A planet of mass M and radius R has no atmosphere. The escape velocity at its surface is ve. An object of mass m is at rest a di

A venturi is constructed of a 10.0 cm pipe with a 2.0 cm diameter throat. Water pressure in the pipe is twice atmospheric pressu

Ostrovityanka [3204]

Response:

$P_t=5066250.0696\ Pa=50\ atm$

Clarification:

Provided:

diameter of the pipe, $d=0.1\ m$

the throat's diameter, $d_t=0.02\ m$

flow velocity, $v=0.4\ m.s^{-1}$

The pressure in the pipe is twice the atmospheric pressure:

$P=2\times 101325=202650\ Pa$

To find the water's flow force:

$F=P\times A$

we now calculate the cross-sectional area of the pipe:

$A=\frac{\pi.d^2}{4}$

$A=\pi \times\frac{0.1}{4}$

$A=0.007854\ m^2$

Thus,

$F=202650\times 0.007854$

$F=1591.6094\ N$

Next, we determine the area at the throat:

$A_t=\frac{\pi.d_t^2}{4}$

$A_t=\frac{\pi\times 0.02^2}{4}$

$A_t=0.000314\ m^2$

So, the pressure at the throat becomes:

$P_t=\frac{F}{A_t}$

$P_t=\frac{1591.6094}{0.000314}$

$P_t=5066250.0696\ Pa=50\ atm$

6 0

1 month ago

A hydraulic lift raises a 2000 kg automobile when a 500 N force is applied to the smaller piston. If the smaller piston has an a

kicyunya [3294]

Answer:

The cross-sectional area of the larger piston is 392cm ^{2}[/tex]

Explanation:

To find the solution, we apply the following equation:

Pascal's principle: F=P*A Formula (1)

F=Force applied to the piston

P: Pressure

A= Area of the piston

Nomenclature:

Fp= Force on the primary piston= 500N

W= weight of the car =m*g=2000kg*9.8m/s2= 19600N

Fs= Force on the secondary piston= W = 19600N

$Ap= Primary piston area=10cm^{2} =10*10^{-4}m^{2}$

As= Area of the secondary piston=?

Pressure applied on one side is distributed to all liquid molecules since liquids are incompressible.

From equation (1)

P=F/A

Pp=Ps

$\frac{Fp}{Ap} = \frac{Fs}{As}$

$As= \frac{Fs*Ap}{Fp}$

$As=\frac{19600*10*10^{-4} }{500}$

$As=0.0392m^{2} =0.0392*10^{4}cm^{2}$

$As=392cm ^{2}$

5 0

2 months ago

Three point charges are positioned on the x axis. If the charges and corresponding positions are +32 µC at x = 0, +20 µC at x =

Sav [3153]

Response:

Magnitude of the electrostatic force acting on the +32 µC charge, $F_{net} = 12 N$

Clarification:

Let q₁ = +32 µC, located at x₁ = 0

q₂ = +20 µC, positioned at x₂ = 40 cm = 0.4 m

q₃ = -60 µC, placed at x₃ = 60 cm = 0.6 m

Define the force magnitude on the +32 µC charge from the +20 µC charge as F₁ (the force on q₁ due to q₂).

$F_{2} = \frac{kq_{1}q_{2} }{x_{2}^2 }$

$F_{2} = \frac{9 * 10^{9} * 32 * 10^{-6} * 20 * 10^{-6} }{0.4^2 }\\F_{2} = 36 N$

Define the force magnitude on the +32 µC charge from the -60 µC charge as F₂ (the force on q₁ due to q₃).

$F_{3} = \frac{kq_{1}q_{3} }{x_{3}^2 }$

$F_{3} = -\frac{9 * 10^{9} * 32 * 10^{-6} * 60 * 10^{-6} }{0.6^2 }\\F_{3} =-48 N$

The resultant electrostatic force on the 32 µC charge is $F_{net} = |F_{2} + F_{3}|$

$F_{net} =| 36 + (-48)| \\F_{net} =|- 12 N| \\ F_{net} = 12 N$

7 0

2 months ago

A coffee company wants to make sure that their coffee is being served at the right temperature. if it is too hot, the customers

Maru [3345]

Response:

The population mean is parameter = 65 c

Explanation:

In the analysis of samples and inferring population behavior, two key elements are essential.

To ascertain the population mean, we typically extract various samples and calculate their average. The average of all these means will serve as an estimate for the population mean. According to the central limit theorem, as sample sizes increase, the average of a sample tends to follow a normal distribution with an estimated mean being the sample mean.

A statistic pertains to a sample, while a parameter refers to the whole population.

In this case, 65 degrees C represents the entire population; thus, it constitutes a parameter.

4 0

2 months ago

Read 2 more answers

A solid conducting sphere carrying charge q has radius a. It is inside a concentric hollow conducting sphere with inner radius b

Softa [3030]

Response:

Clarification:

Refer to the diagram indicating the charges on the specified sphere (see attachment).

The electric field at the stated positions is

E(r) = 0 for r≤a. Equation 1

E(r) = kq/r² for a<r<b. Equation 2

E(r) = 0 for b<r<c. Equation 3

E(r) = kq/r² for r>c. Equation 4.

We understand that electric potential correlates with the electric field through

V = Ed

A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is

E = kQ / r²

Then,

V = Ed,

At d = r = c

Thus,

Vc = (kQ / c²) × c

Vc = kQ / c

As a result, the total charge Q consists of +q, -q, and +q

Hence, Q = q - q + q = q

V = kq / c

B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have

V = kQ/r

V = kQ / b, noting that r = b

So, Q = q

V = kq / b

C. At r = a

Following from equation 1:

E(r) = 0 for r≤a. Equation 1

The electric field at the surface of the solid sphere is 0, E = 0N/C

Thus,

V = Ed = 0 V

Consequently, the electric potential at the solid sphere's surface is 0.

D. At r = 0

The electric potential can be determined by

V = kq / r

As r approaches 0,

V = kq / 0

V approaches infinity.

8 0

3 months ago