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denis-greek [22]

3 years ago

10

Problem 2: (15 pts) A 10-m high cylindrical container is half-filled at the bottom with water of density =1000 kg/m3 while the t

op half is filled with oil of specific gravity of 0.85. What is the pressure difference between the top and bottom of the container?

2 answers:

Step2247 [10]3 years ago

8 0

Answer:

$\Delta p = 90.7 kPa$

Explanation:

specific gravity of oil is $= \frac{\rho_{oil}}{\rho_w}$

$\rho_{oil} = 0.85*1000 = 850 kg/m3$

we know that

change in pressure for oil is given as

$\Delta p = \rho gh$

here density and h is for oil

$\Delta p = 850*5 *9.81 = 41,692.5 kPa$

change in pressure for WATER is given as

$\Delta p = \rho gh$

here density is for water and h is for water

$\Delta p = 1000*5 *9.81 = 49,050 kPa$

pressure change due to both is given as

$\Delta p = 41692.3 + 49050 = 90742.5 N/m2$

$\Delta p = 90.7 kPa$

tamaranim1 [39]3 years ago

5 0

Answer:

I don't get it I don't get it

Explanation:

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if a swimmer is traveling at a constant speed of 0.85 m/s how long would it take to swim the length of 50 meter olymic sized poo

VARVARA [1.3K]

Answer:

58.82 Seconds

Explanation:

50m/0.85m/s=58.82s

7 0

3 years ago

In xray machines, electrons are subjected to electric fields as great as 6.0 x 10^5 N/C. Find

katrin2010 [14]

Answer:

a = 1.055 x 10¹⁷ m/s²

Explanation:

First, we will find the force on electron:

$E = \frac{F}{q}\\\\F = Eq\\$

where,

F = Force = ?

E = Electric Field = 6 x 10⁵ N/C

q = charge on electron = 1.6 x 10⁻¹⁹ C

Therefore,

$F = (6\ x\ 10^5\ N/C)(1.6\ x\ 10^{-19}\ C)\\$

F = 9.6 x 10⁻¹⁴ N

Now, we will calculate the acceleration using Newton's Second Law:

$F = ma\\a = \frac{F}{m}\\$

where,

a = acceleration = ?

m = mass of electron = 9.1 x 10⁻³¹ kg

therefore,

$a = \frac{9.6\ x\ 10^{-14}\ N}{9.1\ x\ 10^{-31}\ kg}\\\\$

<u>a = 1.055 x 10¹⁷ m/s²</u>

5 0

3 years ago

if the volume of a scuba tank filled with air remains constant and its temperature goes down, what happens to its pressure?

Vladimir [108]

Answer:

Decreases

Explanation:

Ideal gas law:

PV = nRT

where P is absolute pressure,

V is volume,

n is number of moles,

R is gas constant,

and T is absolute temperature.

If V is constant and T decreases, then P must decrease.

3 0

3 years ago

At NASA's Zero Gravity Research Facility in Cleveland, Ohio, experimental payloads fall freely from rest in an evacuated vertica

Diano4ka-milaya [45]

Answer:

(a). Energy is 64,680 J

(b) velocity is 51.43m/s

(c) velocity in mph is 115.0mph

Explanation:

(a).

The potential energy $P$ of the payload of mass $m$ is at a vertical distance $h$ is

$P =mgh$ .

Therefore, for the payload of mass $m = 50kg$ at a vertical distance of $h = 132 m$ , the potential energy is

$P = (50kg)(9.8m/s^2)(132m)$

$\boxed{P = 64,680J}$

(b).

When the payload reaches the bottom of the shaft, all of its potential energy is converted into its kinetic energy; therefore,

$mgh= \dfrac{1}{2}mv^2$

$v= \sqrt{2gh}$

$v = \sqrt{2*9.8*135}$

$\boxed{v = 51.43m/s}$

(c).

The velocity in mph is

$\dfrac{51.43m}{s} * \dfrac{3600s}{hr} * \dfrac{1mile}{1609.34m}$

$\boxed{v= 115.0mph}$

5 0

3 years ago

A 1-kilogram mass is attached to a spring whose constant is 21 N/m, and the entire system is then submerged in a liquid that imp

amm1812

Answer:

the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Explanation:

Given that,

mass, m = 1kg

spring constant k = 21N/M

damping force = $-\beta\frac{dx}{dt} = \frac{-10dx}{dt}$

$\beta = 10$

By Newtons second law ,

The diffrential equation of motion with damping is given by

$m\frac{d^2x}{dt^2} = -kx-\beta\frac{dx}{dt}$

substitute the value of m =1kg, k = 21N/M, and $\beta = 10$

$1\frac{d^2x}{dt^2} = -21x=10\frac{dx}{dt}$

$\frac{d^2x}{dt^2} + 10\frac{dx}{dt} + 21x = 0$

suppose the equation of the form $x =e^m^t$ ,

and the auxilliary equation is given by

$m^2 + 10m + 21 = 0\\\\m^2 + 7m+3m+21=0\\\\m(m+7)+3(m+7)=0\\\\(m+7)(m+3)=0\\\\m=-7\\m=-3$

The general solution for the above differential equation is

$x(t) =C_1e^{-3t}+C_2e^{-7t}$

Derivate with respect to t

$x'(t)=-3C_1e^{-3t}-7C_2e^{-7t}$

(a)

since time is 0 then mass is one meter below

so x(0) = 1

Also it start from rest , that implies , velocity is 0 and time is 0

$x'(0) = 0$

substitute the initial condition

$C_1 +C_2 = 1$

$-3C_1-7C_2=0$

Solve the above equation to get C₁ and C₂

$C_1 =\frac{7}{4} and C_2 = -\frac{3}{4}$

substitute for C₁ and C₂ in general solution

$x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

Thus the required value is $x(t) = \frac{7}{4} e^{-3t}-\frac{3}{4} e^{-7t}$

3 0

3 years ago

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