Answer:
0,0560 cal / gºC.
The mass flow rate through a centrifugal compressor is 1 kg/s. If air enters at 1 bar and 288k and leaves at 200 kN/m² and 370k, determine the power of the compressor. Take Cp = 1.103 kJ (kg.K), R = 287 kJ (kg.k)
We have that the power of the compressor is
[tex]H_p=24.242hp[/tex]
From the question we are told that:
Flow rate [tex]W=1kg/s[/tex]
inlet Pressure [tex]P_1=1 bar[/tex]
inlet Temperature [tex]T_1= 288k[/tex]
Outlet Temperature [tex]T_2= 370k[/tex]
Outlet Pressure [tex]P_2=200 kN/m^2=2bars[/tex]
[tex]Cp = 1.103 kJ[/tex]
[tex]R = 287 kJ (kg.k)[/tex]
Generally, the equation for Adiabatic head is mathematically given by
[tex]H=\frac{ZRT_1}{Cp-1/K}[\frac{P_2}{P_1}^{(Cp-1)/Cp}-1][/tex]
Where
[tex]Z=Compressibility\ factor[/tex]
[tex]Z=0.99[/tex]
Therefore
[tex]H=\frac{(0.99)(287)(288)}{(1.103)-1/(1.103)}[\frac{(200)}{(1)}^{((1.103)-1)/(1.103)}-1][/tex]
[tex]H=560925.5958 J/kg[/tex]
[tex]H=5.6*10^5J/kg[/tex]
Generally, the equation for centrifugal compressor power is mathematically given by
[tex]H_p=\frac{WH}{E*33000}[/tex]
Where
E is efficiency (adiabatic)
[tex]E=70\%=0.7[/tex]
Therefore
[tex]H_p=\frac{(1)(5.6*10^5)}{0.7*33000}[/tex]
[tex]H_p=24.242hp[/tex]
In conclusion
The power of the compressor is
[tex]H_p=24.242hp[/tex]
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Consider two closely spaced and oppositely charged parallel metal plates. The plates are square with sides of length L and carry charges Q and -Q on their facing surfaces. What is the magnitude of the electric field in the region between the plates
Answer:
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
Explanation:
In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector
E_ {total} = E₁ + E₂
E_ {total} = 2 E
where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction
to calculate the field created by a plate let's use Gauss's law
Ф = ∫ E . dA = q_{int} /ε₀
As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.
E 2A = q_{int} / ε₀
where the 2 is due to the surface has two faces
indicate that the surface has a uniform charge for which we can define a surface density
σ = q_{int} / A
q_{int} = σ A
we substitute
E 2A = σ A /ε₀
E = σ / 2ε₀
therefore the total field is
E_ {total} = σ /ε₀
let's substitute the density for the charge of the whole plate
σ= Q / L²
E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]
You attach a 2.30 kg weight to a horizontal spring that is fixed at one end. You pull the weight until the spring is stretched by 0.500 m and release it from rest. Assume the weight slides on a horizontal surface with negligible friction. The weight reaches a speed of zero again 0.400 s after release (for the first time after release). What is the maximum speed of the weight (in m/s)
Answer: [tex]3.92\ m/s[/tex]
Explanation:
Given
Mass of the attached object is [tex]m=2.3\ kg[/tex]
Spring is stretched by [tex]A=0.5\ m[/tex]
Speed reaches zero after [tex]t=0.4\ s[/tex]
Speed is zero at the extremities of the S.H.M motion that is
[tex]\Rightarrow \dfrac{T}{2}=0.4\\\\\Rightarrow T=0.8\ s[/tex]
Time period of motion is [tex]0.8\ s[/tex] which can also be given by
[tex]\Rightarrow \omega T=2\pi\\\\\Rightarrow \omega=\dfrac{2\pi }{T}\\\\\Rightarrow \omega =\dfrac{2\pi }{0.8}\\\\\Rightarrow \omega=\dfrac{5\pi }{2}[/tex]
Maximum speed for S.H.M. is [tex]v_{max}=A\omega[/tex]
[tex]\Rightarrow v_{max}=0.5\times 2.5\pi\\\Rightarrow v_{max}=3.92\ m/s[/tex]
The resistance of the light bulb changed as the voltage (and current) changed. Why does this resistance change occur?
