A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5 kg and 0.5 kg, respectively. The piston diameter is 10 cm. The local atmospheric pressure is 1 bar. T
Answers
Answer:
The force acting on the shaft is 1324.75 N
Explanation:
Given that,
Cross sectional area of shaft [tex]A_{sh}=0.8\ cm^{2}[/tex]
Gas pressure [tex]P_{gas}=3\ bar=3\times10^5\ Pa[/tex]
Total mass M=24.5+0.5=25 kg
Diameter of piston, d=10 cm
We need to calculate the cross section area of piston
Using formula of area
[tex]A_{p}=\pi\times\dfrac{d^2}{4}[/tex]
Put the value into the formula
[tex]A_{p}=\pi\times\dfrac{0.1^2}{4}[/tex]
[tex]A_{p}=0.00785\ m^{2}[/tex]
We need to calculate the weight of the piston and shaft
Using formula of weight
[tex]W=mg[/tex]
Put the value into the formula
[tex]W=25\times9.81[/tex]
[tex]W=245.25\ N[/tex]
We need to calculate the force due to gas pressure
Using formula of force
[tex]F_{gas}=P_{gas}\times A_{p}[/tex]
Put the value into the formula
[tex]F_{gas}=3\times10^{5}\times0.00785[/tex]
[tex]F_{gas}=2355\ N[/tex]
We need to calculate the force due to atmospheric pressure,
Using formula of force
[tex]F_{atm}=P_{atm}\times A_{p}[/tex]
Put the value into the formula
[tex]F_{atm}=1\times10^5\times0.00785[/tex]
[tex]F_{atm}=785\ N[/tex]
We need to calculate the force acting on the shaft
from free body diagram
[tex]F_{gas}=F+F_{atm}+W[/tex]
Put the value into the formula
[tex]2355=F+785+245.25[/tex]
[tex]F=2355-785-245.25[/tex]
[tex]F=1324.75\ N[/tex]
Hence, The force acting on the shaft is 1324.75 N
The complete question is :
The figure (attached) shows a gas contained in a vertical piston-cylinder assembly. A vertical shaft whose cross-sectional area is 0.8 cm2 is attached to the top of the piston. Determine the magnitude, F, of the force acting on the shaft, in N, required if the gas pressure is 3 bar. The masses of the piston and attached shaft are 24.5kg and 0.5kg respectively. The piston diameter is 10cm. The local atmospheric pressure is 1 bar. The piston moves smoothly in the cylinder and g=9.81 m/s2.
Solution :
Given
The cross sectional area of the shaft, [tex]$A_{s} = 0.8 \ cm^2$[/tex]
Gas pressure, [tex]$P_{gas} = 3 \ bar = 3 \times 10^5 \ Pa$[/tex]
The total mass, m = 24.5 + 0.5
= 25 kg
Diameter of the piston, d = 10 cm
The cross sectional area of the piston, [tex]$A_{p} = \frac{\pi}{4} \times (0.1)^2$[/tex]
Weight of the piston and shaft, W = mg
= 25 x 9.81
= 245.25 N
Force due to the gas pressure,
[tex]$F_{gas} = P_{gas} \times A_{p}$[/tex]
[tex]$F_{gas} = 3 \times 10^5 \times 0.00785 $[/tex]
= 2355 N
Force due to atmospheric pressure,
[tex]$F_{atm} = P_{atm} \times A_{p}$[/tex]
[tex]$= 1 \times 10^5 \times 0.00785$[/tex]
= 785 N
Now [tex]$F_{gas} = F + F_{atm} + W$[/tex]
⇒ 2355 = F + 785 + 245.25
⇒ F = 1324.75 N
Related Questions
Which waves are electromagnetic and can travel through a vacuum?
Light and heat waves
Longitudinal and transverse waves
Sound waves
Surface waves
Answers
1. When you talk into your paper cup telephone, the person on the other end can feel the bottom
of their telephone vibrate. Why do you think this happens?
Answers
Answer:
The correct answer is - the sound waves make vibration that travels through the string.
Explanation:
When an individual person talks into your paper cup telephone the person on the other end can feel the bottom of their cup vibrate. The sound waves create vibration go through the string that travels through the string to the end of the cup where vibrations can feel.
The sound waves are longitudinal waves that move or travel through different mediums like air, solid, or gas. The waves create vibration in the particles.
