An oil pump is drawing 44kW while pumping oil with a density of 860 kg/m3 at a rate of 0.1 m3/s. The inlet and outlet diameters of the pipe are 8cm and 12cm, respectively. If the pressure increases by 500kPa going through the pump and the motoreffi ciency is 90%, determine the mechanical efficiency of the pump

Answer:

The value is [tex]D = 205.8 \ m [/tex]

Explanation:

The time taken for the column to take a step mathematically represented as

100 steps => 1 minutes => 60 seconds

1 step => t

=> [tex]t = 0.6 \ s [/tex]

Generally the length of the column is mathematically represented as

[tex]D = v * t[/tex]

substituting 343 m/s for v we have

        [tex]D =  343 * 0.6 [/tex]

  =>    [tex]D =  205.8 \  m  [/tex]

Answer:

total work done = -5960.8 J

Explanation:

given data

initial volume v1 = 0.22 m³

initial pressure  p1 = 86 kPa

final pressue p2 = 101.3 kPa

solution

we apply here  isothermal expansion that is express as

p1 × v1 = p2 × v2    ......................1

put here value

86 × 0.22 = 101.3 × v2

v2 = 0.1867 m³  

and

work done will be here

w1 = p1 × v1  × ln([tex]\frac{p1}{p2}[/tex])      ....................2

w1 = 86 × 10³ × 0.22 × [tex]ln(\frac{86}{101.3})[/tex]

w1 = -3.097  × 10³ J

and

it is cooled to initial volume at constant pressure  so here work done will be

w2 = p(v2 - v1)    .................3

w2 =  86 × 10³ × ( 0.1867 - 0.22 )

w2 = -2863.8 J

so

total work done is

total work done = w1 + w2

total work done = -3097 +  -2863.8

total work done = -5960.8 J

Answer:

10 miles per minute.

Answer:

20.1 m/s

Explanation:

Since You are later than usual getting to the stop and see the shuttle pulling away from the stop while you are still 3.9 m behind the bus stop. And In 40.9 m you will reach a barrier and you must catch the shuttle before that point.

Given that the shuttle has a constant acceleration of 4.5 m/s2. 

The total distance to cover is:

Total distance = 40.9 + 3.9 = 44.8 m

Assuming you are starting from rest. Then initial velocity U = 0

Using the 3rd equation of motion to calculate the minimum velocity.

V^2 = U^2 + 2as

V^2 = 0 + 2 × 4.5 × 44.8

V^2 = 403.2

V = sqrt (403.2)

V = 20.1 m/s

Therefore, the minimum velocity you have to run at to catch the bus before it reaches the barrier is 20.1 m/s

Answer:

[tex]x=119m[/tex]

Explanation:

Hello,

In this case, since the hockey puck was moving at 34 m/s and suddenly stopped (final velocity is zero) in 7 seconds, we can first compute the acceleration via:

[tex]a=\frac{v_f-v_o}{t}=\frac{0m/s-34m/s}{7s}\\ \\a=-4.86m/s^2[/tex]

In such a way, we can compute the displacement via:

[tex]x=\frac{v_f^2-v_o^2}{2a}\\ \\x=\frac{0^2-(34m/s)^2}{2*-4.86m/s^2}\\ \\x=119m[/tex]

Best regards.

Answer:

 y = y₀ (1 - ½ g y₀ / v²)

Explanation:

This is a free fall problem. Let's start with the ball that is released from the window, with initial velocity vo = 0 and a height of the window i

          y = y₀ + v₀ t - ½ g t²

          y = y₀ - ½ g t²

for the ball thrown from the ground with initial velocity v₀₂ = v

         y₂ = y₀₂ + v₀₂ t - ½ g t²

in this case y₀ = 0

         y₂2 = v t - ½ g t²

at the point where the two balls meet, they have the same height

         y = y₂

         y₀ - ½ g t² = vt - ½ g t²

         y₀i = v t

         t = y₀ / v

since we have the time it takes to reach the point, we can substitute in either of the two equations to find the height

         y = y₀ - ½ g t²

         y = y₀ - ½ g (y₀ / v)²

         y = y₀ - ½ g y₀² / v²

        y = y₀ (1 - ½ g y₀ / v²)

with this expression we can find the meeting point of the two balls

Answer:

F=200N

a=500m/s2​

Mass=?

