A body floats at a volume of 2/5 when it is placed in oil, how dense the object is ....s=900km/m3

What water pressure must a pump that is located on the first floor supply to have water on the thirteenth of a building with a p

irga5000 [103]

The water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

The given parameters;

Pressure on the 13 th floor, P₁ = 35 PSI
Distance between each floor, d = 10 ft

The vertical pressure of the water is calculated as follows;

$P = \rho gh\\\\\frac{P}{h} = \rho g\\\\\frac{P}{h} = k\\\\\frac{P_1}{h_1} = \frac{P_2}{h_2} \\\\$

The vertical height of the first floor from the 13th floor = 130 ft

The vertical height of the 13 ft floor = 10 ft

$P_1 = \frac{P_2 h_1}{h_2} \\\\P_1 = \frac{35 \times 130}{10} \\\\P_1 = 455 \ PSI$

Thus, the water pressure on the first floor must be 455 PSI in order to push the water to the 13th floor at the given pressure.

Learn more about vertical height and pressure here: brainly.com/question/15691554

3 0

2 years ago

A research submarine has a 20-cm-diameter window that is 9.0 cm thick. The manufacturer says the window can withstand forces up

Ratling [72]

Answer:

The maximum safe depth in salt water is 3758.2 m.

Explanation:

Given that,

Diameter = 20 cm

Radius = 10 cm

Thickness = 9.0 cm

Force $F = 1.2\times10^{6}\ N$

Inside pressure = 1.0 atm

We need to calculate the area

Using formula of area

$A=\pi\times r^2$

Put the value into the formula

$A=\pi\times(10\times10^{-2})^2$

$A=0.0314\ m^2$

We need to calculate the pressure

Using formula of pressure

$P=\dfrac{F}{A}$

Put the value into the formula

$P=\dfrac{1.2\times10^{6}}{0.0314}$

$P=38216560.50\ Pa$

$P=3.8\times10^{7}\ Pa$

We need to calculate the maximum depth

Using equation of pressure

$P=P_{atm}+\rho gh$

$h=\dfrac{P-P_{atm}}{\rho g}$

Put the value into the formula

$h=\dfrac{3.8\times10^{7}-101325}{1029\times9.8}$

$h=3758.2\ m$

Hence, The maximum safe depth in salt water is 3758.2 m.

4 0

3 years ago

Help me please I can't get the final step

inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

Dimensional Analysis

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

$m+n=2$

$-m+2n=2$

Adding both equations:

$3n=4$

Solving:

$n=4/3$

$m=2-4/3=2/3$

Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

6 0

3 years ago

Which of these terms means an object's resistance to changing its motion?

Mila [183]

Answer:

Inertia

F=ma

Action, reaction

All of the above

A heavy object requires more force to push than a lighter object.

8 0

3 years ago

A car is moving at a initial velocity of 20 meters per second accelerates at a rate of 1.5 meters per second squared for 4 secon

Brrunno [24]

Answered using calculus.
Antidifferentiated the acceleration to get velocity. Added variable c as we do not know if there was an extra number there yet.
Knowing that when time is 0, the velocity is 20, we can substitute those numbers into the equation and find that c = 20.
Now we have full velocity equation: v = 1.5t + 20
Now we substitute 4 into t to find out the velocity after 4 seconds. This gives us the final answer of 26m/s

5 0

3 years ago

A body floats at a volume of 2/5 when it is placed in oil, how dense the object is ....s=900km/m3​

A body floats at a volume of 2/5 when it is placed in oil, how dense the object is ....s=900km/m3