A 50-turn coil has a diameter of . The coil is placed in a spatially uniform magnetic field of magnitude so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf, , induced in the coil if the magnetic field is reduced to zero unfiformly
Answer:
EMF = 51.01 Volt
Explanation:
A 50-turn coil has a diameter of 15 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.500~\text{T}0.500 T so that the plane of the coil makes an angle of 30^\circ30 ∘ with the magnetic field. Find the magnitude of the emf induced in the coil if the magnetic field is reduced to zero uniformly in 0.100~\text{s}0.100 s
We have,
Number of turn in the coil, N = 50
The diameter of the coil, d = 15 cm
Radius, r = 7.5 cm = 0.075 m
Initial magnetic field, [tex]B_i=0.5\ T[/tex]
The plane of the coil makes an angle of 30° with the magnetic field.
The magnetic field reduced to zero in 0.1 seconds
We need to find the emf induced in the coil. We know that, emf is equal to the rate of change of magnetic flux. So,
[tex]\epsilon=\dfrac{BNA\cos\theta}{t}\\\\\epsilon=\dfrac{0.5\times 50\times \pi \times 0.075\cos(30)}{0.1}\\\\\epsilon=51.01\ V[/tex]
So, the induced emf in the coil is 51.01 V.
Definition of distance in physics
Answer:
Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.
Explanation:
Hey guys,I hope u r gonna answer this question fast,SI system is extended from of MKS system.Why? I will be waiting for the answer. Good luck thank u
Answer:
Because SI system has fundamental units of MKS System
Answer:
Explanation: the unit of length ,mass , and time are same in both the system , thus, the SI system is the extended from of MKS system.
A 10 n force is applied horizontally on a box to move it 10 m across a frictionless surface. How much work was done to move the box?
Answer:
[tex]\boxed {\boxed {\sf 100 \ J}}[/tex]
Explanation:
We are asked to calculate the work done to move a box.
Work is the product of force and distance or displacement.
[tex]W= F*d[/tex]
A 10 Newton force is applied horizontally on the box. Since the surface is frictionless, there is no force of friction, and the net force is 10 Newtons. The force moves the box 10 meters.
F= 10 N d= 10 mSubstitute the values into the formula.
[tex]W= 10 \ N * 10 \ m[/tex]
Multiply.
[tex]W= 100 \ N*m[/tex]
Let's convert the units. 1 Newton meter is equal to 1 Joule, therefore our answer of 100 Newton meters is equal to 100 Joules.
[tex]W= 100 \ J[/tex]
100 Joules of work was done to move the box.
Liquid plastic is frozen in a physical change that increases its volume. What can be known about the plastic after the change?
(A) Its mass will increase.
(B) Its density will increase.
(C) Its mass will remain the same.
(D) Its density will remain the same.
Answer:
c
Explanation:
Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is option C.
What is the matter?Anything which has mass and occupies space is known as matter ,mainly there are four states of matter solid liquid gases, and plasma.
These different states of matter have different characteristics according to which they vary their volume and shape.
It is known about plastic that its mass will remain the same when liquid plastic is frozen, by increasing its volume.
Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is C.
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A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this tim
moves a distance of
meters.
Answer:
400m
Explanation:
[tex]x = v_{0}t + \frac{at^{2} }{2} \\x = 20*10+\frac{4*10x^{2} }{2} = 400m[/tex]A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be:_______.
A. in the positive y direction
B. in the negative y direction
C. in the positive x direction
D. in the negative x direction
A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be in the negative y direction. Thus, option B is correct.