Explanation:

F=ma

200=m*500

200/500=m

Mass=0.4kg

Answer:

C. Muddy Water

Answer:

   ρ = 4.4 10⁻⁸ Ω m

Explanation:

For this exercise let's start by finding the value of the resistance using Ohm's law

          V = I R

           R = V / I

The voltage is related to the electric field

          V = E L

let's substitute

          R = E L / I

          R = 8.2 10⁻² / 20  

          R = 4.1 10⁻³ L

now we can use the resistance relation

          R = ρ L / A

the area of ​​a circular wire is

          A = π r² = π d² / 4

         ρ = R π d² / (4 L)

let's calculate

        ρ = (4.1 10⁻³ L) π 0.0037² / (4 L) = (4.1 10⁻³) π 0.0037² / (4)

        ρ = 4.4 10⁻⁸ Ω m

The resistivity of the wire is 4.4 10⁻⁸ Ω m.

We know that;

V = I R

R = V / I

The voltage  the electric field can be connected using the formula;

V = E L

Hence;

R = E L /I

R = 8.2 10⁻² / 20  

R = 4.1 10⁻³ L

For resistivity;

R = ρ L / A

the area can be obtained from;

A = π r² = π d² / 4

ρ = R π d² / (4 L)

ρ = (4.1 10⁻³ L) π 0.0037² / (4 L) = (4.1 10⁻³) π 0.0037² / (4)

ρ = 4.4 10⁻⁸ Ω m

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Answer:

8.6602 tons

Explanation:

We first draw the known vector forces.

2fcos30⁰

We have f to be equal to 5tons

Inserting into formula

Σfx = 2(5)cos30⁰

= 8.6602 tons

Σfy is equal to 0, this is because in the y direction, the forces cancel themselves out.

Therefore the resultant force on the ship is equal to 8.6602 tons

I hope this helps!

Please check attachment for diagram.

Answer:

The tension in the light rope must be equal to the weight of the bucket

Explanation:

Given that,

Constant velocity of bucket and direction of bucket in downward

We need to find the tension in the rope

Using given data,

When a bucket  moves downward with a constant velocity then the net force does not applied on the bucket.

So, The weight of the bucket will be equal to the tension in the light rope

In mathematically,

[tex]T=mg[/tex]

Where, T = tension

m = mass of bucket

g = acceleration due to gravity

Hence, The tension in the light rope must be equal to the weight of the bucket.

Answer:

The value is [tex]v_s = 1.394 \ m/s[/tex]

Explanation:

From the question we are told that

The frequency of the second player is [tex]f_2 = 490 \ Hz[/tex]

The beat frequency is [tex]f_b = 2.0 \ Hz[/tex]

The speed of sound is [tex]v_s = 343 \ m/s [/tex]

Generally the frequency of the note played by the first player is mathematically represented as

[tex]f_1 = f_2 + f_b[/tex]

=> [tex]f_1 = 490 + 2.0 [/tex]

=> [tex]f_1 = 492 Hz[/tex]

From the relation of Doppler Shift we have that

[tex]f_1 = \frac{ f_2 (v+ v_o )}{v-v_s }[/tex]

Here [tex] v_o\ is\ the\ velocity\ of\ the\ observer\ with\ value\ 0 \ m/s [/tex]

So

[tex]492 = \frac{ 490 (343+0 )}{343 -v_s }[/tex]

=> [tex]v_s = 1.394 \ m/s[/tex]

Explanation:

Equation for Impact

FΔt = ΔP,

F = force

Δt  = Impact of time

ΔP = Change in momentum

Car steering is engineered to fail in order to maximize the time of contact and hence reduce the initial impact and mitigate the damage incurred.

Road guard railing crumple on contact to maximize impact time and hence reduce impact intensity and mitigate damage.

Road safety containers are loaded with liquid or sand as they improve the period of impact.

