Answer:
(a) m = 0.141 kg/s
(b) t = 47.343 s
(c) t = 143.745 s
Explanation:
Given that:
The volume of air in the tank V = 1.5 m³
The initial temperature in the tank is supposed to be 20° C and not 208 C;
So [tex]T_o = 20^0 C = ( 20 +273) K = 293K[/tex]
The initial pressure in the tank [tex]P_o= 800 \ kPa[/tex]
The throat area [tex]A_t[/tex] = 0.75 cm²
To find the initial mass flow in kg/s.
Lets first recall that:
Provided that [tex]\dfrac{P_{amb}}{P_{tank}}< 0.528[/tex], then the flow is choked.
Then;
[tex]\dfrac{P_{amb}}{P_{tank}}= \dfrac{101.35}{800}[/tex]
[tex]\dfrac{P_{amb}}{P_{tank}}= 0.1266[/tex]
From what we see above, it is obvious that the ratio is lesser than 0.528, therefore, the flow is choked.
Now, for a choked nozzle, the initial mass flow rate is determined by using the formula:
[tex]m = \rho \times A \times V[/tex]
where;
[tex]\rho = \rho_o \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}[/tex]
[tex]\rho =\dfrac{P_o}{RT_o} \bigg ( \dfrac{2}{k+1} \bigg) ^{\dfrac{1}{k-1}}[/tex]
[tex]\rho =\dfrac{800 \times 10^3}{287 \times 293} \bigg ( \dfrac{2}{1.4+1} \bigg) ^{\dfrac{1}{1.4-1}}[/tex]
[tex]\rho =9.51350( 0.8333 ) ^{2.5}[/tex]
[tex]\rho =6.03 \ kg/m^3[/tex]
[tex]T = T_o \bigg ( \dfrac{2}{k+1}\bigg)[/tex]
where;
[tex]T_o = 293 \ K[/tex]
[tex]T = 293 \bigg ( \dfrac{2}{1.4+1}\bigg)[/tex]
[tex]T = 293 \bigg ( \dfrac{2}{2.4}\bigg)[/tex]
[tex]T = 293 ( 0.8333)[/tex]
T = 244.16 K
During a critical condition when Mach No. is equal to one;
[tex]V = a = \sqrt{kRT}[/tex]
[tex]V = \sqrt{1.4 \times 287 \times 244.16}[/tex]
[tex]V = \sqrt{98103.488}[/tex]
V = 313.214 m/s
Thus, the initial mass flow rate [tex]m = \rho \times A \times V[/tex]
m = 6.03 × 0.75 × 10⁻⁴ × 313.214
m = 0.141 kg/s
(b)
The mass balance formula for the control volume surrounding the tank can be expressed as:
[tex]\dfrac{d}{dt}(\rho_o V) = \dfrac{d}{dt} \bigg ( \dfrac{P_o}{RT_o} V\bigg)[/tex]
[tex]= \dfrac{V}{RT_o}\dfrac{dP_o}{dt}= -m[/tex]
When the air mass flow rate is:
[tex]m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}[/tex]
Thus; replacing [tex]m = 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}[/tex] in the previous equation; we have:
[tex]\dfrac{V}{RT_o}\dfrac{dP_o}{dt}= - 0.6847 \dfrac{P_oA}{\sqrt{RT_o}}[/tex]
[tex]\dfrac{dP_o}{P_o}= -0.6847 \dfrac{A\sqrt{RT_o}}{V} \ dt[/tex]
Taking the differential of both sides from 0 → t
[tex]In(P_o)^t_o = -0.6847 \dfrac{A\sqrt{RT}}{V} \times t[/tex]
[tex]In \bigg ( \dfrac{P(t)}{P(0)} \bigg) = -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t[/tex]
[tex]\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )[/tex]
So, when the pressure P = 500 kPa, the time required is:
[tex]\dfrac{500}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )[/tex]
t = 47.343 s
(c)
Let us recall that:
The choking on the nozzle occurred when [tex]\dfrac{P_{amb}}{P_{tank}} = 0.528[/tex]
[tex]\dfrac{P_{amb}}{0.528} = P_{tank}[/tex]
[tex]\dfrac{101.35}{0.528} = P_{tank}[/tex]
[tex]P_{tank}= 191.95 \ kPa \\ \\ P_{tank} \simeq 192 \ kPa[/tex]
From [tex]\dfrac{P(t)}{P(0)} =exp \bigg ( -0.6847 \dfrac{A\sqrt{RT_o}}{V}\times t \bigg )[/tex]; the time required for [tex]P_{tank} \simeq 192 \ kPa[/tex] is:
[tex]\dfrac{192}{800} =exp \bigg ( -0.6847 \dfrac{0.75 \times 10^{-4}\sqrt{287 \times 293}}{1.5}\times t \bigg )[/tex]
By solving:
t = 143.745 s
what is the maximum powr of a module in Watts to the nearest whole Watt?