To determine the force that could generate a torque in the negative z direction, we need to consider the right-hand rule for cross products. The torque vector, denoted by τ, is given by the cross product of the position vector, r, and the force vector, F:
[tex]τ = r × F[/tex]
In this case, the position vector, r, points along the positive x-axis. The negative z-direction torque indicates that the force vector must be perpendicular to both the position vector and the negative z-axis.
Using the right-hand rule, we can determine that the force vector must be in the negative y-direction, which is option B.
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During the data transmission there are chances that the data bits in the frame might get corrupted. This will require the sender to re-transmit the frame and hence it will increase the re-transmission overhead. By considering the scenarios given below, you have to choose whether the packets should be encapsulated in a single frame or multiple frames in order to minimize the re-transmission overhead.
Justify your answer with one valid reason for both the scenarios given below.
Scenario A: Suppose you are using a network which is very prone to errors.
Scenario B: Suppose you are using a network with high reliability and accuracy.
1. Based on Scenario A, multiple frames will minimize re-transmission overhead and should be preferred in the encapsulation of packets.
2. Based on Scenario B, the encapsulation of packets should be in a single frame because of the high level of network reliability and accuracy.
Justification:
There will not be further need to re-transmit the packets in a highly reliable and accurate network environment, unlike in an environment that is very prone to errors. The reliable and accurate network environment makes a single frame economically better.
Encapsulation involves the process of wrapping code and data together within a class so that data is protected and access to code is restricted.
With encapsulation, each layer:
provides a service to the layer above itcommunicates with a corresponding receiving nodeThus, in a reliable and accurate network environment, single frames should be used to enhance transmission and minimize re-transmission overhead. This is unlike in an environment that is very prone to errors, where multiple frames should rather be used to minimize re-transmission overhead.
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1 hallar el trabajo mecanico de un cuerpo que tiene una fuerza de 250 newton y recorre 750 metros
2 hallar la potencia necesaria para levantar un transformador de masa 2500kg,una altura de 4 metros en un tiempo de 30 segundos
porfa es para hoy
Answer: TRACK
Explanation:
Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m. What is the work done on the ball by the 34.4 N tension force in the string during one half-revolution of the ball
Answer:
the work done on the ball is 0
Explanation:
Given the data in the question;
Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m.
circle circumference = 2πr = 2.1 m
radius r will be; r = 2.1 m / 2π = 0.33 m
Tension force = 34.4 N
one half revolution means, displacement of the ball is;
d = 2r = 2 × 0.33 = 0.66 m
Now, Work done = force × displacement × cosθ
we know that, the angle between the tension force on string and displacement of object is always 90.
so we substitute
Work done = 34.4 N × 0.66 m × cos(90)
Work done = 34.4 N × 0.66 m × 0
Work done = 0 J
Therefore, the work done on the ball is 0
A chair of weight 85.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 40.0 N directed at an angle of 35.0deg below the horizontal and the chair slides along the floor.
Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.
Answer:
N = 107.94 N
Explanation:
For this exercise we must use Newton's second law.
Let's set a reference system with the x-axis parallel to the ground and the y-axis vertical
X axis
Fₓ = ma
ej and
N -F_y - W = 0
let's use trigonometry to decompose the applied force
cos -35 = Fₓ / F
sin -35 = F_y / F
Fₓ = F cos -35
F_y = F sin -35
Fₓ = 40.0 cos -35 = 32.766 N
F_y = 40.0 sin -35 = -22.94 N
we substitute
N = Fy + W
N = 22.94 + 85
N = 107.94 N
Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth has proven much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate the difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic meter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about 2.0×1092.0×109 pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the magnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second.
Use two significant figures in your answer, expressed in kilometers.
