Heat rises, and it is warmer at the equator, so I think warm air would rise at the equator and move towards the cooler poles.
Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?
Answer:
225 N
Explanation:
From Pascal's principle,
F/A = f/a ...................... Equation 1
Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.
Making f the subject of the equation,
f = F(a)/A ..................... Equation 2
Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².
Substituting into equation 2
f = 15000(3/200)
f = 225 N.
Hence the downward force that must be applied to small piston = 225 N
Answer:
Twice
Explanation:
From the formula for velocity in a circle
V= 2πr/T
Where V is velocity
r is raduis
T is period
We see that as r increases V increases so if r is doubled V becomes doubled
Answer: Depends
Explanation:
Depends on how much the diver weighs.
I notice that even though we're working with frames of reference
here, you never said which frame the '5 km/hr' is measured in.
In fact ! You didn't even say which frame the '12 km/hr' of his
bike is measured in.
So there are several different ways this could go. I'll do it the way
I THINK you meant it, but that doesn't guarantee anything.
-- Simon is riding his bike at 12 km/hr relative to the sidewalk,
away from Keesha.
-- He throws a ball at Keesha, at 5 km/hr relative to his own face.
-- Keesha sees the ball approaching her at (12 - 5) = 7 km/hr
relative to the ground and to her.
