The net work done per cycle for 1.00 mol of gas initially at 0°C is 4.88 kJ.
An isothermal process is a thermodynamic process in which the temperature of the system remains constant, while other state variables such as pressure and volume may change. An adiabatic process is a thermodynamic process in which there is no exchange of heat between the system and its surroundings, while other state variables such as pressure and volume may change.
a) The net work done by the gas per cycle is equal to the area enclosed by the cycle on the PV diagram. From the given figure, the cycle can be divided into two parts: the first part is an isothermal expansion from volume V0 to volume 5V0 and the second part is an adiabatic compression from volume 5V0 to volume V0.
For the isothermal expansion, the pressure decreases from P0 to P0/4. Using the equation for the work done during an isothermal process, the work done by the gas during this part of the cycle is:
W1 = nRT0 ln(5) = 4RT0 ln(5)
For the adiabatic compression, the pressure increases from P0/4 to P0. Using the equation for the work done during an adiabatic process, the work done by the gas during this part of the cycle is:
W2 = (P0V0^m/P0/4*(5V0)^m)^(1 - m)/1-m - (P0/4*(5V0)^m/P0/4)^(1 - m)/1-m = 4P0V0/3(1 - 1/5^(4/5))
The net work done by the gas per cycle is the sum of the work done in each part of the cycle:
W = W1 + W2 = 4RT0 ln(5) + 4P0V0/3(1 - 1/5^(4/5))
b) Since the cycle is closed, the net energy added to the system per cycle must be equal to the net work done by the gas per cycle:
Q = W = 4RT0 ln(5) + 4P0V0/3(1 - 1/5^(4/5))
c) To obtain a numerical value for the net work done per cycle for 1.00 mol of gas initially at 0°C, we need to substitute the appropriate values for R, T0, P0, V0, n, and m. Assuming the gas is an ideal gas, R = 8.314 J/mol K. At 0°C (273 K), the pressure of 1.00 mol of gas in a volume of V0 = 22.4 L (molar volume of ideal gas at STP) is P0 = 1 atm. Therefore,
W = 4(8.314 J/mol K)(273 K) ln(5) + 4(1 atm)(22.4 L)/3(1 - 1/5^(4/5))) = 4.88 kJ
Therefore, the net work done per cycle for 1.00 mol of gas initially at 0°C is 4.88 kJ.
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Calculate forces and accelerations for a horizontal spring system. A 0.350 kg object attached to a spring of force constant 1.30 x 10 N/m is free to move on the frictionless horizontal surface. If the object is released from rest at x = 0.100 m, find the force on it and its acceleration at x = 0.100 m, x-0.0500m, x = 0. x= -0.0500m and x = -0.100 m.
The force and acceleration alter as the object moves away from its equilibrium position.
How to calculate force and acceleration?The force on the object at any point is given by Hooke's Law:
F = -kx
where F = force, k = force constant of the spring, and x = displacement of the object from its equilibrium position.
The acceleration of the object at any point is given by Newton's Second Law:
a = F/m
where a = acceleration, F = force, and m = mass of the object.
Using these equations, calculate the force and acceleration at each of the specified points:
At x = 0.100 m:
F = -kx = -(1.30 x 10 N/m)(0.100 m) = -0.130 N
a = F/m = (-0.130 N)/(0.350 kg) = -0.371 m/s² (the negative sign indicates that the acceleration is in the opposite direction to the displacement)
At x = 0:
F = -kx = -(1.30 x 10 N/m)(0) = 0 N (the spring is at its equilibrium position)
a = F/m = 0 N/0.350 kg = 0 m/s²
At x = -0.0500 m:
F = -kx = -(1.30 x 10 N/m)(-0.0500 m) = 0.065 N (note that the force is positive because the displacement is negative)
a = F/m = (0.065 N)/(0.350 kg) = 0.186 m/s²
At x = -0.100 m:
F = -kx = -(1.30 x 10 N/m)(-0.100 m) = 0.130 N
a = F/m = (0.130 N)/(0.350 kg) = 0.371 m/s²
So, the force and acceleration change with the displacement of the object from its equilibrium position.
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An electric heater uses 0,5 MJ when connected to a 220 V supply for 30 minutes. Calculate the power rating of the heater and the current drawn from the supply.