Complete question is;
You are looking at a module specification
sheet that has the table of information
below. What is the maximum power of this
module in Watts to the nearest whole Watt?
Value
Polycrystalline si
Characteristic
Cell Type
Cell
Configuration
Voc
160 in series
137.2 V
V_imp: 29.3 V
Ilsc: 8.60 A
I_Imp: 8.02 A
Dimensions (mm/in): 1000 x 1600 x 50 mm / 39.4" x 63" x 2"
Weight: 10 kg / 22 lbs
Answer:
P ≈ 235 Watts
Explanation:
Formula for power is;
P = IV
Now, for maximum power, we will make use of I_imp and V_imp given
Thus, P = I_imp × V_imp
We are given;
V_imp: 29.3 V
I_Imp: 8.02 A
Thus: P = 8.02 × 29.3 = 234.986 Watts
We are to approximate to the nearest whole watt.
Thus: P ≈ 235 Watts
What is the period of a pendulum near Earth’s surface that is 130 cm long?
Question 1 options:
2.3 s
0.83 s
9.1 s
22.9 s
Answer:
A). 2.3 s
Explanation:
Given that,
Length of the string = 130cm or 1.3 m
g near Earth's surface = 9.8 [tex]ms^{-2}[/tex]
To determine the period of a pendulum near Earth’s surface;
T = 2 π [tex]\sqrt{\frac{L}{g} }[/tex]
T = 2π[tex]\sqrt{\frac{1.3}{9.8} }[/tex]
T = 2.3 sec
Therefore, the period of the pendulum near the earth's surface is 2.3 sec.
A rocket takes off against the force of gravity. Consider this a non-isolated system. Derive the rocket equation formula relating velocity to the mass of the rocket and time. Let the velocity of the rocket be v(t) and the relative speedof the ejected fuelrelative to the moving rocket be u. Ignore air drag.
Answer:
Explanation:
The solution of the question has been put in attachment.
Which tool would you choose to create a storyboard, and why?
Answer:
Canva. Although Canva's free storyboard app for iOS and Android is targeted more at video production, it still provides animators and UX designers everything they need.
Explanation:
he mass of the moon is 1/81 of the mass of the earth. Compared to the gravitational force that the earth exerts on the moon, the gravitational force that the moon exerts on the earth is
Answer:
It is the same in magnitude and opposite in direction.
Explanation:
According Newton's Universal Law of Gravitation, the force that any mass exerts on any other one, is proportional to the product of their masses, inversely proportional tho the square of the distance between their centers, and acts along the line that joins both points.At the same time, this force is part of a action-reaction pair, according to Newton's 3rd Law, that states that for a force that one body exerts upon another, there exists an equal and opposite force exerted by the other body on the former one.So, we conclude that the gravitational force that the earth exerts on the moon is equal in magnitude and opposite in direction to the one that the moon exerts on the earth.The gravitational force the moon and the earth exert on each other are equal.
According to Newton's third law of motion, action and reaction are equal and opposite.
The gravitational force the earth exerts on the moon is equal in magnitude but opposite in direction to the gravitational force the moon exerts on the earth.
The magnitude of the gravitational force between the moon and the earth is calculated as follows;
[tex]F = \frac{GM_e M_m}{R^2}[/tex]
Thus, the gravitational force the moon and the earth exert on each other are equal.
Learn more about gravitational force here: https://brainly.com/question/17301500
Martin needs to push a box up a ramp.
Which force will make it hard for him to push the box?
friction
inertia
magnetism
Which of following is true statement of self-efficiency
Explanation:
Which of the following best defines self-efficacy? The belief in one's ability to cause an intended event to happen. ... If Bobby does not value athletic skill, his lack of talent will have less impact on his self-esteem than if he highly values athleticism.
When we walk, our legs alternatively swing forward about the hip joint as a pivot. In this motion the leg is acting approximately as a physical pendulum. Treating the leg as a thin uniform rod of length 0.80 m, find the time it takes for the leg to swing forward.