Explanation:

13+x=43>2x

13+x=43>2x+43>2x

13+x=86>4x

x-4x=86-13

3x=73

x=73/3

x=24.333

x=24.4

Answer:

Explanation:

If Two cylindrical resistors are made from the same material, then their resistivity will be the same.  Formula for calculating resistivity of a material is expressed as;

[tex]\rho = \frac{RA}{L} \ where \ A = \frac{\pi d^2}{4}[/tex] where;

R is the resistance

A is the cross sectional area of the material

L is the length of the material

For the shorter cylinder:

Length = L

diameter = D

[tex]\rho = \dfrac{R_s(\frac{\pi D^2}{4})}{L} \\\\\rho = \dfrac{R_s{\pi D^2}}{4L}[/tex]

For the longer cylinder:

Length = 16L

diameter = 4D

[tex]\rho = \dfrac{R_l(\frac{\pi (4D)^2}{4})}{16L} \\\\\\\rho = \dfrac{R_l(\frac{\pi (16D^2)}{4})}{16L} \\\\\rho = \dfrac{R_l{16\pi D^2}}{16L}\\\\\rho = \dfrac{R_l{\pi D^2}}{L}[/tex]

Since their resistivity are the same then;

[tex]\dfrac{R_s{\pi D^2}}{4L} = \dfrac{R_l{\pi D^2}}{L} \\\\ \dfrac{R_s}{4} = {R_l} \\\\R_s = 4R_l\\\\R_l = \frac{R_s}{4}[/tex]

Hence the resistance of the longer resistor is a quarter of the shorter resistor.

Answer:

The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].

Explanation:

We must notice that basketball depicts a parabolic motion, which consists of combining a constant speed motion in x-direction and free fall motion in the y-direction. The motion is described by the following kinematic formulas:

x-Direction

[tex]x = x_{o} + v_{o}\cdot t \cdot \cos \alpha[/tex]

y-Direction

[tex]y = y_{o} + v_{o}\cdot t\cdot \sin \alpha +\frac{1}{2}\cdot g\cdot t^{2}[/tex]

Where:

[tex]x_{o}[/tex], [tex]y_{o}[/tex] - Initial position of the basketball, measured in meters.

[tex]x[/tex], [tex]y[/tex] - Final position of the basketball, measured in meters.

[tex]v_{o}[/tex] - Initial speed of the basketball, measured in meters per second.

[tex]t[/tex] - Time, measured in seconds.

[tex]\alpha[/tex] - Tilt angle, measured in sexagesimal degrees.

[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.

If we know that [tex]x_{o} = 0\,m[/tex], [tex]y_{o} = 2.20\,m[/tex], [tex]\alpha = 36^{\circ}[/tex], [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], [tex]x = (9.10\pm0.23)\,m[/tex] and [tex]y = 2.70\,m[/tex], the system of equation is reduce to this:

[tex](9.10\pm 0.23)\,m = 0\,m + v_{o}\cdot t \cdot \cos 36^{\circ}[/tex]

[tex]9.10\pm 0.23 = 0.809\cdot v_{o}\cdot t[/tex] (Ec. 1)

[tex]2.70\,m = 2.20\,m + v_{o}\cdot t \cdot \sin 36^{\circ} + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot t^{2}[/tex]

[tex]0.50 = 0.588\cdot v_{o}\cdot t-4.904\cdot t^{2}[/tex] (Ec. 2)

At first we clear [tex]v_{o}\cdot t[/tex] in (Ec. 1):

[tex]v_{o}\cdot t = \frac{9.10\pm 0.23}{0.809}[/tex]

[tex]v_{o}\cdot t = 11.248\pm 0.284[/tex]

(Ec. 1) in (Ec. 2):

[tex]0.5 = 0.588\cdot (11.248\pm 0.284)-4.904\cdot t^{2}[/tex]

Now we clear the time in the resulting expression:

[tex]4.904\cdot t^{2} = 0.588\cdot (11.248\pm 0.284)-0.5[/tex]

[tex]t = \sqrt{\frac{0.588\cdot (11.248\pm 0.284)-0.5}{4.904} }[/tex]