The power rating of the heater is 277.8 W and the current drawn from the supply is 0.07015mA.
The power is obtained from the ratio of work and time and the power is measured in watts(W). The current is defined as the flow of electrons in the circuit and it is measured in ampere(A).
From the given,
The energy of the electric heater = 0.5MJ = 0.5×10⁶ J
Voltage (V) = 220 V
time (t) = 30 minutes = 30×60 s = 1800 s.
To find power (P):
Power = Energy / Time
= 0.5×10⁶ / 1800
= 277.77 W
Power of the electric heater = 277.8 W
To find current from the given:
Power (P) = V×I×t
Current (I) = P/(V×t)
= 277.8 / (220×1800)
= 0.7015 mA
The current drawn from the supply is 0.7015 mA.
Hence, the power rating of the heater is 277.8 W and the current drawn from the supply is 0.7015 mA.
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The hydroelectric dam on the lake has been reintroducing water from its processes before the water has cooled to normal lake temperature, causing a rise in water temperature around the dam. An assessment of the microorganism populations affected by the high water temperature is needed, along with recommendations for the correct ranges of water temperature.
what field of science is this?
This problem is related to environmental science, specifically to the subfield of aquatic ecology.
The process of reintroduction of water from the hydroelectric dam has an influence on the temperature of the water in the nearby lake, which has an impact on the populations of microorganisms. The study of the interactions between the natural world and human society falls under the purview of environmental science, which is relevant to this topic.
Aquatic ecology is a branch of environmental science that focuses on understanding freshwater and marine habitats, as well as the organizms that live there. A thorough analysis of the microorganism populations impacted by the high water temperature is required to solve the issue at hand, as well as suggestions for the suitable ranges of water temperature that will support a healthy ecosystem's functioning.
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A bicycle travels 10.40 km due east in 0.280 h, then 8.80 km at 15.0° east of north in 0.310 h, and finally another 10.40 km due east in 0.280 h to reach its destination. The time lost in turning is negligible. Assume that east is in the +x-direction and north is in the +y-direction.What is the direction of the average velocity for the entire trip? Enter the answer as an angle in degrees north of east.
° north of east
Answer:
19
Explanation:
Need help solving this problem
a) The tension in the rope is 123.9 N.
b) The moment of inertia of the wheel is 0.09 kg m².
c) The angular speed of the wheel 3.50 s after it begins rotating, starting from rest, is 58.5 rad/s.
(a) To determine the tension in the rope, we need to analyze the forces acting on the object. There are two forces: the force of gravity pulling the object down the incline and the tension force pulling the object up the incline.
The force of gravity can be broken down into two components: one parallel to the incline and one perpendicular to the incline.
The parallel component causes the object to accelerate down the incline, while the perpendicular component is balanced by the normal force of the incline.
The tension force is responsible for the object's acceleration down the incline, so we can set up the following equation:
T - mg sin(theta) = ma
where T is the tension force, m is the mass of the object, g is the acceleration due to gravity, theta is the angle of the incline, and a is the acceleration of the object down the incline.
Putting in the given values, we get:
T - (12.5 kg)(9.81 m/s²)(sin(37°)) = (12.5 kg)(2.00 m/s²)
Solving for T, we get:
T = 123.9 N
Therefore, the tension in the rope is 123.9 N.
(b) To determine the moment of inertia of the wheel, we can use the following equation:
I = (1/2)MR²
where I is the moment of inertia, M is the mass of the wheel, and R is the radius of the wheel.
Putting in the given values, we get:
I = (1/2)(12.5 kg)(0.12 m)²
= 0.09 kg m²
Therefore, the moment of inertia of the wheel is 0.09 kg m².
(c) To determine the angular speed of the wheel after 3.50 s, we can use the following equation:
ω = ω₀ + αt
where ω is the final angular speed, ω₀ is the initial angular speed (which is zero in this case), α is the angular acceleration, and t is the time.
We can find the angular acceleration using the following equation:
α = a/R
where a is the acceleration of the object down the incline (which we already know) and R is the radius of the wheel.
