Answer:
0.0119502868 kilocalorie
Explanation:
Answer:
increases by 50
Explanation:
A 1 m3tank containing air at 10oC and 350 kPa is connected through a valve to another tank containing 3 kg of air at 35oC and 150 kPa. Now the valve is opened, and the entire system is allowed to reach thermal equilibrium with the surroundings, which are at 20oC. Determine the volume of the second tank and the final equilibrium pressure of air.
Answer:
- the volume of the second tank is 1.77 m³
- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Explanation:
Given that;
[tex]V_{A}[/tex] = 1 m³
[tex]T_{A}[/tex] = 10°C = 283 K
[tex]P_{A}[/tex] = 350 kPa
[tex]m_{B}[/tex] = 3 kg
[tex]T_{B}[/tex] = 35°C = 308 K
[tex]P_{B}[/tex] = 150 kPa
Now, lets apply the ideal gas equation;
[tex]P_{B}[/tex] [tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex]
[tex]V_{B}[/tex] = [tex]m_{B}[/tex]R[tex]T_{B}[/tex] / [tex]P_{B}[/tex]
The gas constant of air R = 0.287 kPa⋅m³/kg⋅K
we substitute
[tex]V_{B}[/tex] = ( 3 × 0.287 × 308) / 150
[tex]V_{B}[/tex] = 265.188 / 150
[tex]V_{B}[/tex] = 1.77 m³
Therefore, the volume of the second tank is 1.77 m³
Also, [tex]m_{A}[/tex] = [tex]P_{A}[/tex][tex]V_{A}[/tex] / R[tex]T_{A}[/tex] = (350 × 1)/(0.287 × 283) = 350 / 81.221
[tex]m_{A}[/tex] = 4.309 kg
Total mass, [tex]m_{f}[/tex] = [tex]m_{A}[/tex] + [tex]m_{B}[/tex] = 4.309 + 3 = 7.309 kg
Total volume [tex]V_{f}[/tex] = [tex]V_{A}[/tex] + [tex]V_{B}[/tex] = 1 + 1.77 = 2.77 m³
Now, from ideal gas equation;
[tex]P_{f}[/tex] = [tex]m_{f}[/tex]R[tex]T_{f}[/tex] / [tex]V_{f}[/tex]
given that; final temperature [tex]T_{f}[/tex] = 20°C = 293 K
we substitute
[tex]P_{f}[/tex] = ( 7.309 × 0.287 × 293) / 2.77
[tex]P_{f}[/tex] = 614.6211119 / 2.77
[tex]P_{f}[/tex] = 221.88 kPa ≈ 222 kPa
Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
The string will break if the tension in
it exceeds 0.180 N. What is the
smallest possible value of d (in cm)
before the string breaks?
Answer:
define d first?
you need to list more variables
Answer:
list more valuable unit
energy in a stretched spring
Answer:
Elastic potential energy is the potential energy stored by stretching or compressing an elastic object by an external force such as the stretching of a spring. It is equal to the work done to stretch the spring which depends on the spring constant k and the distance stretched.
Hope this helps :)
-ilovejiminssi♡
What are two things that happen to the sugars that are made by the plant during photosynthesis?
I
Answer:
The sugars produced by photosynthesis can be stored, transported throughout the tree, and converted into energy which is used to power all cellular processes. Respiration occurs when glucose (sugar produced during photosynthesis) combines with oxygen to produce useable cellular energy.
Explanation:
I think this is correct lol.
Diagram B D c с Which car has: Ke = 100 PE=0? * 1 point A B C D
Answer:
The car C has KE = 100, PE = 0
Explanation:
The principle of conservation of energy states that although energy can be transformed from one form to another, the total energy of the given system remains unchanged.
The energy that a body possesses due to its motion or position is known as mechanical energy. There are two kinds of mechanical energy: kinetic energy, KE and potential energy, PE.
Kinetic energy is the energy that a body possesses due to its motion.
Potential energy is the energy a body possesses due to its position.