You could create a paper cup telephone but instead of using string, test out different materials and see if those materials will allow sound vibrations to travel through them
Juans mother drives 7.25 miles southwest to her favorite shopping mall. What is the average velocity of her automobile if she arrives at the mark in 20min?
Answers
I got this by multiplying 7.25(3) because I know 20 minutes is 1/3 of 1 he
is c20h12 organic or not organic
Answers
Answer:
.
Explanation:
Calculate the difference in blood pressure between the feet and top of the head for a person who is 1.70 m tall.
Answers
Answer:
[tex]P_2 - P_1 = 1.8 * 10^4\ Pa[/tex]
Explanation:
Given
[tex]Height (h) = 1.70m[/tex]
Required
Determine the difference in the blood pressure from feet to top
This is calculated using Pascal's second law.
The second law is represented as:
[tex]P_2 = P_1 + pgd[/tex]
Subtract P1 from both sides
[tex]P_2 - P_1 = pgd[/tex]
Where
[tex]p = blood\ density = 1.06 * 10^3kg/m^3[/tex]
[tex]g = acceleration\ of\ gravity = 9.8N/kg[/tex]
[tex]d =height = 1.70m[/tex]
P2 - P1 = Blood Pressure Difference
So, the expression becomes:
[tex]P_2 - P_1 = 1.06 * 10^3 * 9.8 * 1.70[/tex]
[tex]P_2 - P_1 = 17659.6Pa[/tex]
[tex]P_2 - P_1 = 1.8 * 10^4\ Pa[/tex]
Hence, the difference in blood pressure is approximately [tex]1.8 * 10^4\ Pa[/tex]
When the torques on the wheel and axle equal zero, the machine has
_____
Answers
Answer:
8
Explanation:
Answer:
rotational equilibrium
Explanation:
EASY!!
A 5000kg truck and a 50 kg person have the same momentum. How is this possible?
Answers
Explanation:
The mass of a truck = 5000 kg
Mass of a person = 50 kg
The momentum of a body is given by :
p = mv
m is mass
If momentum of the truck and the person is same. Let v₁ and v₂ are velocity of the truck and the person.
[tex]5000\times v_1=50\times v_2\\\\\dfrac{v_1}{v_2}=\dfrac{50}{5000}\\\\\dfrac{v_1}{v_2}=\dfrac{1}{100}\\\\v_2=100v_1[/tex]
So, it can only possible if the velocity of the person is 100 times of the velocity of the truck.
Can yall help me please
Answers
Answer:
physical
Explanation:
it is a real life model that was built
Sunita uses a lever of length 5m to lift a piece of stone weighing 1000N.
If the distance of the load from the fulcrum is Im, how much effort does she
apply to lift the piece of stone
Answers
Answer:
F = 250 [N] (Force exerted by Sunita)
Explanation:
First we must understand the use of the lever, in the attached picture we see that it is fulcrum and other features.
The moment that the load exerts with respect to the fulcrum should be calculated, as the distance is 1 [m].
[tex]M=F*d\\M=1000*1\\M=1000[N*m][/tex]
Now Sunita must exert the same moment using the remaining distance (4 [m]).
[tex]1000=F_{sunita}*4\\F_{sunita}=1000/4\\F_{sunita}=250 [N][/tex]
The density of a solid or liquid material divided by the density of water is called
Answers
Answer:
I believe the answer is specific gravity
Explanation:Hope this helps :)
What height would a 4 kg book need to be to have a potential energy of
235.2 J on earth?*
need help!!
Answers
Answer:
5.99 m = 6 m
Explanation:
PE = m*g*h
235.2 J = (4 kg)(9.81 m/s^2)(h)
h = (235.2 J)/(9.81*4)
h = 5.99 m
h = 6 m
A steel cable of diameter 3.0 cm supports a load of 2.0kN. What is the fractional length increase of the cable compared with the length when there is no load if Y=2.0 x 10^11 Pa?