Putting in the given values, we get:
α = 2.00 m/s² / 0.12 m
= 16.7 rad/s²
Putting in the values for α and t, we get:
ω = 0 + (16.7 rad/s²)(3.50 s)
= 58.5 rad/s
Therefore, the angular speed of the wheel 3.50 s after it begins rotating, starting from rest, is 58.5 rad/s.
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16 of 20:
Select the best answer for the question.
16. What happens to a substance at critical temperatures?
O A. The substance changes its state if it continues gaining or losing thermal energy.
O B. The substance can't lose any more thermal energy.
O C. The substance can't change its state, only its temperature.
O D. The substance changes its state only if it gains thermal energy.
At the critical temperatures of the substance, the substance can't lose any more thermal energy. Hence, option B is correct.
At the critical temperature, the properties of the liquid and gas phases become difficult to differentiate, and the substance exhibits particular type of behavior such as infinite compressibility and a lack of surface tension.
Additionally, at the critical temperature, the substance reaches its maximum vapor pressure, and any further increase in temperature and pressure will not cause it to change its state but only its density and hence, it cannot lose any more thermal energy from itself.
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A spring scale hung from the ceiling stretches by 5.9 cm
when a 1.6 kg
mass is hung from it. The 1.6 kg
mass is removed and replaced with a 2.1 kg
mass.
Part A
What is the stretch of the spring?
Express your answer with the appropriate units.
The stretch of the spring is proportional to the weight of the mass hung from it. Since the spring stretches by 5.9 cm when a 1.6 kg mass is hung from it, we can use this information to find the stretch when a 2.1 kg mass is hung from it.
The stretch of the spring is given by:
stretch = (mass x gravity x length) / (spring constant)
where mass is the mass hung from the spring, gravity is the acceleration due to gravity (9.81 m/s^2), length is the stretch of the spring, and the spring constant is a measure of the stiffness of the spring (measured in N/m).
We can rearrange this equation to solve for the stretch of the spring:
stretch = (mass x gravity x length) / (spring constant)
length = (spring constant x stretch) / (mass x gravity)
Substituting the given values, we get:
length = (spring constant x 0.059 m) / (1.6 kg x 9.81 m/s^2)
Simplifying, we get:
length = 0.236 m
Therefore, the stretch of the spring when a 2.1 kg mass is hung from it is 0.236 m.
How much current is in a circuit with a 1.5 V battery and three
2-ohm resistances (bulbs) in series?
Answer: 0.25A
Explanation: To calculate the current in the circuit, we can use Ohm's Law, which states that the current (I) flowing through a circuit is equal to the voltage (V) divided by the resistance (R):
I = V / R
In this case, the total resistance of the circuit is the sum of the individual resistances of the three bulbs in series, which gives us:
R = 2 + 2 + 2 = 6 ohms
The voltage of the battery is given as 1.5V.
So, using Ohm's Law, we can calculate the current in the circuit as:
I = V / R = 1.5 / 6 = 0.25 amps
Therefore, the current in the circuit is 0.25 amps.
The current in the circuit is 0.25 A.
Voltage across the circuit, v = 1.5 V
Resistance in each resistors, R = 2Ω
Since, the resistors are connected in series combination, their effective resistance,
R' = 3R
R' = 3 x 2
R' = 6Ω
According to Ohm's law, if the temperature and all other physical factors remain constant, the voltage across a conductor is directly proportional to the current that is flowing through it.
So, according to Ohm's law,
V = IR
Therefore, current flowing through the given circuit,
I = V/R
I = 1.5/6
I = 0.25 A
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The mass of 12 protons is approximately equal to ?
1 atomic mass unit
12 atomic mass units
The mass of 1 electron
The mass of 12 electrons
Answer:
Option 1, 1 atomic mass unit
a bicycle travels 4.00 km due east in 0.650 h, then 12.10 km at 15.0° east of north in 0.500 h, and finally another 4.00 km due east in 0.650 h to reach its destination. the time lost in turning is negligible. assume that east is in the +x-direction and north is the +y-direction. what is the direction of the average velocity for the entire trip?
please i'm begging someone help me
The direction of the average velocity is 10⁰.
