<span>Ocean currents act much like a conveyer belt,
transporting warm water and precipitation from the equator toward the
poles and cold water from the poles back to the tropics. Thus, currents
regulate global climate, helping to counteract the uneven distribution of solar radiation reaching Earth's surface.</span>
Answer:
I dont know spanish... and i live in Cali... wow....
Explanation:
Answer:
W / A = 39200 kg / m²
Explanation:
For this problem let's use the equilibrium equation of / newton
F = W
Where F is the force of the door and W the weight of water
W = mg
We use the concept of density
ρ = m / V
m = ρ V
The volume of the water column is
V = A h
We replace
W = ρ A h g
On the other side the cylinder cover has a pressure
P = F / A
F = P A
We match the two equations
P A = ρ A h g
P = ρ g h
P = 39200 Pa
The weight of the water column is
W = 1000 9.8 4 A
W / A = 39200 kg / m²
Answer:
a) 1450watts
b) 564watts
c) 1.11
Explanation:
Power consumed = IV
I is the current rating
V is the operating voltage
If a blow-dryer and a vacuum cleaner each operate with a voltage of 120 V and the current rating of the blow-dryer is 12 A, while that of the vacuum cleaner is 4.7 A then their individual power rating is calculated thus;
a) For blow-dryer
Operating voltage = 120V
Its current rating = 12A
Power consumed = IV
= 120×12
= 1440watts
b) For vacuum cleaner:
Operating voltage is the same as that of blow dryer = 120V
Its current rating = 4.7A
Power consumed = IV
= 120×4.7
= 564watts
c) Energy used = Power consumed × time taken
Energy used = Power × time
Energy used by blow dryer = 1440×20×60
= 1,728,000Joules
Energy used up by vacuum cleaner = 564×46×60
= 564×2760
= 1,556,640Joules
Ratio of the energy used by the blow-dryer in 20 minutes to the energy used by the vacuum cleaner in 46 minutes will be 1,728,000/1,556,640 = 1.11
The total flux through the cylinder is zero.
In fact, the electric flux through a surface (for a uniform electric field) is given by:
where
E is the intensity of the electric field
A is the surface
is the angle between the direction of E and the perpendicular to the surface, whose direction is always outwards of the surface.
We can ignore the lateral surface of the cylinder, since the electric field is parallel to it, therefore the flux through the lateral surface of the cylinder is zero (because 
and 
).
On the other two surfaces, the flux is equal and with opposite sign. In fact, on the first surface the flux will be
where r is the radius, and where we have taken 
since the perpendicular to the surface is parallel to the direction of the electric field, so 
. On the second surface, however, the perpendicular to the surface is opposite to the electric field, so 
and 
, therefore the flux is
And the net flux through the cylinder is
