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Monica [59]
4 years ago
7

Sequence the seafloor features as you move from the shoreline outward into the ocean.

Physics
1 answer:
Mademuasel [1]4 years ago
4 0

C) continental shelf - continental slope - abyssal plain

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Explain the application of pascals law<br>​
Marta_Voda [28]

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pplications of Pascal’s Law  

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The construction is such that a narrow cylinder (in this case A) is connected to a wider cylinder (in this case B). They...

Pressure applied at piston A is transmitted equally to piston B without diminishing, on use of an incompressible fluid.

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6 0
4 years ago
Can sound travel through space? Why or why not?
DiKsa [7]

Answer:

I think sound does not travel at all in space. The vacuum of outer space has essentially zero air. Because sound is just vibrating air, space has no air to vibrate and therefore no sound. If you are sitting in a space ship and another space ship explodes, you would hear nothing.

Explanation:

5 0
3 years ago
4. How long does it take a car traveling at 45 km/h to travel 100.0 m?<br> 4500m
Trava [24]

Answer:

8.0s

Explanation:

45 km/h ÷ 3.6  = 12.5 m/s

t.v = d/t

vt/v = d/v

t = d/v = 100.0m /12.5 m/s = 8.0s

Hope this helps!!

8 0
3 years ago
Archimedes supposedly was asked to determine whether a crown made for the king consisted of pure gold (density of gold is 19.3 ×
Vika [28.1K]

Answer:

Explanation:

First, we can find the mass of the object in air

Since W = mg

m = W/g = (8.9)/(9.8) = 0.908 kg

Then, by Archimedes principle, we can find its volume. The volume is found by the weight of the water displaced by the formula

W = Vρg

The Weight is the difference in scale readings. The density of water is 1000 kg/m3

(8.9- 7.8) = V(1000)(9.8)

Thus V = 1 X 10-4 m3

Then, since density is mass / volume

ρ = 0.908/1 X 10-4

ρ = 8254 kg/m3 which is 8 X 103 kg/m3

The density of gold is 19.3 X 103 kg/m3

Since those densities are not the same, the crown is either hollow or not pure gold

8 0
3 years ago
A mountain climber of mass 60.0 kg slips and falls a distance of 4.00 m, at which time he reaches the end of his elastic safety
Sever21 [200]
The energy that the rope absorbs from the climber is Ep=m*g*h where m is mass of the climber, g=9.81m/s² and h is the height the climber fell. h=4 m+2 m because he was falling for 4 meters and the rope stretched for 2 aditional meters. The potential energy stored in the rope is Er=(1/2)*k*x², where k is the spring constant of the rope and x is the distance the rope stretched and it is
x=2 m. So the equation from the law of conservation of energy is:

Ep=Er

m*g*h=(1/2)*k*x²

k=(2*m*g*h)/x² = (2*60*9.81*6)/2² = 7063.2/4 =1765.8 N/m

So the spring constant of the rope is k=1765.8 N/m.
7 0
3 years ago
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