Answers
Answer:
1.429*10^-5 m
Explanation:
From the question, we are given that
Diameter of the cable, d = 3 cm = 0.03 m
Force on the cable, F = 2 kN
Young Modulus, Y = 2*10^11 Pa
Area of the cable = πd²/4 = (3.142 * 0.03²) / 4 = 0.0028 / 4 = 0.0007 m²
The fractional length = Δl/l
Δl/l = F/AY
Δl/l = 2000 / 0.0007 * 2*10^11
Δl/l = 2000 / 1.4*10^8
Δl/l = 1.429*10^-5 m
Therefore, the fractional length is 1.429*10^-5 m long
Problem 1. A tugboat exerts a constant force of 4000N toward the right on a ship, moving it
a distance of 15m. What work is done?
Answers
Answer:
75,000 JExplanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 5000 × 15
We have the final answer as
75,000 JHope this helps you
The correct answer is 40 or - 40??
Answers
Answer:
40 cm
hope it's help you but you can choose if your ans. is -40 cm
A stationary 12.5 kg object is located on a table near the surface of the earth. The coefficient of static friction between the surfaces is 0.50 and of kinetic friction is 0.30. Show all work including units.
A horizontal force of 15 N is applied to the object.
a. Draw a free body diagram with the forces to scale.
b. Determine the force of friction.
c. Determine the acceleration of the object.
Answers
When the object is at rest, there is a zero net force due the cancellation of the object's weight w with the normal force n of the table pushing up on the object, so that by Newton's second law,
∑ F = n - w = 0 → n = w = mg = 112.5 N ≈ 113 N
where m = 12.5 kg and g = 9.80 m/s².
The minimum force F needed to overcome maximum static friction f and get the object moving is
F > f = 0.50 n = 61.25 N ≈ 61.3 N
which means a push of F = 15 N is not enough the get object moving and so it stays at rest in equilibrium. While the push is being done, the net force on the object is still zero, but now the horizontal push and static friction cancel each other.
So:
(a) Your free body diagram should show the object with 4 forces acting on it as described above. You have to draw it to scale, so whatever length you use for the normal force and weight vectors, the length of the push and static friction vectors should be about 61.3/112.5 ≈ 0.545 ≈ 54.5% as long.
(b) Friction has a magnitude of 15 N because it balances the pushing force.
(c) The object is in equilibrium and not moving, so the acceleration is zero.
1. A sailor pulls a boat along a dock using a rope at an angle of 60.0º with the horizontal. How much work does the sailor do if he exerts a force of 255 N on the rope and pulls the boat 3.00 m?
W = FdcosO =255 x 3 x cos 60 =
2. An elephant pushes with 2000 N at an angle of 33o above the horizontal on a load of trees. It then pushes these trees for 150 m. How much work did the elephant do?
3. Hans Full is pulling on a rope to drag his backpack to school across the ice. He pulls upwards and rightwards with a force of 22.9 Newtons at an angle of 35o above the horizontal to drag his backpack a horizontal distance of 129 m. Determine the work done upon the backpack.
4. If 100 N force has 30o angle pulling on a 15 kg block for 5 m. What’s the work?
Answers
Answer:
(1)Work done by snail is 382.5 N
(2)Work done by ELEPHANT is 25160 N
(3)Work done by HANS is 2420 N
(4)Work done on block is 433 N
Explanation:
The work done due to force F applied at an angle θ from the horizontal to the body is give by
Work done = Fscosθ where, s is distance traveled by body
Case 1: F= 255N , s= 3.00m and θ = 60.[tex]0^0[/tex]
Work done = Fscosθ = 255 x 3.00 x cos60.[tex]0^0[/tex] = 382.5N
thus work done by snail is 382.5 N
Case 2: F= 200N , s= 150m and θ = 33.[tex]0^0[/tex]
Work done = Fscosθ = 200 x 150 x cos33.[tex]0^0[/tex] = 25160N
thus work done by elephant is 25160 N
Case 3: F= 22.9N , s= 129m and θ = 35.[tex]0^0[/tex]
Work done = Fscosθ = 22.9 x 129 x cos35.[tex]0^0[/tex] = 2420N
thus work done by Hans is 2420 N
Case 4: F= 100N , s= 5m and θ = 30.[tex]0^0[/tex]
Work done = Fscosθ = 100 x 5.00 x cos30.[tex]0^0[/tex] = 433N
thus work done on block is 433 N
(1) The work done by the sailor is 382.5 J.
(2) The work done by elephant is 25160 J.
(3) The work done by Hans upon the backpack is 2420 J.
(4) The required work done on block is 433 J.