From the principle of conservation of energy, kinetic energy can be transformed into potential energy and vice versa, but in all cases the energy is conserved or constant.
In the diagram above, the cars at various positions of rest or motion are transforming the various forms of mechanical energy, but the total energy is conserved at every point. At the point A, energy is all potential, at B, it is partly potential partly kinetic energy, However, at the point C, all the potential energy has been converted to kinetic energy. At D, some of the kinetic energy has been converted to potential energy as the car climbs up the hill.
Therefore, the car C has KE = 100, PE = 0
The new springs will be identical to the original springs, except the force constant will be 5655.00 N/m smaller. When James removes the original springs, he discovers that the length of each spring expands from 8.55 cm (its length when installed) to 12.00 cm (its length with no load placed on it). If the mass of the car body is 1355.00 kg, by how much will the body be lowered with the new springs installed, compared to its original height
Answer:
Explanation:
For original spring , compression in spring due to a load of 1355 kg is
x = 12 - 8.55 = 3.45 cm = .0345 m
spring constant = W / x
= 1355 x 9.8 / .0345
= 384898.55 N /m
Spring constant of new spring
k = 384898.55 - 5655 = 379243.55 N /m
New compression for new spring
= W / k
= 1355 x 9.8 / 379243.55
= .035 m
= 3.50 cm
Difference of compression = 3.50 - 3.45
= .05 cm .
In later case , car will be more lowered by .05 cm .
If a shopping cart is pushed by a person exerting 50 J of work on it, what is the energy transfer to the shopping cart if it has a mass of 2 kg?
A. 52 J
B. 50 J
C. 25 J
Answer:
B
Explanation:
What current is needed in the solenoid's wires?
A researcher would like to perform an experiment in a zero magnetic field, which means that the field of the earth must be canceled. Suppose the experiment is done inside a solenoid of diameter 1.0 m, length 3.8 m , with a total of 5000 turns of wire. The solenoid is oriented to produce a field that opposes and exactly cancels the 52 μT local value of the earth's field.
What current is needed in the solenoid's wires? Express your answer with the appropriate units.
2. (9 points) A car starts from 10 mph and accelerates along a level road, i.e., no grade change. At 500 ft from its starting point, a radar gun measures its speed as 50 mph. Assuming the car had a constant rate of acceleration, (a) calculate the time elapsed between when the car started at 10 mph to when its speed was measured and (b) what will the speed of the car be another 500 ft downstream of this point
Answer:
a) t = 11.2 s
b) v = 70.5 mph
Explanation:
a)
Since we need to find the time, we could use the definition of acceleration (rearranging terms) as follows:[tex]t = \frac{v_{f} - v_{o}}{a} (1)[/tex]
where vf = 50 mph, and v₀ = 10 mph.However, we still lack the value of a.Assuming that the acceleration is constant, we can use the following kinematic equation:[tex]v_{f} ^{2} - v_{o} ^{2} = 2*a* \Delta x (2)[/tex]
Since we know that Δx = 500 ft, we could solve (2) for a.In order to simplify things, let's first to convert v₀ and vf from mph to m/s, as follows:[tex]v_{o} = 10 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 4.5 m/s (3)[/tex]
[tex]v_{f} = 50 mph*\frac{1609m}{1mi} *\frac{1h}{3600s} = 22.5 m/s (4)[/tex]
We can do the same process with Δx, from ft to m, as follows:[tex]\Delta x = 500 ft *\frac{0.3048m}{1ft} = 152.4 m (5)[/tex]
Replacing (3), (4), and (5) in (2) and solving for a, we get:[tex]a = \frac{v_{f} ^{2} - v_{o}^{2}}{2*\Delta x} = \frac{(22.5m/s) ^{2} - (4.5m/s)^{2}}{2*152.4m} = 1.6 m/s2 (6)[/tex]
Replacing (6) in (1) we finally get the value of the time t:[tex]t = \frac{v_{f} - v_{o}}{a} = \frac{(22.5m/s) - (4.5m/s)}{1,6m/s2} = 11.2 s (7)[/tex]
b)
Since the acceleration is constant, as we know the displacement is another 500 ft (152.4m), if we replace in (2) v₀ by the vf we got in a), we can find the new value of vf, as follows:[tex]v_{f} = \sqrt{v_{o} ^{2} +( 2*a* \Delta x)} = \sqrt{(22.5m/s)^{2} + (2*1.6m/s2*152.4m)} \\ v_{f} = 31.5 m/s (8)[/tex]
If we convert vf again to mph, we have:[tex]v_{f} = 31.5m/s*\frac{1mi}{1609m} *\frac{3600s}{1h} = 70.5 mph (9)[/tex]
Matter is made of small particles to small to be seen. Which of these best describe evidence of this statement? 1. Tara’s crayons melted when she left them under the sun. 2. Kerris glass of water overflowed when she added ice. 3. Sams basketball expands as he pumps air into it. 4. Stephanie dropped a vase and it broke into pieces.