Answer:
26 km
Explanation:
Let's say our "cable" has a cross section of 1 m²
Then each meter of cable would weight 7900(9.8) = 77420 N
A Pascal is a Newton per square meter
2 x 10⁹ / 77420 = 25840 m or about 26 km or about 16 miles
Tay quay OB quay đều quanh trục cố định đi qua O với vận tốc góc không đổi ω. Con lăn A chuyển động trong rãnh thẳng đứng. Tại vị trí trên hình vẽ thì thanh OB thẳng đứng, OA có phương nằm ngang. Hãy xác định vận tốc góc thanh AB, vận tốc của con lăn A; gia tốc góc của thanh AB, gia tốc của con lăn A. Cho ω = 1,5 rad/s, r = 1 m.
A 771.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 300.0 K. How much heat must the smelter produce to completely melt the copper bar? For solid copper, the specific heat is 386 J/kg • K, the heat of fusion is 205 kJ/kg, and the melting point is 1357 K.
Answer:
4.73 × 10^5
Explanation:
The magnetic flux that passes through one turn of a 8-turn coil of wire changes to 5.0 Wb from 8.0 Wb in a time of 0.098 s. The average induced current in the coil is 140 A. What is the resistance of the wire
Answer:
Resistance is 1.75 ohms
Explanation:
Magnetic flux:
[tex]{ \phi{ = NBA}}[/tex]
N is number of turns, N = 8
B is magnetic flux
A is area of projection.
From faradays law:
[tex]E = - \frac{ \triangle \phi}{t} [/tex]
where E is the Electro motive force.
But E = IR
where I is current and R is resistance:
[tex]IR = \frac{( \phi_{1} - \phi _{2}) }{t} \\ \\ 140 \times R = \frac{8 \times (8 - 5)}{0.098} \\ \\ R = \frac{24}{0.098 \times 140} \\ \\ resistance = 1.75 \: ohms[/tex]
In the diagram, the amplitude of the wave is shown by:
A
B
C
D
Answer:
A.
Explanation:
Amplitude measures how much a wave rises or falls. This is illustrated by A.
In the diagram, the amplitude of the wave is shown by A.
What is Amplitude?The amplitude of a periodic variable is a measure of its change in a single period. The amplitude of a non-periodic signal is its magnitude compared with a reference value.There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values.
The amplitude of a variable is simply a measure of change relative to its central position. In contrast, magnitude is a measure of the distance or quantity of a variable irrespective of its direction.
Amplitude is a property that is unique to waves and oscillations.
Therefore, in the diagram, the amplitude of a wave is shown by A.
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Who should you invite to the meeting?
—-
Answer
Irma
Explanation:
You should invite Irma to the meeting because she is in Charlotte and she also visited after 10/24/20
Where is the center of mass of homogeneous body which has a regular
Following the definition of the center of mass, "In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero."
(see explanation below)
what additional load will be required to cause the extension of 2.0cm when an elastic wire extend by 1.0cm when a load of 20g range from it
Answer:
The additional load is 20g
the lamp cord is 85cm long and comprises cupper wire. Calculate the wire‘s resistance?
radius of a wire is 1.8mmm,Use value of resistivity for Cu as 1.75 × 10-8Ωm.
Answer:
R = 0.0015Ω
Explanation:
The formula for calculating the resistivity of a material is expressed as;
ρ = RA/l
R is the resistance
ρ is the resistivity
A is the area of the wire
l is the length of the wire
Given
l = 85cm = 0.85m
A = πr²
A = 3.14*0.0018²
A = 0.0000101736m²
ρ = 1.75 × 10-8Ωm.
Substitute into the formula
1.75 × 10-8 = 0.0000101736R/0.85
1.4875× 10-8 = 0.0000101736R
R = 1.4875× 10-8/0.0000101736
R = 0.0015Ω
Megan accelerates her skateboard from 0 m/s to 8 m/s in 2 seconds. What is the magnitude of the acceleration of the skateboard?
O 8 m/s^2
O 16 m/s^2
O 2 m/s^2
O 4 m/s^2
Answer:
chk picture for eqn
Explanation:
what is the value of x if x-36=5?