Let us solve these questions in parts. All these questions are based on the work done. The work done due to force F applied at an angle θ from the horizontal to the body is give by,
[tex]W =F \times s \times cos \theta[/tex]
Here, s is the distance covered by the body.
(1)
Given data:
F= 255N , s= 3.00m and θ = 60.
The work done is calculated as,
W = Fscosθ
W= 255 x 3.00 x cos60 = 382.5 J.
Thus, the work done by the sailor is 382.5 J.
(2)
Given data:
F= 200N , s= 150m and θ = 33.
The work done by the elephant is calculated as,
W = Fscosθ
W= 200 x 150 x cos33. = 25160 J
Thus, the work done by elephant is 25160 J.
(3)
Given data:
F= 22.9N , s= 129m and θ = 35.
Then the work done upon the backpack is calculated as,
W= Fscosθ
W= 22.9 x 129 x cos35. = 2420 J
Thus, the work done by Hans upon the backpack is 2420 J.
(4)
Given data:
F= 100N , s= 5m and θ = 30.
The work done is calculated as,
W = Fscosθ
W= 100 x 5.00 x cos30. = 433 J
Thus, the required work done on block is 433 J.
Learn more about the work done here:
https://brainly.com/question/13662169
I NEED HELP WITH 8 I WILL GIVE U BRAINLIEST
Answers
Answer:
u will need 1/2 cells
Explanation:
Numerical Problems
A bus is moving with the initial velocity 10 m/s. After 2 seconds, the velocity
becomes 20 m/s. Find the acceleration and distance moved by the bus.
Answers
Answer:
a = 5 [m/s²]
Explanation:
To solve this problem we must use the following equation of kinematics.
[tex]v_{f}=v_{o}+a*t[/tex]
where:
Vf = final velocity = 20 [m/s]
Vo = initial velocity = 10 [m/s]
t = time = 2 [s]
a = acceleration [m/s²]
Now replacing:
[tex]20 =10 +a*2\\10=2*a\\a=5[m/s^{2} ][/tex]
Homeostasis refers to the ability of the body to maintain a stable internal environment despite changes in external conditions.
True
False
Answers
Answer:
True
Explanation:
Which of the following best describes the charge of the nucleus of an atom?
A. The nucleus can have a positive, neutral, or negative charge.
B. The nucleus always has a positive charge.
c. Jahe nucleus always has a neutral charge.
D. The nucleus always has a negative charge.
SUBMIT
Answers
what is the force of gravitational attraction between an object with a mass of 0.5kg and another object that had a mass of 0.33kg and a distance between them of 0.002m
Answers
Answer:
1.1x10^-4 N
Explanation:
F = G[(m_1*m_2)/r^2]
G = Gravitational constant 6.67433x10^-11 (N*m^2)/(kg^2)
m_1 = mass 1
m_2 = mass 2
r = radius between objects
F = G[(0.5kg*0.33kg)/(0.001m)^2]
F = 1.1x10^-4 N
What are the characteristics of supernovae?
release of matter and energy
provide new material for future solar systems
move on to form a black hole or neutron star
might be seen from earth
Answers
Answer:
c
Explanation:
spaceship of mass m travels from the Earth to the Moon along a line that passes through the center of the Earth and the center of the Moon. (a) At what distance from the center of the Earth is the force due to the Earth twice the magnitude of the force due to the Moon
Answers
Answer:
the correct result is r = 3.71 10⁸ m
Explanation:
For this exercise we will use the law of universal gravitation
F = [tex]- \frac{m_{1} m_{2} }{r^2}[/tex]
We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m
rocket force -Earth
F₁ = - \frac{m' M }{r^2}
rocket force - Moon
F₂ = - \frac{m' m }{(d-r)^2}
in the problem ask for what point the force has the relation
2 F₁ = F₂
let's substitute
2 [tex]2 \frac{M}{r^2} = \frac{m}{(d-r)^2}[/tex]
(d-r) ² = [tex]\frac{m}{2M}[/tex] r²
d² - 2rd + r² = \frac{m}{2M} r²
r² (1 -\frac{m}{2M}) - 2rd + d² = 0
Let's solve this quadratic equation to find the distance r, let's call
a = 1 - \frac{m}{2M}
a = 1 - [tex]\frac{7.36 10^{22} }{2 \ 5398 10^{24}}[/tex] = 1 - 6.15 10⁻³
a = 0.99385
a r² - 2d r + d² = 0
r = [tex]\frac {2d \frac{+}{-} \sqrt{4d^2 - 4 a d^2}} {2a}[/tex]
r = [2d ± 2d [tex]\sqrt{1-a}[/tex]] / 2a
r = [tex]\frac{d}{a}[/tex] (1 ± √ (1.65 10⁻³)) = [tex]\frac{d}{a}[/tex] (1 ± 0.04)
r₁ = \frac{d}{a} 1.04
r₂ = \frac{d}{a} 0.96
let's calculate
r₁ = [tex]\frac{3.84 10^8}{0.99385}[/tex] 1.04
r₁ = 401.8 10⁸ m
r₂ = \frac{3.84 10^8}{0.99385} 0.96
r₂ = 3.71 10⁸ m
therefore the correct result is r = 3.71 10⁸ m
Explain why crinoid fossils can be found on the shore of Lake Michigan, which is a fresh water lake.