What is the direction of the average velocity?The direction of the average velocity of the bicycle is calculated as follows;
The sum of the horizontal component of the velocity is calculated as;
Vx = 4 km x cos(0) / 0.65h + 12 km x cos (15) / 0.5h + 4 km x cos(0) / 0.65 h
Vx = 6.154 km/h + 23.18 km/h + 6.154 km/h
Vx = 35.488 km/h
The sum of the vertical component of the velocity is calculated as;
Vy = 4 km x sin(0) / 0.65h + 12 km x sin (15) / 0.5h + 4 km x sin(0) / 0.65 h
Vy = 0 km/h + 6.21 km/h + 0 km/h
Vy = 6.21 km/h
The direction of the average velocity is calculated as;
θ = tan⁻¹ (Vy/Vx)
θ = tan⁻¹ (6.21/35.488)
θ = 9.9⁰ ≈ 10⁰
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In 7.5 s, 4200 waves pass a given point. What is the frequency of the waves?
In 7.5 s, 4200 waves will pass a given point. Then, the frequency of the waves is 560 Hz.
Frequency is the number of cycles or oscillations of a wave that occur in one second, measured in Hertz (Hz). It is the rate at which a wave completes a full cycle, which is usually measured as the number of wavelengths that pass a fixed point in a unit of time.
The formula for frequency is;
frequency = waves/time
Where "waves" is the number of waves that pass a given point and "time" is the time it takes for those waves to pass.
In this case, we know that 4200 waves pass a given point in 7.5 seconds, so we can plug those values into the formula;
frequency = 4200/7.5
Simplifying;
frequency = 560
Therefore, the frequency of the waves is 560 Hz.
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A torque of T= 4Nm is required to rotate the intermediate cylinder at W= 30 rad/min .
Calculate the viscosity of the
oil. All cylinders are 450 mm long. Neglect the end effects.
R=0.15m, t=0.003 m.
Result: 1.258 N.s/m
The viscosity of the oil with a torque 4Nm and a rotational speed is 30 rpm is 0.2003 N.s/m.
From the given,
torque = 4Nm
rotational speed = 30 rpm = (30 × 3.14)/60 = 3.14 rad/sec
radius = 0.15 m
thickness (h) = 0.003 m
To find linear velocity,
V = R×ω
= 0.15×3.14
V =0.471 m/s
The dragging force,
F = 2μA(V/h) (V is linear velocity and h is the thickness )
Area = area of cylinder = 2πRH, R is radius and H is the height of the cylinder.
F = 2μ(2πRH) (V/h)
= 2μ(2×3.14×0.15×0.45) (0.471/0.003)
= 133.10 μ (μ is the viscosity of the oil)
F = 133.1μ
Torque (τ) = Force × radius
4 Nm = 133.1μ × 0.15
μ = 4 / (133.1×0.15)
= 0.2003 N.s/m
The viscosity of the oil is 0.2003 N.s/m.
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A 104 kg man is skiing across level ground at a speed of 8.7 m/s when he comes to the small slope 1.2 m higher than ground level shown in the following figure. 1.2 m (a) If the skier coasts up the hill, what is his speed (in m/s) when he reaches the top plateau? Assume friction between the snow and skis is negligible. m/s (b) What is his speed (in m/s) when he reaches the upper level if a 75 N frictional force acts on the skis? m/s †
The speed of skier's when he reaches the top plateau is 5.26 m/s, and the speed of skier's when he reaches the upper level with a frictional force of 75 N is 4.23 m/s.
We can use the conservation of mechanical energy to solve this problem. Initially, the skier has kinetic energy, and at the top of the slope, he will have both potential and kinetic energy.
Since friction is negligible, the only force acting on the skier is the force of gravity. The work done by this force will be equal to the change in the skier's potential energy as he climbs up the slope. Therefore;
mgh = (1/2)mv²
where m = 104 kg is mass of the skier, g = 9.8 m/s² is the acceleration due to gravity, h = 1.2 m is height of the slope, and v is the skier's velocity when he reaches the top.
Solving for v, we get;
v = √(2gh) = √(29.81.2) = 5.26 m/s
Therefore, the skier's speed when he reaches the top plateau is 5.26 m/s.
In this case, there is also a frictional force acting on the skier, which does negative work on the skier as he moves up the slope. The work done by the frictional force is equal to the force of friction multiplied by the distance traveled;
W = Fd = μmgd
where μ = F/N is the coefficient of kinetic friction, N is the normal force acting on the skier (equal to the skier's weight), and d is the distance traveled up the slope (equal to the height of the slope, 1.2 m).
