Answer:
Explanation:
For the problem, we should have same reynolds number
ρvd/mu = constant
1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600
d = 25.66 cm
The combustion of propane (C3H8) in the presence of excess oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) → 3CO2 (g) + 4H2O (g) When 2.0 mol of O2 are consumed in this reaction, ________ mol of CO2 are produced.
Answer:
1.2
Explanation:
2.0 mol O₂ × (3 mol CO₂ / 5 mol O₂) = 1.2 mol CO₂
Three rocks of equal mass are thrown with identical speeds from the top of the same building (identical height). Rock X is thrown vertically downward, rock Y is thrown vertically upward, and rock Z is thrown horizontally.Required:Which rock has the greatest speed just before it hits the ground? Assume air resistance is negligible.
Answer:Three rocks of equal mass are thrown
Explanation:
The phenomena: Three rocks of equal mass are thrown with identical speeds from the top of the same building (identical height). Rock X is thrown vertically downward, rock Y is thrown vertically upward, and rock Z is thrown horizontally.
An alarm clock is plugged into a 120 volt outlet and has a resistance of 15,000 ohms. How much power does it use?
Answer:
The power used is 0.96 watts.
Explanation:
Recall the formula for electric power (P) as the product of the voltage applied times the circulating current:
[tex]P=V\,\,I[/tex]
and recall as well that the circulating current can be obtained via Ohm's Law as the quotient of the voltage applied divided the resistance:
[tex]V=I\,\,R\\I=\frac{V}{R}[/tex]
Then we can re-write the power expression as:
[tex]P=V\,\,I=V\,\,\frac{V}{R} =\frac{V^2}{R}[/tex]
which in our case becomes:
[tex]P=\frac{V^2}{R}=\frac{120^2}{15000} =0.96\,\,watts[/tex]
Which scientist's work led to our understanding of how planets move around
the Sun?
A. Albert Einstein
B. Lord Kelvin
C. Johannes Kepler
D. Edwin Hubble
Answer:
Johannes KeplerExplanation:
He made rules about planetary motion.The scientist Johannes Kepler was a German astronomer.He found out that the planets evolved around the Sun.He also made the laws of planetary motion.Hope this helped,
Kavitha
5. Two men, Joel and Jerry, push against a car that has stalled, trying unsuccessfully to get it moving. Jerry stops after 10 min, while Joel is able to push for 5.0 min longer. Compare the work they do on the car
Answer:
The work done by both Joel and Jerry is equal to 0 J.
Explanation:
The work done on a body by an external agency is the product of the force applied on the body and the distance through which the body moves. Therefore,
W = F.d
where,
W = Work Done on the Body
F = Force Applied on the Body
d = displacement covered by the body
In the given case of both Joel and Jerry, they are unable to move the car. It means that the displacement covered by the car is zero. Hence,
W = F(0)
W = 0 J (For both Joel and Jerry)
According to the law of conservation of energy, if the ocean water cools, then something else should warm. What is it that warms?
Answer:
The answer is air
Explanation:
A circuit element maintains a constant resistance. If the current through the circuit element is doubled, what is the effect on the power dissipated by the circuit element
Answer:
This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.Explanation:
The formula for calculating the power expended in a circuit is P = I²R... 1
i is the current (in amperes)
R is the resistance (in ohms)
If a circuit element maintains a constant resistance and the current through the circuit element is doubled, then new current I₂ = 2I
New power dissipated P₂ = (I₂)²R
P₂ = (2I)²R
P₂ = 4I²R ... 2
Dividing equation 2 by 1 will give;
P₂/P = 4I²R/I²R
P₂/P = 4
P₂ = 4P
This shows that the power dissipated by the circuit element is four times its original power if the current is doubled.
