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Ber [7]
2 years ago
13

the flow energy of 124 L/min of a fluid passing a boundary to a system is 108.5 kJ/min. Determine the pressure at this point

Physics
1 answer:
Andreyy892 years ago
5 0

Answer:

The pressure at this point is 0.875 mPa

Explanation:

Given that,

Flow energy = 124 L/min

Boundary to system P= 108.5 kJ/min

P=1.81\ kW

We need to calculate the pressure at this point

Using formula of pressure

P=F\times v

P=A_{1}P_{1}\times v_{1}

Here, A_{1}v_{1}=Q_{1}

Where, v = velocity

Put the value into the formula

1.81 =P_{1}\times0.124\times\dfrac{1}{60}

P_{1}=\dfrac{1.81\times60}{0.124}

P_{1}=875.80\ kPa

P_{1}=0.875\ mPa

Hence, The pressure at this point is 0.875 mPa

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A developer wants to build a sprawling two-story office complex. The developer's and architect's vision is something low, modern
Lera25 [3.4K]

Answer:

a. directive zoning

Explanation:

Directive zoning is an instrument used in master plans, whereby the city is divided into areas on which differentiated land use and land use guidelines apply, especially urban indexes. Directive zoning acts primarily by controlling two main elements: the use and size (or size) of lots and buildings. It is therefore assumed that the end result achieved through individual actions is in line with the municipality's objectives, which include proportionality between occupation and infrastructure, the need to protect fragile areas and / or cultural interest, the harmony from the volumetric point of view, etc.

4 0
2 years ago
Anna applies a force of 19.5 newtons to push a book placed on a table. If the normal force of the book is 51.7 newtons, what is
GarryVolchara [31]

that would be given by

[email protected]

@ representing coefficient of kinetic friction.

thus 19.5/51.7 = 0.377

6 0
2 years ago
1) An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and
Lady_Fox [76]

Answer: Capacitor

Explanation:

Given the following ;

At time, t = T/4 ; I = 0

At time, t = T/2 ; I = - Imax

Where T = period

We can confirm what the unknown element is by using the relation, such that the parameters satisfy the stated condition ;

Recall;

I = Iosin(wt + π/2) [for CAPACITOR]

w = 2π/T

At t = T/4

I = Iosin(2π/T(T/4) + π/2)

I = Iosin(π/2 + π/2)

I = Iosin(0)

I = 0

At t = T/2

I = Iosin(2π/T(T/2) + π/2)

I = Iosin(π + π/2)

I = Iosin(0)

I = Iosin(3π/2)

I = Iosin(540/2)

I = Iosin(270)

I = -Io

Note : Io = Imax

Both conditions are met, Hence, the unknown element is a CAPACITOR.

6 0
2 years ago
A soccer ball kicked with a force of 13.5 n accelerates at 6.5 m/s^2 to the right. what is the mass of the ball?
Art [367]

Answer:

2.08 kg

Explanation:

We can solve the problem by using Newton's second law:

F=ma

where

F is the net force acting on an object

m is the mass of the object

a is its acceleration

In this problem, the ball is kicked with a force of F=13.5 N, and its acceleration is a=6.5 m/s^2, therefore we can re-arrange the equation to find the mass of the ball, m:

m=\frac{F}{a}=\frac{13.5 N}{6.5 m/s^2}=2.08 kg

7 0
2 years ago
Read 2 more answers
Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much l
alex41 [277]

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:

a)

The magnitude of the gravitational force between two objects is given by the formula

F=G\frac{m_1 m_2}{r^2}

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A - Star A, we have:

m_1=M_p\\m_2 = M_s\\r=R

So the force is

F_A=G\frac{M_p M_s}{R^2}=F_0

For the system planet B - Star B, we have:

m_1 = 4 M_p\\m_2 = M_s\\r=R

So the force is

F=G\frac{4M_p M_s}{R^2}=4F_0

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C - Star C, we have:

m_1 = M_p\\m_2 = 4M_s\\r=R

So the force is

F=G\frac{M_p (4M_s)}{R^2}=4F_0

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

b)

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

G\frac{mM}{r^2}=m\frac{v^2}{r}

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

v=\sqrt{\frac{GM}{r}}

For System A,

M=M_s\\r=R

So the tangential speed is

v_A=\sqrt{\frac{GM_s}{R}}

For system B,

M=M_s\\r=R

So the tangential speed is

v_B=\sqrt{\frac{GM_s}{R}}=v_A

So, the speed of planet B is the same as planet A.

For system C,

M=4M_s\\r=R

So the tangential speed is

v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A

So, the speed of planet C is twice the speed of planet A.

3 0
2 years ago
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