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tamaranim1 [39]
3 years ago
9

009 10.0 points

Physics
1 answer:
Ilya [14]3 years ago
3 0

The density is 1106 kg/m^3

Explanation:

The bulk modulus of a liquid is defined as

B=\rho_0 \frac{\Delta p}{\Delta \rho}

where

\rho_0 is the density at the reference level

\Delta p is the change in pressure

\Delta \rho is the change in density

The equation can be re-arranged as

\Delta \rho = \frac{\rho_0 \Delta p}{B}

In this problem, we have:

\rho_0 = 1100 kg/m^3 (density at the surface)

B=2.3\cdot 10^9 N/m^2 (bulk modulus of seawater)

\Delta p = 126 - 1 = 125 atm = (125\cdot 1.01\cdot 10^5) Pa (pressure difference between the point where p = 126 atm and the pressure at sea level, which is 1 atm)

Therefore,

\Delta \rho = \frac{(1100)(125\cdot 1.01\cdot 10^5)}{2.3\cdot 10^9}=6.0 kg/m^3

So, the density at the depth where the pressure is 126 atm is

\rho = \rho_0 +\Delta \rho = 1100+6=1106 kg/m^3

Learn more about pressure in fluids:

brainly.com/question/4868239

brainly.com/question/2438000

#LearnwithBrainly

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Una barra de aluminio que esta a 78 GRADOS CENTIGRADOS entra en contacto con una barra de cobre de la misma longitud y área que
stiks02 [169]

Answer:

Al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.

Explanation:

Dado que una barra de aluminio que está a 78 grados centígrados entra en contacto con una barra de cobre de la misma longitud y área que esta a 28 grados centígrados, y posteriormente se lleva acabo la transferencia de energía entre ambas barras llegando a su equilibrio térmico, para determinar la temperatura a la que ambas barras llegarán se debe realizar el siguiente cálculo:

(78 + 28) / 2 = X

106 / 2 = X

53 = X

Por lo tanto, al llegar a su equilibrio térmico ambas barran tendrán una temperatura de 53 grados centígrados.

8 0
2 years ago
a ball rolls along the floor with a constant velocity of 3 m/s. How far will it have gone after 234 seconds?
sergij07 [2.7K]

Answer:

The ball travel for 702m

Explanation:

distance = speed × time

speed = 3m/s

time = 234s

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Mr. Rudman drives his race car for 4 hrs at 150<br> miles/hr. How far will he travel?
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An object thrown vertically upward from the surface of a celestial body at a velocity of 36 ​m/s reaches a height of sequalsminu
Anarel [89]

Answer:

v = -1.8t+36

20 seconds

360 m

40 seconds

36 m/s

The object speed will increase when it is coming down from its highest height.

Explanation:

s=-0.9t^2+36t

Differentiating with respect to time we get

\frac{ds}{dt}=-1.8t+36\\\Rightarrow v=-1.8t+36

a) Velocity of the object after t seconds is v = -1.8t+36

At the highest point v will be 0

0=-1.8t+36\\\Rightarrow t=\frac{-36}{-1.8}\\\Rightarrow t=20\ s

b) The object will reach the highest point after 20 seconds

s=-0.9t^2+36t\\\Rightarrow s=-0.9\times 20^2+36\times 20\\\Rightarrow s=360\ m

c) Highest point the object will reach is 360 m

s=-0.9t^2+36t\\\Rightarrow 360=-0.9t^2+36t\\\Rightarrow -0.9t^2+36t-360=0\\\Rightarrow -9t^2+360t-3600=0

\frac{-360\pm \sqrt{0}}{2\left(-9\right)}\\\Rightarrow t=20\ s

d) Time taken to strike the ground would be 20+20 = 40 seconds

[tex]v=u+at\\\Rightarrow v=0+0.9\times 2\times 20\\\Rightarrow v=36\ m/s

Acceleration will be taken as positive because the object is going down. Hence, the sign changes. 2 is multiplied because the expression is given in the form of s=ut+\frac{1}{2}at^2

e) The velocity with which the object strikes the ground will be 36 m/s

f) The speed will increase when the object has gone up and for 20 seconds and falls down for 20 seconds. The object speed will increase when it is coming down from its highest height.

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Djdjjdjdjdjjfjfhshdhehe
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