Answer:
Explanation:
I think the answer is statement no 3.
Hope it helps.
Answer:
1 Tara's crayons melted when she left them under the sun
What is the correct coefficient for 2H2 + O2 →2H2O
Explanation:
2forH2,1for02,and2forH20
Question 15 of 25
What is the period of a wave that has a frequency of 30 Hz?
Answer:
0.033 seconds
Explanation:
Period = 1/30 = 0.033 seconds
Answer:
The answer is 0.03 s
Explanation:
A.P.E.X.
g Suppose that you seal an ordinary 60W lightbulb and a suitable battery inside a transparent enclosure and suspend the system from a very sensitive balance. (a) Compute the change in the mass of the system if the lamp is on continuously for one year at full power. (b) What difference, if any, would it make if the inner surface of the container were a perfect reflector
Answer:
kekemeeimdeiddnekem
Explanation:
mdjdjdiddmjd jjeneeiej
Using a simply pulley/rope system, a crewman on an Arctic expedition is trying to lower a 6.17-kg crate to the bottom of a steep ravine of height 23.8 meters. The 55.6-kg crewman is walking along holding the rope, being careful to lower the crate at a constant speed of 1.50 m/s. Unfortunately, when the crate reaches a point 13.2 meters above the ground, the crewman steps on a slick patch of ice and slips. The crate immediately accelerates toward the ground, dragging the hapless crewman across the ice and toward the edge of the cliff.
If we assume the ice is perfectly slick (that is, no friction between the crewman and the ice once he slips and falls down), at what speed will the crate hit the ground? Assume also that the rope is long enough to allow the crate to hit the ground before the crewman slides over the side of the cliff. At what speed will the crewman hit the bottom of the ravine?
Answer:
a. Vc = 5.06 m/s
b. Vp = 22.18 m/s
Explanation:
The acceleration of the pulley-mass system is as follows:
a = [tex]\frac{mg}{m + M}[/tex]
Solving for acceleration, we get:
a = [tex]\frac{6.17 *9.8}{6.17 + 55.6}[/tex]
a = 0.97
So, for the part a:
Calculate the velocity of the crewman by using the following equation:
Vc = [tex]\sqrt{Vi^{2} + 2ay}[/tex]
Substituting the values into the equation, we get:
Vc = [tex]\sqrt{1.50^{2} + 2*0.97*13.2}[/tex]
Vc = 5.06 m/s
Now, for part b:
Calculate the final velocity of the pulley by using the following expression:
Vp = [tex]\sqrt{Vi^{2} +2gy }[/tex]
Just plugging in the values.
Vp = [tex]\sqrt{5.06^{2} +2*9.8*23.8 }[/tex]
Vp = 22.18 m/s
Potential energy is energy due to motion.
True or False?