Answer:
0.3 seconds
Explanation:
When we walk, our legs alternatively swing forward about the hip joint as a pivot. In this motion the leg is acting approximately as a physical pendulum. Treating the leg as a thin uniform rod of length 0.80 m,
To find the time it takes for the leg to swing forward, where g = 9.8 m/s^2
Let's use the simple pendulum formula which says that:
T = 2π( sqrt ( L/g))
T = 2π[sqrt (0.8/9.8)]
T = 2π × 0.082
T = 0.5129
For the leg to swing forward only, the number of complete oscillations will be half.
Period T = time t ÷ number of complete oscillations. That is,
T = t ÷ 1/2
0.5129 = t ÷ 1/2
t = 0.5129 × 1/2
Time t = 0.256 seconds.
Therefore, the time it takes for the leg to swing forward is 0.3 seconds approximately
the maximum value of magnetic in an electric field 3.2 *10^4
Answer:
the answer is 12 because if your magnetic value and Electric field is 3.2 the answer will be 12
Hi so this is 15 points :D
The layers of Earth are the crust, mantle, and core, with the core being divided into inner and outer layers. Which of the following does NOT describe the layers of the solid Earth?
(A) The core makes up the majority of Earth's volume.
(B) The mantle is composed of rocks known as silicates.
(C) The crust is extremely thin when compared to the core or mantle.
(D) The core is made up of dense elements, such as iron and nickel.
Answer:
A. The core makes up the majority of Earth's volume.
Explanation:
Although the core and mantle are about equal in thickness, the core actually forms only 15% of earth's volume whereas the mantle takes up 84%
An 8.00- W resistor is dissipating 100 watts. What are the current through it and the difference of potential across it?
Answer:
I= 3.5 amps
Explanation:
Step one:
given data
rating of resistor R= 8 ohms
power P= 100W
Required
The current I
Step two
Yet this power is also given by
[tex]P = I^2R[/tex]
make I subject of the formula we have
[tex]I= \sqrt{\frac{P}{R} }[/tex]
substitute
[tex]I= \sqrt{\frac{100}{8} }\\\\I=\sqrt{12.5}\\\\I= 3.5 amps[/tex]
What is the minimum number of 20-amp 277-volt branch circuits required where 810 incandescent luminaires are to be installed
Complete Question
What is the minimum number of 20-amp 277-volt branch circuits required where 810 incandescent luminaries are to be installed?The lamps to be installed are rated 75 watts at 277 volts. This is NOT a continuous load so each circuit can carry 20 amps maximum
Answer:
The number is [tex]k = 11[/tex]
Explanation:
From the question we are told that
The current is [tex]I = 20 \ A[/tex]
The voltage is [tex]V = 277 \ volt[/tex]
The power rating of each bulb is [tex]P = 75 \ W[/tex]
The total number of incandescent luminaries is [tex]n = 810[/tex]
Generally the amount of current each lamp takes is mathematically represented as
[tex]I_e = \frac{P}{ V}[/tex]
=> [tex]I_e = \frac{75}{ 277}[/tex]
=> [tex]I_e = 0.270 \ A[/tex]
Generally the number of bulb fitting(i.e bulbs ) a 20 A current can supply (i.e the bulbs the branch ) to is
[tex]N = \frac{I}{I_e}[/tex]
=> [tex]N = \frac{20 }{ 0.270 }[/tex]
=> [tex]N = 73.86[/tex]
Generally the number of 20-amp 277-volt branch circuits required is mathematically represented as
[tex]k = \frac{n}{N}[/tex]
=> [tex]k = \frac{810}{73.86}[/tex]
=> [tex]k = 11[/tex]
i need help with the one someone?
Answer:A road sign gives drivers informaton about traffic on busy highways. One sign shows 15 miles in 20 minutes. What is this speed in miles per hour?
Explanation:
A standing wave with a fundamental frequency 1.25x10 2 Hz is set up on a string of length 0.750 meters. What is the frequency of the third harmonic?
Answer:
500Hertz
Explanation:
Given the fundamental frequency of the string given as 1.25x10²Hz
Harmonics are integral multiples of the fundamental frequency.
In strings, the third harmonic F3 is expressed in terms of the fundamental frequency expressed as;
F3 = 4F0
F0 is the fundamental frequency;
Given
F0 = 1.25x10²Hz
F3 = 4(1.25x10²)
F3 = 4(125)
F3 = 500Hz
Hence the frequency of the third harmonic is 500Hertz
An object has more momentum if it has which of the following?
