A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 58.9 cm ( 0.589 m) and the flow speed of the petroleum is 12.1 m/s. At the refinery, the petroleum flows at 6.29 m/s. What is the volume flow rate of the petroleum along the pipe and what is the pipe's diameter at the refinery?

Answer:

eating instrument must be: HARDNESS,  INERT, NOT TOXIC

eating tools: digested by the body

Explanation:

An eating instrument must be able to contain food, so it must have a good HARDNESS, besides it must be poorly absorbent of heat and the most important must be INERT, not react with food or be NOT TOXIC to humans.

Additionally, for a spoon to be edible, it must be able to be digested by the body, in general they are made with a starch base, so that the non-digestible parts of it have not been toxic to the body and can be eliminated from it.

There are numerous features of a spoon but these properties must an edible spoon

have

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Answer:

7.14 s

93.2 m/s, 24.9° below the horizontal

Explanation:

Given in the y direction:

Δy = -30 m

v₀ = 90 m/s sin 20° ≈ 30.8 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

-30 m = (30.8 m/s) t + ½ (-9.8 m/s²) t²

4.9t² − 30.8t − 30 = 0

t = [ 30.8 ± √((-30.8)² − 4(4.9)(-30)) ] / 2(4.9)

t = 7.14 s

Find: vᵧ

v² = v₀² + 2aΔy

vᵧ² = (30.8 m/s)² + 2 (-9.8 m/s²) (-30 m)

vᵧ = -39.2 m/s

The magnitude of the velocity is:

v² = vₓ² + vᵧ²

v² = (90 m/s cos 20°)² + (-39.2 m/s)²

v = 93.2 m/s

The direction of the velocity is:

tan θ = vᵧ / vₓ

tan θ = (-39.2 m/s) / (90 m/s cos 20°)

θ = -24.9°

Answer:

use the distance formula d=v*t

d= distance

v = velocity

t = time = t1 + t2 = 2h +3h = 5 hours (total time given)

At 65 km/h distance (d1) answer:

d1 = (65 km/h)*( 2h) = 130 km

Total Distance:

so d1 + d2 = Dtotal

Dtotal = 130 km + (78 km/h)*3 = 130 km + 234 km = 364 km

Answer:

1)  the two objects reach the floor at the same time.

2)the book reaches the floor much earlier than the foil

In conclusion, the difference in motion between the two systems subjected to the same acceleration depends on the weight of the body and friction force, when the body has less weight, the friction of the air affects it more

Explanation:

This interesting experiment has the following results

1) first case. Sheet on top of book

In this case the two objects reach the floor at the same time.

This shows that the acceleration in the two objects is the same and we call it the acceleration of gravity.

The speed of the body increases as it goes down linearly.

This occurs because the book that receives air resistance is much heavier, so the resistance has almost no effect on its movement, the sheet does not have the air resistance because it goes down next to the book.

2) second case. Book and sheet next to each other.

In this case the book reaches the floor much earlier than the foil.

This is because the resisting force of the air has almost no effect on the book and its movement is little affected by this force.

In the case of the blade, it has very little weight, therefore as its speed increases, the resistance force of the air rapidly equals the weight of the blade.

           W_sheet - fr = 0

so after this, since the acceleration is zero, it goes down at constant speed, this speed is called the terminal velocity.

In conclusion, the difference in motion between the two systems subjected to the same acceleration depends on the weight of the body and friction force, when the body has less weight, the friction of the air affects it more.

Answer:

Some protons align with the field and some align opposite to it.

Explanation:

Majority align to the field because these protons tend to act like small magnets under the effect of this external field

On account of external magnetic field, the protons will align with the magnetic field. Hence, option (a) is correct.

The given problem is based on the concept of magnetic field. The region where the magnetic force is experienced is known as magnetic field. Generally, the protons are the charged entities carrying the positive polarity and are one of the major constituents of modern atomic structure.

Thus, we can conclude that on account of external magnetic field, the protons will align with the field.