Answer:
true
Explanation:
Answer:
true
Explanation:
please give brainlest need 1
Which of the following does NOT have a positive impact on your position on the
health continuum?
avoiding risk behaviors
having a positive social environment
eating nutritious foods
O having a chronic disease
Answer:
Having a chronic disease
Explanation:
The steepness of a line on a graph is called the
O A. rise
OB. slope
C.
run
D. verticle axis
Answer:
slope
Explanation:
The slope is how how steep the line is.
a passenger on cruise between San Juan, Puerto Rico and Miami, Florida accidentally drops a souvenir metal cube over the side of the boat, into the water. Each side of the metal cube measures 1 meter. The cube promptly sinks to the deepest part of the Puero Rico Trench. Once at the bottom, what pressure does the cube experience? Neglect Atmospheric Pressure. Use wikipedia to see depth of Trench!
Answer:
P = 84.1 MPa
Explanation:
The pressure at the bottom of column of of salt water of height h, is given by the following expression:[tex]P = \rho * g * h (1)[/tex]
where ρ = density of salt water (in Kg/m³),
g = acceleration due to gravity (in m/s²)
h = height of the column of water.
Replacing by their values in (1):[tex]P = \rho * g * h = 1023.6kg/m3*9.8m/s2*8380m = 84.1 MPa (2)[/tex]
Neglecting the atmospheric pressure, the pressure on the cube at the bottom of Puerto Rico Trench is given by (2):P = 84.1 MPa.Plzz help me with this
I’ll give brainliest
Answer:
B. Objects with more mass have more gravitational force acting upon them.
Answer:
Should be A but it can be B as well.
A 0.15 kg baseball collides with a 1.0 kg bat. The ball has a velocity of 40 m/s immediately before the collision. The center of mass of the bat also has a velocity of 40 m/s, but in the opposite direction, just before the collision. The coefficient of restitution between the bat and the ball is 0.50. Estimate how fast the baseball is moving as it leaves the bat following the collision.
Answer:
The final velocity of the baseball as it leaves the bat is 40 m/s
Explanation:
The given parameters of the baseball and bat are;
The mass of the baseball = 0.15 kg
The mass of the bat = 1.0 kg
The velocity of the ball before collision, v₁ = 40 m/s
The velocity of the bat before collision, v₂ = -40 m/s
The coefficient of restitution, e = 0.50
Let, 'v₃', and 'v₄' represent the final velocity of the ball and the bat respectively after collision, we have;
Taking the final velocity of the bat, v₄ = 0 m/s
According to Newton's Law of restitution
e = (v₃ - v₄)/(v₁ - v₂)
∴ 0.5 = (v₃ - v₄)/(40 - (-40))
80 × 0.5 = 40 = (v₃ - v₄)
v₃ - v₄ = 40
v₃ = 40 + v₄ = 40 + 0 = 40
The final velocity of the baseball as it leaves the bat, v₃ = 40 m/s.
How does altitude from the surface of earth affect the time period of a simple pendulum
Answer:
because the strength of Earth's gravitational field is not uniform everywhere, a given pendulum swings faster, and thus has a shorter period, at low altitudes and at Earth's poles than it does at high altitudes and at the Equator.
A car starts from rest at a stop sign. It accelerates at 2.0 m/s2 for 6.2 seconds, coasts for 2.5 s , and then slows down at a rate of 1.5m/s2 for the next stop sign. How far apart are the stop signs
Answer:
the distance between the stop signs is 120.7 m.
Explanation:
The car moved in three stages;
(1) It accelerates from rest at 2.0 m/s² for 6.2 seconds
(2) it moved at a constant speed for 2.5 s
(3) it finally decelerate at the rate of 1.5m/s²
(1) The distance moved by the car during the first stage;
s₁ = ut + ¹/₂at²
s₁ = 0 + ¹/₂ (2)(6.2)²
s₁ = 38.44 m
(2) The distance moved by the car during the second stage;
calculate the constant speed of the car,
v = u + at
v = 0 + 2 x 6.2
v = 12.4 m/s
The distance moved by the car as it coasts for 2.5s: s₂ = vt
s₂ = 12.4 x 2.5
s₂ = 31 m
(3) The distance moved by the car during the third stage;
When the car stops, the final velocity is zero.
v² = u² + 2as₃
a = -1.5 m/s², since the car slowed down or decelerated.