Answer:
C.) A high velocity and Large mass.
Explanation:
Momentum of any object is defined by following formula
Here : m = mass of object
v = velocity of object
now we know that since momentum is product of mass and velocity
So in order to have more momentum we need the value of this product to be more. So this product will me large is both the physical quantity will be more in magnitude. So if mass is large and velocity will be more then the product of them will be large and hence the momentum of object will be more. Btw I had that question too.
What is the relationship between a planet’s distance from the sun and the length of its year? Explain.
Answer:
Planets that are close to the sun have shorter years, and planets that are farther have longer years, because of how fast/slow it orbits the sun.
Explanation:
An elf pushes a sleigh with force of 9N over a distance of 11m. How much work did the elf. Do on the sleigh
Answer:
99 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 9 × 11
We have the final answer as
99 JHope this helps you
5.
The diagram below shows two forces applied to a 2.0-kilogram block on a frictionless, horizontal surface.
F = 8.0 N
2.0 kg
F2 = 3.0 N
Frictionless Surface
What is the acceleration of the block?
A 4.0 m/s2 to the left
B.O 2.5 m/s2 to the right
C.O 2.5 m/s2 to the left
D. 1.5 m/s2 to the right
Answer:2.5m/s^2 to the left
Explanation:
The acceleration of the block is "2.5 m/s² to the left"
Given,
Force,
[tex]F_1 = 8 N[/tex][tex]F_2 = 3 \ N[/tex]Mass,
[tex]m = 2 \ kg[/tex]Considering equilibrium of forces on block,
then,
→ [tex]F_1 -F_2 = ma[/tex]
By substituting the values, we get
→ [tex]8-3 = 2a[/tex]
→ [tex]5 = 2a[/tex]
[tex]a = \frac{5}{2}[/tex]
[tex]a = 2.5 \ m/s^2[/tex] (to the left)
Thus the above answer i.e., option C is right.
Learn more:
https://brainly.com/question/21517582
What is the speed of a helicopter that traveled 1500 miles in 7 hours?
A solid cube of aluminum (density of 2.7 g/cm³) has a volume of 0.9 cm³. how many atoms are contained in the cube?
Answer:
[tex]0.542*10^{23}\ Aluminum\ Atoms[/tex]
Explanation:
[tex]We\ are\ given:\\Density\ of\ aluminum=2.7\ g/cm^3\\Volume\ of\ aluminum-cube=0.9\ cm^3\\Hence,\\As\ we\ know\ that,\\Density=\frac{Mass}{Volume}\\Mass=Density*Volume\\Hence,\ here,\\Mass\ of\ the\ solid\ iron\ cube=2.7*0.9=2.43\ g\\Now,\\We\ also\ know\ that,\\Gram\ Atomic\ mass\ of\ Aluminum = 26.98 \approx 27\ g\\Hence,\\No.\ of\ particles=\frac{Mass}{GAM}*Avagadro's Constant\\Hence,\ here\\No.\ of\ Aluminum\ atoms=\frac{2.43}{27}*6.022*10^{23} \approx 0.542*10^{23}\ Aluminum\ Atoms[/tex]
ball a 0.604 kg moving right at 11.6 m/s makes a head-on collision with ball B at rest. after, ball A moves right at 2.09 m/s, and ball B moves right at 5.03 m/s. what is the mass of ball B? unite=kg
Answer:
1.142
Explanation:
Help me! Btw he’s playing golf.
Answer:
Muscular energy
Explanation:
Hope it helps!!!!
What are electromagnetic waves?
Answer:
Electromagnetic waves or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field. In other words, EM waves are composed of oscillating magnetic and electric fields. They are also perpendicular to the direction of the EM
Explanation:
I WILL GIVE YOU BRIANLIEST PLEASE HELP!!!!!!
A small plane starts from rest and accelerates uniformly to the east to a takeoff velocity of 70 m/s in 5 seconds. What is the plane’s acceleration?
who win tug of war with what net force
Answer:
The force on rope is equal for both of them at any time.
For winning the game the force on ground is responsible.
Explanation:
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. The proton must impact the nucleus with a kinetic energy of 2.30 MeV. Assume the nucleus remains at rest. With what speed must the proton be fired toward the target?