Answers
Answer:
Explanation:
Crinoid fossils can be found on the shore of Lake Michigan because during the Carboniferous period, all of what is now the United States, except for a small part of the upper Midwest, and all of the states along the East Coast, except for Florida, was covered by a warm, shallow inland sea(seawater), and since crinoid fossils live in sea water, they washed up on many places like Lake Michigan.
Answer:
Explanation:
Crinoid fossils can be found on the shore of Lake Michigan because during the Carboniferous period, all of what is now the United States, except for a small part of the upper Midwest, and all of the states along the East Coast, except for Florida, was covered by a warm, shallow inland sea(seawater), and since crinoid fossils live in sea water, they washed up on many places like Lake Michigan.
A north magnetic pole brought near a south magnetic pole will...
A)
cancel the south pole.
B)
attract the south pole.
C)
repel the south pole.
Answers
What does a light-year measure? A. volume B. weight C. circumference D. distance
Answers
Answer:
Distance. A light-year is the distance a single photon of light can travel in empty space in a year.
Answer:
D-Distance
Explanation:
Revolution + tilt = what
Answers
Example
A 1.8 m tallman stand in an elevator accelerating upward at 12 m/s?,
what is the blood pressure in the brain and foot.
Take the height difference between the heart and the brain to be
0.35 m?
13.3 x 103 Pa] & [Pblood = 1060 kg.m-3]
Note: [: P Heart
Answers
Answer:
Explanation:
From the given information; the diagram below shows a clearer understanding.
The blood pressure in the brain [tex]P_{brain} = P_{heart} - \delta ( a - g ) h[/tex]
= 13300 - 1060 (12-9.81) 0.35
= 13300 - 1060 (2.19) 0.35
= 13300 - 812.49
= 12487.51 Pa
The blood pressure in the feet [tex]P_{feet} = P_{heart} + \delta (a + g) h[/tex]
= 13300 + 1060 (12 + 9.81) 1.45
= 13300 + 1060( 21.81 ) 1.45
= 13300 + 33521.97
= 46821.97 Pa
Answer:
The blood pressure in the brain = [tex]12487.51 pa[/tex]The blood pressure in the feet = [tex]46821.97pa[/tex]Explanation:
[tex]P_brain = P_heart - y(a - g)h\\\\P_brain = 13300 - 1060 (12-9.81)0.35\\\\P_brain = 12487.51 pa\\\\[/tex]
[tex]P_feet = P_heart + y(a+g)*h_r\\\\P_feet = 13300 + 1060(12+9.81)*(1.8-0.35)\\\\P_feet = 46821.97pa[/tex]
For more information, visit
https://brainly.com/question/20534740
The x-axis of a position-time graph represents
Answers
Answer:
The y-axis represents position relative to the starting point, and the x-axis represents time.
Explanation:
1. What is the momentum of a 1550 kg car that is traveling leftward at a velocity of 15 m/s?
Answers
Answer:
Momentum, p = 23250 kg m/s
Explanation:
Given that
Mass of a car, m = 1550 kg
Speed pf car, v = 15 m/s
We need to find the momentum of the car. The formula for the momentum of an object is given by :
p = mv
Substituting all the values in the above formula
p = 1550 kg × 15 m/s
p = 23250 kg m/s
So, the momentum of the car is 23250 kg m/s.