0 = 12.4² + (2 x - 1.5)s₃
0 = 153.76 - 3s₃
3s₃ = 153.76
s₃ = 153.76 / 3
s₃ = 51.253 m
The total distance moved by the car from the start to stop = s₁ + s₂ + s₃
d = 38.44 m + 31 m + 51.253 m
d = 120.7 m
Therefore, the distance between the stop signs is 120.7 m.
a girl whose mass is 40kg walk up a flight of 20steps each 15mm hight in 10seconds.find power developed by the girl showing the solution
Answer: Approximately 11.76 joules per second
=========================================================
Work Shown:
Mass = 40 kg
Force pulling down = (mass)*(gravity) = 40*9.8 = 392 newtons
Roughly 392 newtons of force are pulling down on her.
To climb the steps, she must apply 392 newtons of force upward.
---------------
Displacement = 20*(15 mm) = 300 mm = 0.3 m
Work = Force*Displacement
Work = 392*0.3
Work = 117.6 joules of energy
---------------
Power = (Work)/(Time)
Power = (117.6 joules)/(10 seconds)
Power = (117.6/10) joules per second
Power = 11.76 joules per second, which is approximate
The combination of an applied force and a friction force produces a constant torque of 36.0 N⋅m on a wheel rotating on a fixed axis. While the force acts for 6.00 s, the angular velocity of the wheel increases from 0 to 10.0 rad/s. The force is removed and the wheel comes to rest in 60.0 s
a. Find the moment of inertia of the wheel.
b. Find the magnitude of the torque due to friction.
c. Find the total number of rotations during the 66.0 s.
d. Find the Kinetic energy of the wheel at 6.00 s when the force is removed.
Answer:
D. Find the kinetic energy of the wheel at 6.00 z when the force is removed.
a man is trying to pull a box a distance of 3 m with a force of 20 N that makes a 35º with the horizontal.
Answer:
34.4Joules
Explanation:
Complete question
a man is trying to pull a box a distance of 3 m with a force of 20 N that makes a 35º with the horizontal. Find the workdone
Work done = Fdsin theta
Force F = 20N
distance d = 3m
theta = 35 degrees
Substitute
Workdone = 20(3)sin 35
Workdone = 60sin35
Workdone = 34.4Joules
Hence the workdone by the man is 34.4Joules
. A 13-g goldfinch has a speed of 8.5 m/s. What is its kinetic energy?
The kinetic energy is 0.469 Joule.
The Kinetic energy is given as,
[tex]K.E=\frac{1}{2} m*v^{2} [/tex]
Where m is mass and v is speed of object.
Given that, [tex]m=13g=\frac{13}{1000}=0.013Kg [/tex] and [tex]v=8.5m/s[/tex]
Substitute values in above expression.
[tex]K.E=\frac{1}{2}*0.013*(8.5)^{2}=0.469Joule [/tex]
Hence, The kinetic energy is 0.469 Joule.
Learn more:
https://brainly.com/question/12337396
Imagine a third particle, which we will call a cyberon. It has three times the mass of an electron (3_m). It has a positive charge that is three times the magnitude (3_(qe)) of the charge on an electron. What is the ratio of the speed v_c that the cyberon would have when it reaches the upper plate after being released from rest at position h_0 to the speed ve that the electron would have?
Answer:
The answer is "The last choice".
Explanation:
Please find the complete question in the attachment.
In an external electric field, its electrical energy at positive charge becomes directed to just the electrical domain. Therefore it will speed towards its base plate whenever cyber one is released to rest at h0. It was never going to reach the top plate. Thus, the last choice corrects because in this the cyber-on never reaches its upper stage.
Help plz I’ll mark brainliest
Answer:
The second option- a substance that a wave can travel through.