A rigid massless rod is rotated about one end in a horizontal circle. There is a particle of mass m1 attached to the center of the rod and a particle of mass m2 attached to the outer end of the rod. The inner section of the rod sustains a tension that is 5 times as great as the tension that the outer section sustains. Find the ratio m1/m2
Answer:
m₁/m₂ =8
Explanation:
Writing the centripetal force equation: where T₁ is inner tension and T₂ is outer tension
T₂ = m₂ × ω²R....equation 1
T₁-T₂ = m₁ × ω²R/2
Also T₁=5T₂
5T₂-T₂ = m₁ × ω²R/2
4 × 2T₂ = m₁ × ω²R
8T₂ = m₁ × ω²R
dividing it by equation 1
8 = m₁/m₂
Hence m₁/m₂ =8
Suppose that the voltage of the battery in the circuit is 3.9 V, the magnitude of the magnetic field (directed perpendicularly into the plane of the paper) is 1.1 T, and the length of the rod between the rails is 0.22 m. Assuming that the rails are very long and have negligible resistance, find the maximum speed attained by the rod after the switch is closed.
Answer:
v = 16.11 m/s
Explanation:
In order to calculate the maximum speed of the rod, you use the following formula:
[tex]\epsilon=vBLsin\theta[/tex] (1)
ε = voltage of the circuit = 3.9V
v: maximum speed of the rod = ?
B: magnitude of the magnetic field = 1.1T
L: length of the rod = 0.22m
θ: angle between the direction of motion of the rod and the direction of the magnetic field = 90°
You solve the equation (1) for v and replace the values of the other parameters:
[tex]v=\frac{\epsilon}{BLsin\theta}=\frac{3.9V}{(1.1T)(0.22m)sin90\°}\\\\v=16.11\frac{m}{s}[/tex]
The maximum speed of the rod is 16.11 m/s
A 32-cm-long solenoid, 1.8 cm in diameter, is to produce a 0.30-T magnetic field at its center. If the maximum current is 4.5 A, how many turns must the solenoid have?
Answer:
16,931 turnsExplanation:
The magnetic field produced is expressed using the formula
[tex]B = \frac{\mu_0NI}{L}[/tex]
B is the magnetic field = 0.30T
I is the current produced in the coil = 4.5A
[tex]\mu_0[/tex] is the magnetic permittivity in vacuum = 1.26*10^-6Tm/A
L is the length of the solenoid = 32 cm = 0.32 m
N is the number of turns in the solenoid.
Making N the subject of the formula from the equation above;
[tex]B = \frac{\mu_0NI}{L}\\\\BL = \mu_0NI\\\\Dividing\ both\ sides \ by \ \mu_0I\\\\\frac{BL}{\mu_0I} =\frac{\mu_oNI}{\mu_0I} \\\\[/tex]
[tex]N = \frac{BL}{\mu_0I}[/tex]
Substituting the give values to get N;
[tex]N = \frac{0.3*0.32}{1.26*10^{-6} * 4.5}\\\\N = \frac{0.096}{0.00000567} \\\\N = 16,931.21[/tex]
The number of turns the solenoid must have is approximately 16,931 turns
The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 m, and a baseball has a mass of 145 g.(a) What force did he produce on the ball during this record-setting pitch? (b) Draw free-body diagrams of the ball during the pitch and just after it left the pitcherâs hand.
Answer:
Explanation:
F ×1 = 0.5×0.145×47×47
F = 160.15 N
What is the equivalent temperature in kelvin if you have a metal at 50°F?
Answer:
The required temperature is 283 K.
Explanation:
[tex]T\:=\:\left(50-32\right)\times \frac{5}{9}+273\\\\T=283\:K[/tex]
Best Regards!
An 75-kg hiker climbs to the summit of Mount Mitchell in western North Carolina. During one 2.0-h period, the climber's vertical elevation increases 540 m. Determine the change in gravitational potential energy of the climber-Earth system.
Answer:
The change in gravitational potential energy of the climber-Earth system is [tex]\Delta PE = 396900 \ J[/tex]
Explanation:
From the question we are told that
The mass of the hiker is [tex]m = 75 \ kg[/tex]
The time taken is [tex]T = 2 \ hr = 2 * 3600 = 7200 \ s[/tex]
The vertical elevation after time T is [tex]H = 540 \ m[/tex]
The change in gravitational potential is mathematically represented as
[tex]\Delta PE = mgH[/tex]
here g is the acceleration due to gravity with value [tex]g = 9.8 \ m/s^2[/tex]
substituting values
[tex]\Delta PE = 75 * 9.8 * 540[/tex]
[tex]\Delta PE = 396900 \ J[/tex]
What is the equivalent resistance between the points A and B of the network?
