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malfutka [58]
3 years ago
12

When a submarine dives to a depth of 5.0 × 102 m, how much pressure, in pa, must its hull be able to withstand? how many times l

arger is this pressure than the pressure at the surface?
Physics
2 answers:
nordsb [41]3 years ago
5 0
<span>Depth = 5.0 Ă— 10^2 m
 Density of sea water = 1.025 x 10^3
 Pd = Po + pgh
Atmospheric pressure is standard Patm = 1.01325 x 10^5 Pa
  Since the normal pressure is retained in the hull, no need to bother about Po Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
 So now Pd / Patm = 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
 So it is 49.56 times larger.</span>
mote1985 [20]3 years ago
5 0
Pressure is force over surface.
 p=F/S.
 In this case the force is given by the weight of the water G, so
 p = G/S; 
<span> G=m*g;
 m=ρ*V,
 V=S*h
 since we are interested only in the volume of water which presses on the surface of the submarine S, at depth h.
</span><span> So:
</span><span> p= ρ*S*h*g/S = ρ*g*h.
 This is the pressure at a certain depth h, exerted only by the water so we must add the atmospheric pressure p0.
</span> <span>Therefore the total pressure is pt = p0 + ρ*g*h.
 We can see that when h=0, pt=p0 which is correct since at the surface we have only atmospheric pressure.
</span><span> The ratio pt/p0 = 1+ ρ*g*h/p0
</span><span> p0=101325 Pa 
</span><span> h = 500 m 
</span><span> g= 9.8 m/s^2 
</span><span> ρ = 1.025*10^3 kg/m^3. 
</span><span> We substitute in the equation and get pt/p0 = 50.56 .
 So the pressure at 500m depth is 50.56 times greater than at the surface.
 </span>
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A river flows due south with a speed of 2.0 m/s. You steer a motorboat across the river; your velocity relative to the water is
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Answer:

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Explanation:

solution:

we have river speed v_{r}=2 m/s

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a) v_{m}=v_{r}+v_{m/r}

solving graphically

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note :

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Calculate a. The heat that must be supplied to a 500.0 g copper kettle containing 400.0 g of water to raise its temperature from
uysha [10]

Explanation:

<h2>For Copper</h2>

dH copper = mCdT copper

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dH = 500 g (0.385 J/g C) (78 C rise)

dH = 15,015 Joules

<h2> For Water</h2>

dH water =  m C dT water

<em>(Specific Heat of copper=</em>4.184 J/g-C<em>)   </em>

<em />

dH = 400 g (4.184 J/g-C) (78 C rise)

dH = 130,540 Joules

total heat = 15,015 + 130,540 = 145,555 Joules

<h2>Percentage for Water  </h2>

(130,540 Joules  / 145,555 Joules) x 100 = 89.7 %

If we consider that we have 3 Significant Figures,

then, the answers become ,

15.0 KJ must be added for Copper

130.5 KJ must be added for Water

and the total of 145 KJ must be added in the kettle with the water

89.7 %  of heat goes to the Water

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