Explanation:
Hope This Helps!!
(brainliest please)
Question 2 of 25
Which three statements are true about the wave shown?
A. The wave is a longitudinal wave.
B. The wave could be an electromagnetic wave.
C. The wave could be a sound wave.
D. The wave is a mechanical wave.
Answer:
ACD
Explanation:
according to it's description above
The wave is a mechanical wave and can also be sound wave also the wave is a longitudinal wave. The correct option is A, C, and D.
What is mechanical wave?A mechanical wave is one whose energy cannot be transmitted via a vacuum. To transfer their energy from one place to another, mechanical waves need a medium.
A mechanical wave is something like a sound wave. A vacuum can not be traversed by sound waves.
Waves like mechanical waves require a medium to travel through. Non-mechanical waves are those that can travel through any medium.
Mechanical waves include, but are not limited to, sound waves, water waves, and seismic waves.
The three claims are accurate as follows: The wave is a longitudinal wave and can also be a mechanical wave or a sound wave.
Thus, the correct option is A, C, and D.
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Your question seems incomplete, the missing image is:
Estimate the constant rate of withdrawal (in m3 /s) from a 1375 ha reservoir in a month of 30 days during which the reservoir level dropped by 0.75 m in spite of an average inflow into the reservoir of 0.5 Mm3 /day. During the month, the average seepage loss from the reservoir was 2.5 cm, total precipitation on the reservoir was 18.5 cm and the total evaporation was 9.5 cm
Answer:
Explanation:
1 ha = 10⁴ m²
1375 ha = 1375 x 10⁴ m² = 13.75 x 10⁶ m²
In flow in a month = .5 x 10⁶ x 30 m³ = 15 x 10⁶ m³
Net inflow after all loss = 18.5 - 9.5 - 2.5 cm = 6.5 cm = .065 m
Net inflow in volume = 13.75 x 10⁶ x .065 m³= .89375 x 10⁶ m³
Let Q be the withdrawal in m³
Q - 15 x 10⁶ - .89375 x 10⁶ = 13.75 x 10⁶ x .75 = 10.3125 x 10⁶
Q = 26.20 x 10⁶ m³
rate of withdrawal per second
= 26.20 x 10⁶ / 30 x 24 x 60 x 60
= 26.20 x 10⁶ / 2.592 x 10⁶
= 10.11 m³ / s
The rate of withdrawal is 10.11 cubic meter per second
Given-
Total reservoir is 1375 hectare. which is equal to [tex]1375\times 10^4[/tex] meter square.
The average seepage loss from the reservoir is 2.85 cm or 0.0285 m.
Total precipitation on the reservoir is 18.5 cm or 0.185 m.
Total evaporation is 9.5 cm or 0.085 m.
The average inflow into the reservoir is [tex]0.5\times10^6[/tex] cubic meter per day.
The total inflow in a month can be calculate is
[tex]=0.5\times 30=1500\times10^4[/tex]
Net inflow is equal to the total precipitation on the reservoir subtract by all the losses.It can be represent as,
[tex]Q_{net}=0.185 - 0.095 - 0.025[/tex]
[tex]Q_{net}=0.065[/tex]
Total volume inflow is equal to the product of net inflow and total reservoir,
[tex]V_{net} =1375\times 10^4\times Q_{net}[/tex]
[tex]V_{net} =1375\times 10^4\times 0.065[/tex]
[tex]V_{net} =89.375[/tex]
The constant rate of withdrawal in cubic meter can be calculated by adding the net inflow in a month, total volume inflow and the reservoir.
[tex]Q=1375\times 10^4\times 0.75+89.375\times 10^4+1500\times10^4[/tex]
[tex]Q=2620\times10^4[/tex]
For per second withdrawal,
[tex]Q=\dfrac{2620\times10^4}{30\times24\times60\times60}[/tex]
[tex]Q=10.11[/tex]
Hence, the rate of withdrawal is 10.11 cubic meter per second.
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