Explanation:
First, simplify the circuit. Then calculate the parallel and consecutive resistances to find the answer.
Which describes any compound that has at least one element from group 17? Halide;noble gas; metalliod; transition metal
Answer:
Halide
Explanation:
It has at least one element from the halogen group (17)
Halide describes any compound that has at least one element from group 17, therefore the correct option is option A.
What are halides?When the elements belonging to group 17 of the periodic table form ionic compounds with other electropositive elements, then these compounds are known as halides.
These elements from group 17 are also known as halogens. Generally, these halides have very high electronegativity as they reside on the right side of the periodic table.
Generally, the valency of the halogens element involved in the halide compound is one and they form ionic compounds with the alkali and alkaline earth metals.
Thus, halides are compounds that have at least one element from group 17.
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A load of 223,000 N is placed on an aluminum column 10.2 cm in diameter. If the column was originally 1.22 m high find the amount that the column has shrunk.
Answer:
0.4757 mm
Explanation:
Given that:
Load P = 223,000 N
the length of the height of the aluminium column = 1.22 m
the diameter of the aluminum column = 10.2 cm = 0.102 m
The amount that the column has shrunk ΔL can be determined by using the formula:
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
where;
A = πr²
2r = D
r = D/2
r = 0.102/2
r = 0.051
A = π(0.051)²
A = 0.00817
Also; the young modulus of aluminium [tex]E_{Al}[/tex] is:
[tex]E_{Al}= 7*10^{10} \Nm^{-2}[/tex]
[tex]\Delta L = \dfrac{PL}{AE_{Al}}[/tex]
[tex]\Delta L = \dfrac{223000* 1.22}{0.00817* 7*10^{10}}[/tex]
ΔL = 4.757 × 10⁻⁴ m
ΔL = 0.4757 mm
Hence; the amount that the column has shrunk is 0.4757 mm
A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediately after the impact, a 320-g piece moves along the x-axis with a speed of 2.00 m/s and a 355-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction relative to the +x-axis does the third piece move?
(a) 39.8º from the +x-axis
(b) 36.9° from the +x-axis
(c) 39.9° from the +x-axis
(d) 216.9° from the +x-axis
(e) 219.8° from the +X-axis
Answer:
M1 Vx1 + M2 Vx2 + M3 Vx3 = 0 conservation of momentum in x direction
Vx3 = -(M1 Vx1 + M2 Vx2 ) / M3
Vx3 = - 320 * 2 / 100 = -6.4 m/s M2 has no x-component of momentum
Likewise:
Vy3 = -(M1 Vy1 + M2 Vy2 ) / M3
Vy3 = - 355 * 1.5 / 100 = -5.33 m/s
tan theta = -5.33 / -6.4 = .833 where theta is in the third quadrant and measured from the negative x-axis
theta = 39.8 deg
180 + 39.8 = 219.8 from the positive x-axis
In an undergraduate physics lab, a simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min. What are the period and length of the pendulum
Explanation:
We have
A simple pendulum is observed to swing through 71 complete oscillations in a time period of 1.80 min.
The frequency of a pendulum is equal to the no of oscillation per unit time. so,
[tex]f=\dfrac{N}{t}\\\\f=\dfrac{71}{1.8\times 60}\\\\f=0.65\ Hz[/tex]
Tim period is reciprocal of frequency. So,
[tex]T=\dfrac{1}{0.65}\\\\T=1.53\ s[/tex]
The time period of a pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
l is length of pendulum
[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(1.53)^2\times 9.8}{4\pi ^2}\\\\l=0.58\ m[/tex]
So, the period and length of the pendulum are 1.53 s and 0.58 m respectively.
