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Lady bird [3.3K]
3 years ago
12

What effects do oceans have on climate? A. Ocean winds bring rain and fog and often bring warm water that keeps the climate mild

. B. Ocean winds usually bring snow and freezing temperatures. C. Ocean winds often deliver very hot air and dry weather.What effects do oceans have on climate?
A.
Ocean winds bring rain and fog and often bring warm water that keeps the climate mild.

B.
Ocean winds usually bring snow and freezing temperatures.

C.
Ocean winds often deliver very hot air and dry weather.
Physics
1 answer:
nalin [4]3 years ago
7 0
A. Ocean winds bring rain and fog and often bring warm water that keeps the climate mild
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Help what is the answer pls
Vladimir [108]
I think the answer should be: “100.4957 N”
7 0
3 years ago
A person is lying on a diving board 3.50 m above the surface of the water in a swimming pool. She looks at a penny that is on th
sergij07 [2.7K]

Answer:

1.995 m

Explanation:

Distance of penny as seen by the person = 5 m

Height of person from water surface = 3.50 m

Apparent depth of penny = 5 - 3.50 = 1.5 m

refractive index of water, n = 1.33

real depth / apparent depth = n

real depth = 1.33 x 1.5 = 1.995 m

Thus, the actual depth of water at that point is 1.995 m.

4 0
3 years ago
An airplane flies with a constant speed of 840km/h. how far can it travel in 1 hour?
arlik [135]
Distance = speed X time

In this example, the speed of the airplane = 840km. The time (that the question is asking)is how far can it travel in 1 hour.

So just plug in your numbers.

Distance = 840km X 1 hour = 840km/hour or 840km for short.
3 0
3 years ago
Momentum,
Serggg [28]

Answer:

THE WALL MOVES AWAY FROM THE BALL

Explanation:

NEWTON'S THIRD LAW STATES THAT THERE IS A OPPOSITE REACTION

5 0
3 years ago
A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.
Zolol [24]
These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is 45^{\circ}.

Calling F the force of the wind, and d=15~km=15000~m the distance covered by the boat, the work done by the wind is:
W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J

The total time of the motion is t=1~h=3600~s and therefore the power of the wind is
P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55

Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
\epsilon =  \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
P= \frac{W}{t}
But we can rewrite the work as
W=Fdcos\theta
where F is the force applied, d the displacement of rock and \theta=60^{\circ] is the angle between the direction of the force and the displacement (3 m). 
Therefore we can rewrite the power as
P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta
where v=d/t=5~m/s is the velocity, Using the data of the exercise, we can then find the force, F:
F= \frac{P}{v cos\theta} =   \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N

and now we can also calculate the work, which is 
W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J
3 0
3 years ago
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