There are two solutions:

[tex]t_{1} = \sqrt{\frac{0.588\cdot (11.248- 0.284)-0.5}{4.904} }[/tex]

[tex]t_{1} \approx 1.101\,s[/tex]

[tex]t_{2} = \sqrt{\frac{0.588\cdot (11.248+ 0.284)-0.5}{4.904} }[/tex]

[tex]t_{2}\approx 1.131\,s[/tex]

The initial velocity is cleared within (Ec. 2):

[tex]v_{o}=\frac{0.50+4.904\cdot t^{2}}{0.588\cdot t}[/tex]

The bounds of the range of initial speed is determined hereafter:

[tex]t_{1} \approx 1.101\,s[/tex]

[tex]v_{o} = \frac{0.50+4.904\cdot (1.101)^{2}}{0.588\cdot (1.101)}[/tex]

[tex]v_{o} = 9.954\,\frac{m}{s}[/tex]

[tex]t_{2}\approx 1.131\,s[/tex]

[tex]v_{o} = \frac{0.50+4.904\cdot (1.131)^{2}}{0.588\cdot (1.131)}[/tex]

[tex]v_{o} = 10.185\,\frac{m}{s}[/tex]

The range of initial speeds allowed to make the basket is: [tex]9.954\,\frac{m}{s}\leq v \leq 10.185\,\frac{m}{s}[/tex].

Answer:

The critical radius of the plastic insulation is 0.72 inches.

Explanation:

Given that,

Diameter = 0.091 in

Thickness = 0.02 in

Initial temperature = 90°F

Final temperature = 50°F

Heat transfer coefficient = 2.5 Btu/h.ft²°F

Material conductivity = 0.075 Btu/h.ft °F

We need to calculate the critical radius of the plastic insulation

Using formula of critical radius

[tex]r_{cr}=\dfrac{2K}{h}[/tex]

Where, k = Material conductivity

h = Heat transfer coefficient

Put the value into the formula

[tex]r_{cr}=\dfrac{2\times0.075}{2.5}[/tex]

[tex]r_{cr}=0.06\ ft[/tex]

[tex]r_{cr}=0.72\ inches[/tex]

Hence, The critical radius of the plastic insulation is 0.72 inches.

Answer:

The value is [tex]d = 31.45 \ m [/tex]

Explanation:

Generally the relative speed at which you are moving with respect to the person ahead of you is mathematically represented as

[tex]v_r = v_s - v_c[/tex]

substituting 1.05 m/s for [tex] v_c [/tex] and 2.75 m/s for [tex]v_s[/tex]

So

[tex]v_r = 2.75 - 1.05[/tex]

=> [tex]v_r = 1.7 \ m/s [/tex]

Generally the distance by which the person is ahead of you is mathematically represented as

[tex]d = v_r * t[/tex]

substituting 18.5 s for [tex] t [/tex]

       [tex]d =  1.7  * 18.5[/tex]

=>      [tex]d  =  31.45 \  m [/tex]

Answer:

30.56 m/s^2

Explanation:

Given that In order to attain orbit around earth, the ATLAS V rocket must accelerate up to a speed of about 7700 meters per second in about 4.2 minutes.

The average acceleration that is required to accomplish this will be

Average acceleration = change in velocity / time

Average acceleration = 7700/ 4.2 × 60

Average acceleration = 7700/252

Average acceleration = 30.56 m/s^2

Answer:

44 years

Explanation:

Use half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.25 A₀ = A₀ (½)^(t / 22)

0.25 = (½)^(t / 22)

t / 22 = 2

t = 44

44 sec is half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value.

The period of time it takes for one-half of a radioactive isotope to decay is known as the half-life. A given radioactive isotope's half-life is constant; it is unaffected by external factors and independent of the isotope's starting concentration.

Use half life equation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

0.25 A₀ = A₀ (½)^(t / 22)

0.25 = (½)^(t / 22)

t / 22 = 2

t = 44 sec

44 sec is half life of a given sample of radium is 22 years the sample will reduce to 25% of its original value.