A wheel has a rotational inertia of 16 kgm2. Over an interval of 2.0 s its angular velocity increases from 7.0 rad/s to 9.0 rad/s. What is the average power done by the torque
Answer:
128.61 WattsExplanation:
Average power done by the torque is expressed as the ratio of the workdone by the toque to time.
Power = Workdone by torque/time
Workdone by the torque = [tex]\tau \theta[/tex] = [tex]I\alpha * \theta[/tex]
I is the rotational inertia = 16kgm²
[tex]\theta = angular\ displacement[/tex]
[tex]\theta = 2 rev = 12.56 rad[/tex]
[tex]\alpha \ is \ the\ angular\ acceleration[/tex]
To get the angular acceleration, we will use the formula;
[tex]\alpha = \frac{\omega_f^2- \omega_i^2}{2\theta}[/tex]
[tex]\alpha = \frac{9.0^2- 7.0^2}{2(12.54)}\\\alpha = 1.28\ rad/s^{2}[/tex]
Workdone by the torque = 16 * 1.28 * 12.56
Workdone by the torque = 257.23 Joules
Average power done by the torque = Workdone by torque/time
= 257.23/2.0
= 128.61 Watts
How much work is needed to move an object from one position to another when both positions are located the same distance from the center of the earth
Answer:
The product of the object's weight and the horizontal distance between the two positions.
Explanation:
Work is the product of force and the distance through which this force is moved. The distance moved can be vertical, or horizontal. For two bodies located the same distance from the center of the earth, the work done will be the product of the weight of the product and the horizontal distance between the two positions. If the vertical work is needed, then the work is zero, since there is no height gradient between them.
B. Write short notes on:
1. Horticulture
2. Pisciculture
3. Aviculture
4. Veterinary science
5. Intensive farming.
1. Horticulture is the agriculture of plants, mainly for food, materials, comfort and beauty for decoration.
2.Pisciculture also known as fish farming is the rearing of fish for food in enclosures such as fish ponds or tanks.
3.Aviculture is the practice of keeping and breeding birds, especially of wild birds in captivity. Aviculture is generally focused on not only the raising and breeding of birds, but also on preserving avian habitat, and public awareness campaigns.
4. Veterinary medicine is the branch of medicine that deals with the prevention, control, diagnosis, and treatment of disease, disorder, and injury in animals. Along with this, it also deals with animal rearing, husbandry, breeding, research on nutrition and product development.
5. Intensive agriculture, also known as intensive farming and industrial agriculture, is a type of agriculture, both of crop plants and of animals, with higher levels of input and output per cubic unit of agricultural land area.
Hope this helps.
During a move, Jonas and Matías carry a 115kg safe to the third floor of a building, covering a height of 6.6m.
1) = what work do they do?
2) = what power do they develop if the work is done in 5.5 minutes?
Answer:
work is =7590joules
power = 23watts
Answer:
1) 7590 Joules
2) 23 Watts
Explanation:
1) Work = force × distance
W = mgh
W = (115 kg) (10 m/s²) (6.6 m)
W = 7590 J
2) Power = work / time
P = W / t
P = (7590 J) / (330 s)
P = 23 W
The upward velocity of a 2540kg rocket is v(t)=At + Bt2. At t=0 a=1.50m/s2. The rocket takes off and one second afterwards v=2.00m/s. Determine the constants A and B with units.
Answer:
The value of A is 1.5m/s^2 and B is 0.5m/s^³
Explanation:
The mass of the rocket = 2540 kg.
Given velocity, v(t)=At + Bt^2
Given t =0
a= 1.50 m/s^2
Now, velocity V(t) = A*t + B*t²
If, V(0) = 0, V(1) = 2
a(t) = dV/dt = A+2B × t
a(0) = 1.5m/s^²
1.5m/s^² = A + 2B × 0
A = 1.5m/s^2
now,
V(1) = 2 = A× 1 + B× 1^²
1.5× 1 +B× 1 = 2m/s
B = 2-1.5
B = 0.5m/s^³
Now Check V(t) = A× t + B × t^²
So, V(1) = A× (1s) + B× (1s)^² = 1.5m/s^² × 1s + 0.5m/s^³ × (1s)^² = 1.5m/s + 0.5m/s = 2m/s
Therefore, B is having a unit of m/s^³ so B× (1s)^² has units of velocity (m/s)
Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/kg. The final specific volume is 0.027 m^3/kg. Find the specific work in the process.