Answer:
[tex]V_1= 3.4*10^7m/s[/tex]
Explanation:
From the question we are told that
Nucleus diameter [tex]d=5.50-fm[/tex]
a 12C nucleus
Required kinetic energy [tex]K=2.30 MeV[/tex]
Generally initial speed of proton must be determined,applying the law of conservation of energy we have
[tex]K_2 +U_2=K_1+U_1[/tex]
where
[tex]K_1[/tex] =initial kinetic energy
[tex]K_2[/tex] =final kinetic energy
[tex]U_1[/tex] =initial electric potential
[tex]U_2[/tex] =final electric potential
mathematically
[tex]U_2 = \frac{Kq_pq_c}{r_2}[/tex]
where
[tex]r_f[/tex]=distance b/w charges
[tex]q_c[/tex]=nucleus charge [tex]=6(1.6*10^-^1^9C)[/tex]
[tex]K[/tex]=constant
[tex]q_p[/tex]=proton charge
Generally kinetic energy is know as
[tex]K=\frac{1}{2} mv^2[/tex]
Therefore
[tex]U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2} mv_1^2 +U_1[/tex]
Generally equation for radius is [tex]d/2[/tex]
Mathematically solving for radius of nucleus
[tex]R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})[/tex]
[tex]R=2.75*10^-^1^5m[/tex]
Generally we can easily solving mathematically substitute into v_1
[tex]q_p[/tex][tex]=6(1.6*10^-^1^9C)[/tex]
[tex]K_1=9.0*10^9 N-m^2/C^2[/tex]
[tex]U_1= 0[/tex]
[tex]R=2.75*10^-^1^5m[/tex]
[tex]K=2.30 MeV[/tex]
[tex]m= 1.67*10^-^2^7kg[/tex]
[tex]V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }[/tex]
[tex]V_1= 3.4*10^7m/s[/tex]
Therefore the proton must be fired out with a speed of [tex]V_1= 3.4*10^7m/s[/tex]
A softball player swings a bat, accelerating it from rest to 2.2 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 2.3-kgkg uniform rod of length 0.96 mm , and compute the torque the player applies to one end of it.
Answer:
83.168 Nm
Explanation:
I=mass of bat* ( length )^2 /3
Moment of inertia from one end = 2.2 * (0.96)^2 / 3.
Ang. acc. = 2.2×2 pi / 0.2 = 22pi
Torque = 2.3 * (0.96)^2 / 3 * 22 pi = 83.168 Nm SI units
What would be harder to accelerate- a 4kg shot put ball or a 2,000 kg boulder
Answer:
2,000 kg boulderExplanation:t
The Inertial is the resistance of a body at rest to be in motion
Given the following item
1. 4kg shot put ball
2. 2,000 kg boulder
The body of greater mass has greater inertia, hence would be harder to accelerate.
An object with a small inertial mass will accelerate more than an object with a large inertial mass when acted upon by the same force.
Of the 2 items in the list, the 2,000 kg boulder has the largest mass, hence it also has the greater inertial and more difficult to accelerate.
Which of these quantities impact centripetal force?
A.mass and speed only
B.mass and radius only
C.speed and radius only
D.mass, speed, and radius all impact centripetal force
Answer: D) Mass, speed and radius all impact the centripetal force
Explanation:
If we change either the mass, speed, and/or radius, then the centripetal force will be different. Though on rare occasions we could change the variables together in a way to have the centripetal forces line up, but in general the before and after forces will likely be different. Changing one variable at a time only will definitely result in a different centripetal force.
This is because the centripetal force F is determined by the formula
F = (m*v^2)/r
where,
m = mass
v = velocity
r = radius
A neutron star consists of neutrons at approximately nuclear density. Estimate, for a 10-km-diameter neutron star its mass number, its mass, the acceleration of gravity at its surface.
Answer:
a) 7.2*10^55
b) 1.2*10^29 kg
c) 3.2*10^11 m/s
Explanation:
Given a diameter of 10 km, that's a radius of 5 km.
We first attempt to find its mass number
A = (r/1.2*10^-15)³
A = (5/1.5*10^-15)³
A = 7.2*10^55
This gotten mass number is then multiplied by the mass of each neutron in the star. With each neutron being, 1.7*10^-27
M = 7.5*10^55 * 1.7*10^-27
M = 1.2*10^29 kg
And finally, using the formula
g = GM/r², we can find its acceleration due to gravity.
g = (6.67*10^-11 * 1.2*10^29) / 5000²
g = 8*10^18 / 2.5*10^7
g = 3.2*10^11 m/s