The net work done on the skier will be equal to the change in his mechanical energy;
Wnet = ΔK + ΔU = (1/2)m[tex]V_{f}[/tex]² - (1/2)m[tex]V_{i}[/tex]² + mgh
where vi = 8.7 m/s is the skier's initial velocity, [tex]V_{f}[/tex] is his final velocity at the top of the slope, and ΔK and ΔU are the changes in kinetic and potential energy, respectively.
Since the net work done on the skier is equal to the work done by the gravitational force minus the work done by the frictional force, we have;
Wnet = mgh - μmgd
Substituting the expressions for Wnet and mgh, we get:
(1/2)m[tex]V_{f}[/tex]² - (1/2)m[tex]V_{i}[/tex]² = μmgd
Solving for [tex]V_{f}[/tex], we get:
[tex]V_{f}[/tex] = √([tex]V_{i}[/tex]² + 2μgd) = √(8.7² + 20.729.8×1.2) = 4.23 m/s
Therefore, the skier's speed when he reaches the upper level with a frictional force of 75 N is 4.23 m/s.
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A gas expands from I to F in the figure. The energy added to the gas by heat is 465 J when the gas goes from I to F along the diagonal path. What is the change in internal energy of the gas?
The gas's internal energy that expands from I to F changes by 1015 J.
How to determine internal energy?Since the gas is expanding and energy is added to it by heat, the change in internal energy of the gas can be calculated using the first law of thermodynamics, which states that:
ΔU = Q - W
where ΔU = change in internal energy,
Q = heat added to the system, 465 J and
W = work done by the system.
Assuming that the process is quasi-static use approximation known as the "staircase approximation."
By adding up the work done in each step, find that the total work done by the gas is approximately -550 J. The negative sign indicates that work is done on the gas.
Therefore, using the first law of thermodynamics, we can calculate the change in internal energy of the gas as:
ΔU = Q - W
ΔU = 465 J - (-550 J)
ΔU = 1015 J
Therefore, the change in internal energy of the gas is 1015 J.
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Humans can hear thunder at different frequencies. These correspond to wavelengths ranging from 2.75 to 16.5 meters. If the speed of
sound is approximately 330 m/s, what frequencies of thunder can humans hear? (1 point)
O 907.5 to 5,445 hertz
O 2.75 to 16.5 hertz
O 20 to 120 hertz
O 0.0083 to 0.05 hertz
Humans can hear thunder at frequencies ranging from 20 Hz to 120 Hz.
option C.
What is the frequencies heard?
The frequency of a sound wave is given by the equation:
f = v / λ
where;
v is the speed of soundλ is the wavelengthFor the lower frequency of thunder, λ = 16.5 m,
f = 330 m/s / 16.5 m
f = 20 Hz
For the higher frequency of thunder, λ = 2.75 m,
f = 330 m/s / 2.75 m
f = 120 Hz
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3. Ana is skating down the ice at the Anoka vs Blaine hockey game. She is skating at 10 meters/second while driving towards the net. She accelerates (shoots) the puck to a final speed of 40 meters/second over a time of .5 seconds (how long it is on the stick). What was the puck's
The acceleration of the puck was 60 m/s².
What was the puck's acceleration?The acceleration of the puck is calculated by applying the formula for acceleration.
a = (v - u) / t
Where;
a is accelerationv is final velocityu is initial velocityt is time.The acceleration of the puck is calculated as follows;
a = (40 m/s - 10 m/s) / 0.5 s
a = 60 m/s²
Thus, the acceleration of the puck is determined from the change in velocity of the puck to change in time of motion of the pcuk.
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The missing question;
What was the puck's acceleration?
On a warm summer day, a large mass of air (atmospheric pressure 1.01×105Pa) is heated by the ground to a temperature of 25.0 ∘C and then begins to rise through the cooler surrounding air. Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.70×104 Pa. Assume that air is an ideal gas, with γ=1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 ∘C per 100 m of altitude, is called the dry adiabatic lapse rate.)
The temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.70×10⁴ Pa is approximately 14.3°C.