Answer:
The pressure is constant, and it is P = 150kpa.
the specific volumes are:
initial = 0.062 m^3/kg
final = 0.027 m^3/kg.
Then, the specific work can be written as:
[tex]W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.[/tex]
The fact that the work is negative, means that we need to apply work to the air in order to compress it.
Now, to write it in more common units we have that:
1 kPa*m^3 = 1000J.
-5.25 kPa*m^3/kg = -5250 J/kg.
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A patch of mud has stuck to the surface of a bicycle tire as shown. The stickiness of
the mud is the centripetal or tension force that keeps the mud on the tire as it spins.
Has work been done on the mud as the tire makes one revolution, if the mud stays
on the tire? Explain.
Answer:
Yes, work has been done on the mud.
Explanation:
Work is done on a body, when a force is applied on the body to move it through a certain distance. In the case of the mud, the tire exerts a centripetal force on the mud. The centripetal force moves the mud along a path that follows the circle formed by the tire in one revolution of the tire. The total distance traveled is the circumference of the circle formed. The work done on the mud is therefore the product of the centripetal force on the mud from the tire, and the circumference of the circle formed by the tire, usually expressed in radian.
Suppose that 300 keV X-ray photons are aimed at a zinc cube (Zinc, Z = 30). According to the chart below, what effect will predominate when the X-rays hit the metal?
a) Photoelectric Effect 3
b) Compton Effect 3
c) Pair Production
Answer:
the answer is option A = photoelectric effect
Explanation:
If the threshold frequency of a metal is lower than the energy of X-rays, then photoelectric effect will happen.
The magnetic field strength at the north pole of a 2.0-cm-diameter, 8-cm-long Alnico magnet is 0.10 T. To produce the same field with a solenoid of the same size, carrying a current of 1.8 A , how many turns of wire would you need
Answer:
The number of turns of the solenoid is 3536 turns
Explanation:
Given;
magnetic field of the solenoid, B = 0.1 T
current in the solenoid, I = 1.8 A
length of the solenoid, L = 8cm = 0.08m
The magnetic field near the center of the solenoid is given by;
B = μ₀nI
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
n is number of turns per length
I is the current in the coil
The number of turns per length is calculated as;
n = B / μ₀I
n = (0.1 ) / (4π x 10⁻⁷ x 1.8)
n = 44203.95 turns/m
The number of turns is calculated as;
N = nL
N = (44203.95)(0.08)
N = 3536 turns
Therefore, the number of turns of the solenoid is 3536 turns
A conventional current of 8 A runs clockwise in a circular loop of wire in the plane, with center at the origin and with radius 0.078 m. Another circular loop of wire lies in the same plane, with its center at the origin and with radius 0.03 m. How much conventional current must run counterclockwise in this smaller loop in order for the magnetic field at the origin to be zero
Answer:
I2 = 3.076 A
Explanation:
In order to calculate the current in the second loop, you take into account that the magnitude of the magnetic field at the center of the ring is given by the following formula:
[tex]B=\frac{\mu_oI}{2R}[/tex] (1)
I: current in the wire
R: radius of the wire
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
In the case of the two wires with opposite currents and different radius, but in the same plane, you have that the magnitude of the magnetic field at the center of the rings is:
[tex]B_T=\frac{\mu_oI_1}{2R_1}-\frac{\mu_oI_2}{2R_2}[/tex] (2)
I1: current of the first ring = 8A
R1: radius of the first ring = 0.078m
I2: current of the second ring = ?