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Answer:

kinetic energy = 0.1168 J

Explanation:

From Hooke's law, we know that ;

F = kx

k = F/x

We are given ;

Mass; m = 1.95 kg

Spring stretch; d = x = 0.0865

So, Force = mg = 1.95 × 9.81

k = 1.95 × 9.81/0.0865 = 221.15 N/m

Now, initial energy is;

E1 = mgL + ½k(x - L)²

Also, final energy; E2 = ½kx² + ½mv²

From conservation of energy, E1 = E2

Thus;

mgL + ½k(x - L)² = ½kx² + ½mv²

Making the kinetic energy ½mv² the subject, we have;

½mv² = mgL + ½k(x - L)² - ½kx²

We are given L=0.0325 m

Plugging other relevant values, we have ;

½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 - 0.0325)² - ½(221.15 × 0.0865²)

½mv² = 0.62170875 + 0.3224367 - 0.82734979375

½mv² = 0.1168 J

Answer:

10kg

Explanation:

Let PE=potential energy

PE=196J

g(gravitational force)=9.8m/s^2

h(change in height)=2m

m=?

PE=m*g*(change in h)

196=m*9.8*2

m=10kg

Answer:

20 m/s.

Explanation:

Data obtained from the question include the following:

Initial position (d1) = 20 m

Final position (d2) = 60 m

Initial time (t1) = 1 sec

Final time (t2) = 3 secs

Velocity (v) =?

Next, we shall determine the change in position (Δd). This can be obtained as follow:

Initial position (d1) = 20 m

Final position (d2) = 60 m

Change in position (Δd) =?

Change in position (Δd) = d2 – d1

Change in position (Δd) = 60 – 20

Change in position (Δd) = 40 m

Next, we shall determine the change in time (Δt). This can be obtained as follow:

Initial time (t1) = 1 sec

Final time (t2) = 3 secs

Change in time (Δt) =?

Change in time (Δt) = t2 – t1

Change in time (Δt) = 3 – 1

Change in time (Δt) = 2 secs.

Finally, we shall determine the velocity of the object as follow:

Change in position (Δd) = 40 m

Change in time (Δt) = 2 secs.

Velocity (v) =?

Velocity (v) = Change in position (Δd) /Change in time (Δt)

v = Δd/Δt

v = 40/2

v = 20 m/s

Therefore, the velocity of the object between 1 and 3 secs is 20 m/s

Answer:

0.96 m

Explanation:

First, convert km/h to m/s.

162.3 km/h × (1000 m/km) × (1 hr / 3600 s) = 45.08 m/s

Now find the time it takes to move 20 m horizontally.

Δx = v₀ t + ½ at²

20 m = (45.08 m/s) t + ½ (0 m/s²) t²

t = 0.4436 s

Finally, find how far the ball falls in that time.

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.4436 s) + ½ (-9.8 m/s²) (0.4436 s)²

Δy = -0.96 m

The ball will have fallen 0.96 meters.

Answer:

1840:1

Explanation:

If m is the mass of the electron, and 1840m is the mass of the proton, then:

p₁ = p₂

m₁v₁ = m₂v₂

m v₁ = 1840m v₂

v₁ = 1840 v₂

The kinetic energy of the electron is:

KE₁ = ½ m₁ v₁²

KE₁ = ½ m (1840 v)²

KE₁ = 1692800 mv²

The kinetic energy of the proton is:

KE₂ = ½ m₂ v₂²

KE₂ = ½ (1840m) v₂²

KE₂ = 920 mv²

The ratio of the kinetic energies is:

KE₁ / KE₂

(1692800 mv²) / (920 mv²)

1840:1

Answer:

0.228 nm

Explanation:

Atomic mass number of copper = 64

but an atomic mass unit = 1.66 x 10^-27 kg

therefore, the mass of the copper atom m = 64 x 1.66 x 10^-27 kg = 1.06 x 10^-25 kg

The number of atoms in this mass n = ρ/m

where ρ is the density of copper = 8.96 x 10^3 kg/m^3

==> n = (8.96 x 10^3)/(1.06 x 10^-25) = 8.45 x 10^28 atoms/m^3

We know that the volume occupied by this amount of atoms n = [tex]a^{3}[/tex]

where a is the lattice constant

equating, we have

8.45 x 10^28 = [tex]a^{3}[/tex]

a = 4.389 x 10^9

we also know that

d =  1/a

where d is the smallest distance between the two copper atom.

d = 1/(4.389 x 10^9) = 2.28 x 10^-10 m

==> 0.228 nm

Answer:

They have different wavelengths.

They have different frequencies.

They propagate at different speeds through non-vacuum media depending on both their frequency and the material in which they travel.

