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True [87]
3 years ago
6

What number will be shown by hydrometer if sunk in pure water ?​

Physics
2 answers:
adoni [48]3 years ago
8 0
1000kg/meters cubed if it’s sunk in pure water
serious [3.7K]3 years ago
7 0

The hydrometer displays the density of the liquid you stick it in, COMPARED TO WATER.

If you stick it in pure water, it should read 1.00 . This is a good way to check your hydrometer and see whether it's reading correctly ("calibrated").

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Does anyone here watch the flash show? And if so, who’s your favorite character?
Kazeer [188]

Answer:

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Explanation:

8 0
3 years ago
Read 2 more answers
Electric charge is distributed over the disk x2 + y2 ≤ 4 so that the charge density at (x, y) is rho(x, y) = 4x + 4y + 4x2 + 4y2
maw [93]

Answer:

Q=185.84C

Explanation:

We have to take into account the integral

Q=\int \rho dV

In this case we have a superficial density in coordinate system.

Hence, we have for R: x2 + y2 ≤ 4

Q=\int_{-2}^2\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\rho dydx

but, for symmetry:

Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C

HOPE THIS HELPS!!

8 0
3 years ago
9. Which of the following characteristics of water is most responsible for the sinking of the Titanic? A. Water is liquid at roo
lozanna [386]

Answer:

D. Frozen water is less dense than liquid water.

Explanation:

Titanic sank in the North Atlantic Ocean in the year 1912. There were more than 2000 passengers on board from which about 1500 died.

Frozen water that is ice has less density than liquid water. The density of ice is 0.9167 g/cm^3 and the density of the water is 0.9998 g/cm^3.

Due to this very slight difference in density, ice floats on water. Titanic, on the night of April 14, 1912, crashed into an iceberg around midnight which caused opening of 5 of its watertight compartments. The water filled into the ship and it eventually sank in.

8 0
4 years ago
Question 9 please. How do you find instantaneous velocity and how do I sketch a velocity vs time graph from a position vs time o
pickupchik [31]

How do you find instantaneous velocity

Select a point on a distance-time curve graph. Draw a tangent to the curve at that point. Tangent -> hypotenuse of right angled triangle. Opp/adjacent in graph units is vel at that point -> in distance and/or time

3 0
4 years ago
Por una tubería de 0.06 m de diámetro circula agua con una velocidad desconocida, al llegar a la parte estrecha de la tubería de
Vesnalui [34]

Answer:

La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 \frac{m}{s}

Explanation:

La ecuación de continuidad es simplemente una expresión matemática del principio de conservación de la masa.  Este principio establece que la masa de un objeto o colección de objetos nunca cambia con el tiempo.

La ecuación de continuidad es la relación que existe entre el área y la velocidad que tiene un fluido en un lugar determinado y dice que el caudal de un fluido es constante a lo largo de un circuito hidráulico.

En otras palabras, la ecuación de continuidad se basa en que el caudal (Q) del fluido ha de permanecer constante a lo largo de toda la conducción. Cuando un fluido fluye por un conducto de diámetro variable, su velocidad cambia debido a que la sección transversal varía de una sección del conducto a otra.

Entonces, siendo el caudal es el producto de la superficie de una sección del conducto por la velocidad con que fluye el fluido,  en dos puntos de una misma tubería se cumple:

Q1=Q2

A1*v1= A2*v2

donde:

  • A es la superficie de las secciones transversales de los puntos 1 y 2 del conducto.
  • v es la velocidad del flujo en los puntos 1 y 2 de la tubería.

Siendo A=pi*r^{2} =pi*(\frac{D}{2} )^{2} =\frac{pi*D^{2} }{4} , donde pi es el número π, r es el radio del conducto y D el diámetro del conducto, entonces:

\frac{pi*D1^{2} }{4}*v1=\frac{pi*D2^{2} }{4}*v2

En este caso:

  • D1: 0.06 m
  • v1: ?
  • D2: 0.04 m
  • v2: 2.6 m/s

Reemplazando:

\frac{pi*(0.06m)^{2} }{4}*v1=\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}

Resolviendo:

v1=\frac{\frac{pi*(0.04m)^{2} }{4}*2.6\frac{m}{s}}{\frac{pi*(0.06m)^{2} }{4}}

v1=\frac{(0.04m)^{2} }{(0.06m)^{2}  }*2.6\frac{m}{s}

v1= 1.156 \frac{m}{s}

<u><em>La velocidad con la que se desplaza el agua antes de llegar a la parte estrecha de la tubería es 1.156 </em></u>\frac{m}{s}<u><em></em></u>

8 0
3 years ago
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