R2: radius of the first second = 0.03m
To find the values of the current of the second ring, which makes the magnitude of the magnetic field equal to zero, you solve the equation (2) for I2:
[tex]\frac{\mu_oI_2}{2R_2}=\frac{\mu_oI_1}{2R_1}\\\\I_2=I_1\frac{R_2}{R_1}=(8A)\frac{0.03m}{0.078m}=3.076A[/tex]
The current of the second ring is 3.076A and makes that the magntiude of the total magnetic field generated for both rings is equal to zero.
A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.
Complete question:
Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. In contrast, a passenger not wearing a seat belt or using an air bag decelerates over a distance of 5mm.
(a) A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s . Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.
Answer:
The net force on the person as the air bad deploys is -6750 N backwards
Explanation:
Given;
mass of the passenger, m = 60 kg
velocity of the car at impact, u = 15 m/s
final velocity of the car after impact, v = 0
distance moved as the front of the car crumples, s = 1 m
First, calculate the acceleration of the car at impact;
v² = u² + 2as
0² = 15² + (2 x 1)a
0 = 225 + 2a
2a = -225
a = -225 / 2
a = -112.5 m/s²
The net force on the person;
F = ma
F = 60 (-112.5)
F = -6750 N backwards
Therefore, the net force on the person as the air bad deploys is -6750 N backwards
A meat baster consists of a squeeze bulb attached to a plastic tube. When the bulb is squeezed and released, with the open end of the tube under the surface of the basting sauce, the sauce rises in the tube to a distance h, as the drawing shows. Using 1.013 105 Pa for the atmospheric pressure and 1200 kg/m3 for the density of the sauce, find the absolute pressure in the bulb when the distance h is (a) 0.15 m and (b) 0.10 m.
Answer:
(a) P = 103064 Pa = 103.064 KPa
(b) P = 102476 Pa = 102.476 KPa
Explanation:
(a)
First we need to find the gauge pressure:
Gauge Pressure = Pg = (density)(g)(h)
Pg = (1200 kg/m³)(9.8 m/s²)(0.15 m)
Pg = 1764 Pa
So, the absolute Pressure is:
Absolute Pressure = P = Atmospheric Pressure + Pg
P = 1.013 x 10⁵ Pa + 1764 Pa
P = 103064 Pa = 103.064 KPa
(b)
First we need to find the gauge pressure:
Gauge Pressure = Pg = (density)(g)(h)
Pg = (1200 kg/m³)(9.8 m/s²)(0.1 m)
Pg = 1176 Pa
So, the absolute Pressure is:
Absolute Pressure = P = Atmospheric Pressure + Pg
P = 1.013 x 10⁵ Pa + 1176 Pa
P = 102476 Pa = 102.476 KPa
The absolute pressure in the bulb is approximately 1.031 x 10⁵ Pa when h = 0.15 m and 1.025 x 10⁵ Pa when h = 0.10 m.
Absolute pressure is the total pressure exerted by a fluid, including both the pressure from the fluid itself and the atmospheric pressure. It is the sum of the gauge pressure, which is the pressure above atmospheric pressure, and the atmospheric pressure. Absolute pressure is measured relative to a complete vacuum, where the pressure is zero.
In fluid mechanics, absolute pressure is important for determining the forces and behaviors of fluids in various systems. It is commonly expressed in units such as pascals (Pa), atmospheres (atm), pounds per square inch (psi), or torr.
The absolute pressure in the bulb can be calculated using the following formula:
P = P₀ + ρgh
where:
P is the absolute pressure in the bulb,
P₀ is the atmospheric pressure (1.013 x 10⁵ Pa),
ρ is the density of the sauce (1200 kg/m³),
g is the acceleration due to gravity (9.8 m/s²), and
h is the height of the sauce in the tube.
(a) When h = 0.15 m:
P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.15 m)
P ≈ 1.013 x 10⁵ Pa + 1764 Pa
P ≈ 1.031 x 10⁵ Pa
(b) When h = 0.10 m:
P = 1.013 x 10⁵ Pa + (1200 kg/m³) x (9.8 m/s²) x (0.10 m)
P ≈ 1.013 x 10⁵ Pa + 1176 Pa
P ≈ 1.025 x 10⁵ Pa
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