Explanation:

The complete question is

Consider the following:

a) radio waves emitted by a weather radar system to detect raindrops and ice crystals in the atmosphere to study weather patterns;

b) microwaves used in communication satellite transmissions;

c) infrared waves that are perceived as heat when you turn on a burner on an electric stove;

d) the multicolor light in a rainbow;

e) the ultraviolet solar radiation that reaches the surface of the earth and causes unprotected skin to burn; and

f) X rays used in medicine for diagnostic imaging.

Which of the following statements correctly describe the various forms of EM radiation listed above?

check all that apply to the above

They have different wavelengths.

They have different frequencies.

They propagate at different speeds through a vacuum depending on their frequency.

They propagate at different speeds through non-vacuum media depending on both their frequency and the material in which they travel.

They require different media to propagate.

All the above phenomena are due the electromagnetic wave spectrum. Electromagnetic waves travel at a constant speed of 3 x 10^8 m/s in a vacuum. Within the spectrum, the different types of electromagnetic waves exists in different band range of frequencies and wavelengths unique to each of the waves, and the energy they carry. When these waves enter a non-vacuum medium, their speed change, depending on the nature of the material of the medium, and the frequency or the wavelength of the incoming wave.

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, [tex]P_{gauge \ pressure}[/tex] = 2.95 atm

Atmospheric pressure, [tex]P_{atm}[/tex] = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

[tex]P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2[/tex]

where;

P₁ = [tex]P_{gauge \ pressure} + P_{atm \ pressure}[/tex]

P₂ = [tex]P_{atm}[/tex]

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

[tex]P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 = P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 = P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + \rho gz_1 = \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} + \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} + \rho gz_1)}{\rho} }[/tex]

where;

[tex]\rho[/tex] is the density of seawater = 1030 kg/m³

[tex]v_2 = \sqrt{ \frac{2(2.95*101325 \ + \ 1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s[/tex]

Therefore, the water is flowing at the rate of 28.04 m/s.

Answer:

C. effusion because there is a movement of a gas through a small opening into a larger volume

Explanation:

Edge2020

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Answer:

400 N

Explanation:

Pressure is equal on both pistons.

P = P

F / A = F / A

F / (πd²/4) = F / (πd²/4)

F / d² = F / d²

1600 N / (8.0 cm)² = F / (4.0 cm)²

F = 400 N

The force that should be applied to the smaller piston is 400 N.

Given that,

Based on the above information, the calculation is as follows:

[tex]1600 N \div (8.0 cm)^2 = F \div (4.0 cm)^2[/tex]

F = 400 N

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Answer:

 n (a sin θ) =  m λ₀

n> 1, therefore the fringes move away from each other

Explanation:

The diffraction experiment the constructive interference fringes is described by

          a sin θ = m λ₀

in this equation it is assumed that the experiment emptied the air n = 1

When the same experiment is performed in water, the wavelength changes

           λₙ = λ₀ / n

execution for constructive interference

            a sin θ = m λₙ

we substitute

           a sin θ = m λ / n

           n (a sin θ) =  m λ₀

the refractive index of water is n = 1.33, so for the same wavelength the separation of the spectrum is multiplied by n> 1, therefore the fringes move away from each other

Answer:

32.1

Explanation:

NOTE: You did not state the angle of incidence, and thus, I will be using 45° as my angle of incidence, all you need to do is replace it with your own value if it's different.

To solve this question, we are going to be using Snell's Law.

Snell's law describes the relationship between the angles of incidence and refraction, when referring to light or other waves passing through a boundary between two different isotropic media, such as water, glass, or air.

Snell's law is mathematically given as

sin(A1)/sin(A2) = n2/n1, where

n1 = incidence index

n2 = refracted index

A1 = incidence angle

A2 = refracted angle

The refraction index of oil is 1.15, and that of water is 1.33, so

if we take oil first,

sin A2 = (n1.sinA1)/n2

sin A2 = (1 * sin 45)/1.15

sin A2 = 0.7071/1.15

sin A2 = 0.6149

A2 = sin^-1 0.6149

A2 = 37.9°

Then

sin A3 = (1.15 * sin 37.9) / 1.33

sin A3 = (0.6149 * 1.15) / 1.33

sin A3 = 0.7071 / 1.33

sin A3 = 0.5317

A3 = sin^-1 0.5317

A3 = 32.1

Answer:

The velocity of the particle will be downward.