Answer:
Therefore, the value of x is 41
Explanation:
x=5+36
x=41
A rectangular loop of wire with sides 0.129 and 0.402 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). The magnetic field has a magnitude of 0.888 T and is directed parallel to the normal of the loop's surface. In a time of 0.172 s, one-half of the loop is then folded back onto the other half, as indicated in part b of the drawing. Determine the magnitude of the average emf induced in the loop.
Answer:
[tex]0.2677\ \text{V/m}[/tex]
Explanation:
A = Area of loop = [tex]0.129\times0.402[/tex]
B = Magnetic field = [tex]0.888\ \text{T}[/tex]
t = Time taken = [tex]0.172\ \text{s}[/tex]
Electric field is given by
[tex]E=B\dfrac{dA}{dt}\\\Rightarrow E=0.888\times\dfrac{0.129\times 0.402}{0.172}\\\Rightarrow E=0.2677\ \text{V/m}[/tex]
The emf induced is [tex]0.2677\ \text{V/m}[/tex].
A cycle track is 500 metres long. A cyclist completes 10 laps (that is, he rides completely round the track 10 times).
a) How many kilometres has the cyclist travelled?
b) On average it took the cyclist 50 second to complete one lap (that it, to ride round just one).
(i) What was the average speed of the cyclist?
(ii) How long in minutes and seconds dit it take the cyclist to complete the 10 laps?
c) Near the end of the run the cyclist put on a spurt. During this spurt it took the cyclist 2 seconds to increase speed from 8 m/s to 12 m/s. What was the cyclist's acceleration during this spurt?
Explanation:
a) D_t = 500m*10laps
D_t = 5000m or 5km
b)
I) v = d/t
v = 500m/50s
v = 10m/s
ii) 10m/s = 5000m/t
t = 5000m/10m/s
t = (500s)*(1min/60s)
t = 8'20" or 8 mins and 20 sec
c) v_f = v_0 + a*t
v_f-v_0 = a*t
a = (v_f-v_0)/t
a = (12m/s-8m/s)/2s
a = (4m/s)/2s
a = 2m/s²
Part(a),
The distance travelled by the cyclist is 5 km.
Part(b),
(i) The average speed is 10 m/s
(ii) The time taken to cover 10 laps is 8 minutes and 20 seconds.
Part(c),
The acceleration is 2 m/s²
What is speed?Speed is defined as the ratio of the time distance travelled by the body to the time taken by the body to cover the distance. Speed is the ratio of the distance travelled by time. The unit of speed in miles per hour.
a) The distance will be calculated as
D = 500m*10laps
D = 5000m or 5km
b) The average speed is calculated as,
I) v = d/t
v = 500m/the 50s
v = 10m/s
The time will be calculated as,
ii) 10m/s = 5000m/t
t = 5000m/10m/s
t = (500s)*(1min/60s)
t = 8'20" or 8 mins and 20 sec
The acceleration is calculated as,
c) vf = v0 + at
vf-v0 = at
a = (vf-v0)/t
a = (12m/s-8m/s)/2s
a = (4m/s)/2s
a = 2m/s²
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00
Use base units to check whether the following equations are balance
(a) pressure = depth x density gravitational field strength,
(b) energy = mass x (speed of light).
dod to molt solid
Answer:
In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards.
The pressure at a depth in a fluid of constant density is equal to the pressure of the atmosphere plus the pressure due to the weight of the fluid, or p = p 0 + ρ h g , p = p 0 + ρ h g , 14.4
Granite: 2.70 × 10 32.70 × 10 3
Lead: 1.13 × 10 41.13 × 10 4
Iron: 7.86 × 10 37.86 × 10 3
Oak: 7.10 × 10 27.10 × 10 2
27. The part of the Earth where life exists .
Mesosphere
Stratosphere
Troposphere
Biosphere
Answer:
Biosphere is the part of the earth where life exists.
