PHYSICS COLLEGE

G Water enters a house through a pipe 2.40 cm in diameter, at an absolute pressure of 4.10 atm. The pipe leading to the second-floor bathroom, 5.20 m above, is 1.20 cm in diameter. The flow speed at the inlet pipe is 4.75 m/s a) What is the algebraic expression for flow speed in the bathroom?
b) Calculate the flow speed in the bathroom.
c) What is algebraic expression for the pressure in the bathroom?
d) Calculate the water pressure in the bathroom. Report your answer in the (atm) unit.

Answers

Answer 1

Answer:

A) A₁ V₁ = A₂V₂

B) V₂ = 19 m /s

C) P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg

D) P₂ = 1.88 atm

Explanation:

A) From the piaget's theory of conservation of volume, we can calculate the rate of flow of water from;

A₁ V₁ = A₂V₂

Where;

A₁ and A₂ are area of cross section V₁ and V₂ are velocity of flow at two places along pipe.

B) Using the formula given in A above, we obtain;

π x 1.2² x 4.75 = π x 0.6² x V₂

V₂ x 0.36 = 6.84

V₂ = 6.84/0.36

V₂ = 19 m /s

c ) To find pressure we shall apply Bernoulli's theorem in fluid dynamics;

P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg

Where;

P₁ and P₂ are pressure at ground and second floor respectively

v₁ and v₂ are velocity at ground and second floor respectively

h₁ and h₂ are height at ground and second floor respectively ρ is density of water.

Thus, plugging in the relevant values to obtain;

4.1 x 10⁵ + (1/2 x 1000 x 4.75²) = P₂ + (1/2 x 1000 x 19²) + (5.2 x 1000 x 9.8)

(4.1 x 10⁵) + (0.11 x 10⁵) = P₂ + (1.8 X 10⁵) + (0.51 X 10

P₂ = 1.9 X 10⁵ N/m² = 1.88 atm

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Answers

Answer:

Uses of concave mirror:

Shaving mirrors.

Head mirrors.

Ophthalmoscope.

Astronomical telescopes.

Headlights.

Solar furnaces.

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please give me full points.

You wish to buy a motor that will be used to lift a 10-kg bundle of shingles from the ground to the roof of a house. The shingles are to have a 1.5-m/s2 upward acceleration at the start of the lift. The very light pulley on the motor has a radius of 0.17 m . Part A Determine the minimum torque that the motor must be able to provide. Express your answer with

Answers

To lift a 10-kg bundle of shingles with an upward acceleration of 1.5 m/s², a motor with a pulley of radius 0.17 m must provide a torque of at least 19 N.m.

We wish to use a motor to lift a 10-kg (m) bund of shingles with an upward acceleration of 1.5 m/s² (a).

The resulting force (F = m.a) is the difference between the tension (T) of the pulley and the weight (w = m.g) of the shingles.

$T-m.g=m.a\nT = m.g+m.a = m(g+a) = 10 kg (9.8m/s^(2)+1.5m/s^(2) )=113 N$

where,

g: gravity

Then, we can calculate the minimum torque (τ) that the motor must apply using the following expression.

$\tau = r * T = 0.17m * 113N = 19N.m$

where,

r: radius of the pulley

To lift a 10-kg bundle of shingles with an upward acceleration of 1.5 m/s², a motor with a pulley of radius 0.17 m must provide a torque of at least 19 N.m.

Learn more: brainly.com/question/19247046

Answer:

$\tau=19.21\ N-m$

Explanation:

It is given that,

Mass of bundle of shingles, m = 10 kg

Upward acceleration of the shingles, $a=1.5\ m/s^2$

The radius of the motor of the pulley, r = 0.17 m

Let T is the tension acting on the shingles when it is lifted up. It can be calculated as :

$T-mg=ma$

$T=m(g+a)$

$T=10* (9.8+1.5)$

T = 113 N

Let $\tau$ is the minimum torque that the motor must be able to provide. It is given by :

$\tau=r* T$

$\tau=0.17* 113$

$\tau=19.21\ N-m$

So, the minimum value of torque is 19.21 N-m. Hence, this is the required solution.

A young man walks daily through a gridded city section to visit his girlfriend, who lives m blocks East and nblocks North of where the young man resides. Because the young man is anxious to see his girlfriend, his route to her never doubles back—he always approaches her location. In terms of m and n, how many different routes are there for the young man to take?

Answers

Answer:

The man ate eggs.

Explanation:

He should brush his teeth before seeing his girlfriend.

a mass of .4 kg is raised by a vertical distance of .450 m in the earth's gravitational field. what is the change in its gravitational potential energy

Answers

Answer:

E = 1.76 J

Explanation:

Given that,

Mass of an object, m = 0.4 kg

It moves by a vertical distance of 0.45 m in the Earth's gravitational field.

We need to find the change in its gravitational potential energy. It can be given by the formula as follow :

$E=mgh\n\nE=0.4* 9.8* 0.45\n\nE=1.76\ J$

So, the change in its gravitational potential energy is 1.76 J.

What allows two or more atoms to be held together? shared protons shared electrons shared energy shared neutrons

Answers

It is shared electrons.

The following information should be considered:

In the case when two or more atoms can be together at the time when they share electrons with each other.
By sharing, they create a covalent bond and that way the atoms can be stable.

learn more: brainly.com/question/2514933?referrer=searchResults

Answer:

try electrons i hope this helps!!

Explanation:

A lead ball is dropped into a lake from a diving board 6.10 mm above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 ss after it is released. How deep is the lake?

Answers

Answer:

D=1.54489 m

Explanation:

Given data

S=6.10 mm= 0.0061 m

To find

Depth of lake

Solution

To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation

$S=v_(1)t+(1/2)gt^(2) \n 0.0061m=(0m/s)t+(1/2)(9.8m/s^(2) )t^(2)\n t^(2)=(0.0061m)/(4.9m/s^(2) )\n t=\sqrt{1.245*10^(-3) }\n t=0.035s$

So ball takes 0.035sec to hit the water

As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as

$v_(f)=v_(i)+gt\nv_(f)=0+(9.8m/s^(2) )(0.035s)\n v_(f)=0.346m/s$

Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake

So the depth of lake given as:

$D=|vt|\nD=|0.346m/s*4.465s|\nD=1.54489m$

Answer: d = 1.54m

The depth of the lake is 1.54m

Explanation:

The final velocity of the ball just before it hit the water can be derived using the equation below;

v^2 = u^2 + 2as ......1

Where ;

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled.

Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:

v^2 = 2gs

v = √2gs ......2

g = 9.8m/s^2

s = 6.10mm = 0.0061m

substituting into equation 2

v = √(2 × 9.8× 0.0061)

v = 0.346m/s

The time taken for the ball to hit water from the time of release can be given as:

d = ut + 0.5gt^2

Since u = 0

d = 0.5gt^2

Making t the subject of formula.

t = √(2d/g)

t = √( 2×0.0061/9.8)

t = 0.035s

The time taken for the ball to reach the bottom of the lake from the when it hits water is:

t2 = 4.5s - 0.035s = 4.465s

And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;

depth d = velocity × time = 0.346m/s × 4.465s

d = 1.54m

The depth of the lake is 1.54m

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