Explanation:

Given that,

The velocity of a particle is v₁. It is accelerated from 1 to 2.

If it moves away from point 2 on its way then

We need to find the velocity of particle

According to figure,

A particle moves downward from 1 to 2 with the velocity v₁ and after that the particle moves downward from 2 to 3 with the velocity v₂.

Hence, The velocity of the particle will be downward.

Answer:

Statement 1 and statement 2 are correct and statement 2 is the correct explanation of statement 1

Explanation:

Both the velocity and kinetic energy of a gas molecule depends on its relative molecular mass according to Graham's law of diffusion in gases. Hence, the greater the relative molecular mass of the gas, the lesser its average velocity and kinetic energy.

Hence we can see that statement 2 vividly explains the postulation of statement 1 and makes the points more easily comprehensible.

Answer:

Power, P = 307.31 W

Explanation:

It is given that,

Kasek climb at an angle of 6° with the horizontal at a steady speed of 4.0 m/s.

The total mass of bicycle and Kasek is 75 kg

We need to find the Kasek's power output to climb the same hill at the same speed. The angle is made with the horizontal. It means that,

F = F sinθ

So,

Power output is given by :

[tex]P=mg\sin\theta\times v\\\\P=75\times 9.8\times \sin(6)\times 4\\\\P=307.31\ W[/tex]

So, Kasek's power output to climb the same hill is 307.31 W.

Explanation:

T = 2π√(m/k)

T = 2π√(16 kg / 4 N/m)

T = 4π s

T ≈ 12.6 s

C.) Alpha Centauri

Explanation:

Due to it being the closest planetary system to earth.

About 4.367 light years away.

Answer: not enough information given

Explanation:

Explanation:

Density = mass / volume

ρ = 11.1 g / 44.5 mL

ρ = 0.249 g/mL

Answer:

Gravitational potential energy is converted into kinetic energy

Explanation:

During an avalanche, the  gravitational potential

energy of the snow on the mountain is converted into

kinetic energy as the snow cascades down.

The potential energy stored by the snow collected high in the mountain under the gravitational field created by our Earth, breaks loose and as it comes down acquiring velocity, it is converted into kinetic energy due to its accelerated motion

Answer:

The force of the radiation on the surface is  3.33 X 10⁻¹⁰ N

Explanation:

Given;

intensity of light, I = 1 kw/m²

area of the surface, A = 1 cm² = 1 x 10⁻⁴ m²

Power of the incident light, P = I x A

Power of the incident light, P = (1 kw/m²) x (1 x 10⁻⁴ m²)

Power of the incident light, P = 1 x 10⁻⁴ kW = 0.1 W

Power of the incident light is given by;

P = Fv

where;

F is the force of the radiation on the surface

v is the speed of light = 3 x 10⁸ m/s

F = P/ v

F = (0.1) / (3 x 10⁸)

F = 3.33 X 10⁻¹⁰ N

Therefore, the force of the radiation on the surface is  3.33 X 10⁻¹⁰ N

Answer:

is the equation for magnetic force on a length l of wire carrying a current I in a uniform magnetic field B, as shown in Figure 2. If we divide both sides of this expression by l, we find that the magnetic force per unit length of wire in a uniform field is F l=IBsinθ.

Explanation:

Answer:

The  value is  [tex]d = 0.000293 \ m[/tex]

Explanation:

From the question we are told that

   The wavelength is  [tex]\lambda = 633 \ nm = 633 *10^{-9} \ m[/tex]

    The  distance of the screen is  [tex]D = 2.5 \ m[/tex]

    The  order of the bright fringes is  [tex]n = 10[/tex] (10  fringe + central maximum = eleven bright fringes )

      The distance between the fringe is  [tex]y = 54 \ mm = 0.054 \ m[/tex]

Generally the condition for constructive interference is  

        [tex]d sin \theta = n * \lambda[/tex]

=>     [tex]d = \frac{n * \lambda}{sin \theta}[/tex]

Now  from the SOHCAHTOA rule the angle  [tex]sin \theta[/tex] is mathematically represented as

      [tex]sin (\theta) = \frac{y}{D}[/tex]

So  

          [tex]d = \frac{n * \lambda}{\frac{y}{D} }[/tex]

=>       [tex]d = \frac{10 * 633 *10^{-9}}{\frac{0.054}{ 2.5} }[/tex]

=>      [tex]d = 0.000293 \ m[/tex]

Answer:

Explanation:

Magnitude of force per unit length of wire on each of wires

= μ₀ x 2 i₁ x i₂ / 4π r    where i₁ and i₂ are current in the two wires , r is distance between the two and  μ₀ is permeability .