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Answer:C

Explanation:

I did the test

a sound wave has a frequency of 85 Hz and a wavelength of 4.2 m. Then the wave speed of the sound wave is 20.23 m/s.  Hence option D is correct.

Wave is is a disturbance in a medium that carries energy as well as momentum . wave is characterized by amplitude, wavelength and phase. Amplitude is the greatest distance that the particles are vibrating. especially a sound or radio wave, moves up and down. Amplitude is a measure of loudness of a sound wave. More amplitude means more loud is the sound wave.

Wavelength is the distance between two points on the wave which are in same phase. Phase is the position of a wave at a point at time t on a waveform. There are two types of the wave longitudinal wave and transverse wave.

The wave speed is given by,

c = νλ

where λ is wavelength, ν is frequency.

Given,

frequency ν = 85 Hz

wavelength λ = 4.2 m

The speed of the wave,

c = 85 × 4.2 = 20.23 m/s

Hence option D is correct.

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Answer:

FALSE

Explanation:

Answer:

0.55 hz

Explanation:

Given that the plane fly through the turbulent boundary layer of the atmosphere at a speed of 50 m/sec. And the velocity fluctuations in the atmosphere are of order 0.5 m/sec, the length scale of the large eddies is about 100 m.

The maximum speed attained will be

Maximum speed = 50 + 0.5 = 5.5 m/s

The Length = 100m

Speed = FL

Where F = frequency

Substitute speed and distance length into the formula

55 = 100F

F = 55/100

F = 0.55 Hz

Therefore, the highest frequency the anemometer will encounter will be 0.55 Hz

Answer:

40079 Hz

Explanation:

1. The first step is to calculate energy dissipation ∈=u^3/l and here u is fluctuating velocity

∈=(0.5^3)/100 = 0.00125 m^2/s^3

2. Find out the length scale of the small eddies

η=(viscosity/∈)^1/4

η=(1.470e-5/0.00125)^1/4 = 0.00126 m

3. The frequency associated with these small-scale eddies will be the greatest frequency the anemometer will encounter, thus:

u_max=f_max * η

u_max = u + u' = 50+0.5=50.5 m/s

f_max = u_max/η = 50.5/0.00126 = 40079 Hz

This is the heighest frequency the anemometer will encounter.

Answer:

Hello!

Sorry you haven't put up an image of your question! Without it we can't answer your question!

Explanation:

Maybe put up another one and it'll be answered!

:D

Answer:

The hight of the building is 38.16 m

Explanation:

These two pieces of information given, first, the watermelon is 30 m  above the ground and after 1.50 s the watermelon has been spotted. Now we are required to find the height of the building.

Use the below formula to find the height of buildings.

S = ut + ½ gt^2

30  =1.5u + (1/2) × 9.8 (1.5)^2

u = 12.65 m/sec

v^2 – u^2 = 2gs

(12.65)^2 = 2×9.8 s’

S’ = 8.16 m  

h = s + s’

h = 30 + 8.16 = 38.16 m

The hight of the building is 38.16 m.

The height of the building is 38.16 m.

Given data:

The height above the ground is, h = 30.0 m.

The time interval after observation of first spot is, t = 1.50 s.

We need to find the height of building. And since two pieces of information given, first, the watermelon is 30 m  above the ground and after 1.50 s the watermelon has been spotted. So, using the second kinematic equation of motion as,

[tex]h = ut + \dfrac{1}{2}gt^{2}[/tex]

Here, u is the initial speed. Solving as,

[tex]30 = (u \times 1.50) + \dfrac{1}{2} \times 9.8 \times (1.50)^{2}\\\\u =12.65 \;\rm m/s[/tex]

Now landing distance (s') is calculated using the third kinematic equation of motion as,

[tex]v^{2} =u^{2}+2(-g)s\\\\0^{2} =12.65^{2}+2(-9.8)s\\\\s = 8.16 \;\rm m[/tex]

Then the height of building is given as,

H = h + s

H = 30 m + 8.16 m

H = 38.16 m

Thus, we can conclude that the height of the building is 38.16 m.

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