom of the ladder
at the beginning of the slide. Anybody who wants to slide faster can
borrow a set of teflon pants, put them on before he uses the slide, and
return them when he's ready to go home from the park.
To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.
In other words the acceleration can be described as

Where
G = Gravitational Universal Constant
M = Mass of Earth
r = Radius of Earth
This equation can be differentiated with respect to the radius of change, that is


At the same time since Newton's second law we know that:

Where,
m = mass
a =Acceleration
From the previous value given for acceleration we have to

Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:




But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:


Therefore there is a weight loss of 0.3N every kilometer.
Answer:
d) What is the force if we doubled both the masses AND we doubled the distance