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.
Take east to be the positive direction. Then the resultant force from adding <em>F</em>₁ and <em>F</em>₂ is
<em>F</em>₁ + <em>F</em>₂ = (-45 N) + 63 N = 18 N
which is positive, so it's directed east.
To this we add a third force <em>F</em>₃ such that the resultant is 12 N pointing west, making it negative, so that
18 N + <em>F</em>₃ = -12 N
<em>F</em>₃ = -30 N
So <em>F</em>₃ has a magnitude of 30 N and points west.
Answer:
Explanation:
24 - gauge wire , diameter = .51 mm .
Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m
R = ρ l / s
1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]
= 8.42 x 10⁻² ohm
= .084 ohm
B ) Current required through this wire
= 12 / .084 A
= 142.85 A
C )
Let required length be l
resistance = .084 l
2 = 12 / .084 l
l = 12 / (2 x .084)
= 71.42 m
Explanation:
well there is nothing there and it could be different by diffrent objects, idk
Answer:
Pressure, P = 32666.66 Pa
Explanation:
It is given that,
Surface area of foot of Bimaba is 150 cm² or 0.015 m².
Her weight is 50 kg
We need to find the pressure does she exert on the ground, as she stands on her one foot. The force acting per unit area is called pressure. It can be given by :
So, the pressure is 32666.66 Pa.