Putting the values ,

force per unit length = 10⁻⁷ x 2 x i x 2i / ( 6 x 10⁻³ )

= .67 i² x 10⁻⁴

force on 3 m length

= 3 x .67 x 10⁻⁴ i²

Given ,

8 x 10⁻⁶ = 3 x .67 x 10⁻⁴ i²

i²  = 3.98 x 10⁻²

i = 1.995 x 10⁻¹

= .1995

=  0.2 A approx .

2 i = .4 A Ans .

The greater current is 0.4 A

Since the two long wires are parallel, the magnetic force, F on each wire is given by

F = μ₀I₁I₂L/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₁ = current in first wire, I₂ = current in second wire, L = length of section of wires = 3.0 m and d = separation distance of the wires = 6.0 mm = 6.0 × 10⁻³ m.

Given that F = 8.0 μN = 8.0 × 10⁻⁶ N and I₂ = 2I₁ (the current in one wire is twice the current in the other wire), we have

F = μ₀I₁I₂L/2πd

F = μ₀ 2I₁I₁L/2πd

F = μ₀I₁²L/πd

Since we require the current in the wire, we make I₁ subject of the formula.

So, I₁ = √(Fπd/μ₀L)

Substituting the values of the variables into the equation, we have

I₁ = √(Fπd/μ₀L)

I₁ = √[8.0 × 10⁻⁶ N × π × 6.0 × 10⁻³ m/(4π × 10⁻⁷ H/m × 3.0 m)]

I₁ = √[48.0π × 10⁻⁹ Nm/12π × 10⁻⁷ H]

I₁ = √[4 × 10⁻² Nm/H]

I₁ = 2 × 10⁻¹ A

I₁ = 0.2 A

Since I₂ is the greater current and I₂ = 2I₁,

I₂ = 2 × 0.2 A

I₂ = 0.4 A

So, the greater current is 0.4 A

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Answer:

160 Hz

Explanation:

Fundamental frequency, [tex]f_{0}[/tex], is the lowest frequency that can be obtained from the stretched string. While higher frequencies are termed harmonics or overtones.

Since the string has three equal segments, the frequency generated, [tex]f_{2}[/tex], is the second harmonic but third overtone.

From the relationship between [tex]f_{0}[/tex] and [tex]f_{2}[/tex], we have;

                          [tex]f_{2}[/tex] = 3[tex]f_{0}[/tex]

⇒                     480 = 3[tex]f_{0}[/tex]

                           [tex]f_{0}[/tex]  = 160

The fundamental frequency of vibration for the string is 160 Hz.

Answer:

initial velocity (u) = 200m/s

final velocity (v) = 350 m/s

time (t) = 15s

acceleration (a) = ?

NOW,

a=v-u/t

a= 350-200/15

a= 50/15

a= 3.3333

Explanation:

it's too easy just u need to understand the question . and go according to it's content .

main thing to memorize is it's simple formula.

I HAVE SOLVE THIS QNA. IN VERY

SIMPLE AND UNDERSTANDABLE FORM.

I høpë u hađ uņdērstøöď ťhìs şølutîóñ

:verý ×wəłł.

Explanation:

Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

Initially, both the astronaut and the tool are at rest, so u₁ = u₂ = 0 m/s.

After throwing the tool, the tool has a velocity of v₂ = 3.20 m/s.

(68.5 kg) (0 m/s) + (2.25 kg) (0 m/s) = (68.5 kg) v + (2.25 kg) (3.20 m/s)

0 = (68.5 kg) v + 7.2 kg m/s

v = -0.105 m/s

The astronaut moves at a speed of 0.105 m/s in the opposite direction.

The speed and the direction of the astronaut is 0.105 m/s in opposite direction to the motion of the tool.

Note: The momentum of the astronaut is equal and opposite to the momentum of the tool

To calculate the speed and the direction of the astronaut, we use the formula below.

Formula:

Where:

make V the subject of the equation

From the question,

Given:

Substitute these values into equation 2

Hence, The speed and the direction of the astronaut is 0.105 m/s in opposite direction to the motion of the tool.

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