Using the ideal gas law, we can write: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the absolute temperature. Since the mass of air is not changing, we can write: PV = constant.
Applying this to the situation where the air mass rises to a level where the pressure is 8.70×10⁴ Pa, we get:
(1.01×10⁵ Pa)×V = (nR/T1)×T1(8.70×10⁴ Pa)×V = (nR/T2)×T2Dividing the second equation by the first and using the fact that γ=Cp/Cv=1.40 for air, we get:
(T2/T1) = [(P2/P1)^(γ-1)/γ] = [(8.70×10⁴ Pa)/(1.01×10⁵ Pa)]^(1.4/1.4) = 0.813Solving for T2, we get:
T2 = T1×(P2/P1)^(γ-1)/γ = (25+273) K×0.813 ≈ 287.3 K ≈ 14.3°CThus, the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.70×10⁴ Pa is approximately 14.3°C.
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find a recent (less than 5 years old) article addressing some aspect of
environmental science that we have covered so far this year.
"Greenland ice sheet melting at alarming rate of 8,300 tonnes per second" is a recent article addressing some aspect of environmental science that we have covered so far this year.
This article, published in The Guardian in January 2021, addresses the melting of the Greenland ice sheet, which is a topic covered in environmental science. The article highlights a new study that found the ice sheet is melting at a rate of 8,300 tonnes per second, which is seven times faster than in the 1990s.
The melting of the Greenland ice sheet contributes significantly to rising sea levels and could have disastrous consequences for coastal communities worldwide. The article discusses the causes and potential consequences of this rapid melting, as well as the urgent need for action to mitigate climate change and its effects. This topic relates to the study of climate change, the cryosphere, and the impacts of human activity on the environment.
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Children and parents spend less time together in middle childhood than in the preschool years.
a. True
Ob. False
E
C
A
Apr 30
Children and parents spend less time together in middle childhood than in the preschool years. Yes this true. Hence option A is correct.
Children are people who are between the ages of birth and puberty, or between the early years of development and puberty. It may also be a reference to a human embryo. The term "child" as used in law often refers to a minor, also known as a person under the age of majority. In general, children are subject to fewer rights and obligations than adults. They are considered incapable of making important judgements.
In their own species, a parent is someone who cares for their children. In humans, a parent is a child's primary carer. A individual whose gamete produced a child—a male through the sperm and a female through the ovum—is said to be the biological parent. First-degree relatives and sharing 50% genetic material are the biological parents.
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A 2000 kg car moves at a speed of 30 m/s. To reach this speed, it was necessary to burn 0.1 l of gas. Burning gas provides 30 MJ/l of source energy. Determine the energy efficiency (in %) of this car.
The energy efficiency of the car is approximately 16.7%.
The energy efficiency of a car is the ratio of the useful work output (in this case, the kinetic energy of the car) to the total energy input (in this case, the energy released by burning the gasoline). The equation for energy efficiency is:
Efficiency = Useful work output / Total energy inputThe useful work output can be calculated as the kinetic energy of the car using the equation:
KE = 0.5mv²where m is the mass of the car and v is its velocity.
Substituting the given values:
KE = 0.5 x 2000 kg x (30 m/s)² = 900,000 JThe total energy input is the energy released by burning 0.1 L of gasoline, which is:
Total energy input = 0.1 L x 30 MJ/L = 3 MJ = 3,000,000 JSubstituting these values into the equation for efficiency:
Efficiency = (900,000 J / 3,000,000 J) x 100% = 0.3 x 100% = 16.7%Therefore, the energy efficiency of the car is approximately 16.7%.
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3. Ana is skating down the ice at the Anoka vs Blaine hockey game. She is skating at 10
meters/second while driving towards the net. She accelerates (shoots) the puck to a final speed
of 40 meters/second over a time of .5 seconds (how long it is on the stick). What was the puck's
acceleration?
The acceleration of the puck was 60 m/s².
What was the puck's acceleration?The acceleration of the puck is calculated by applying the formula for acceleration.
a = (v - u) / t
Where;
a is accelerationv is final velocityu is initial velocityt is time.The acceleration of the puck is calculated as follows;
a = (40 m/s - 10 m/s) / 0.5 s
a = 60 m/s²
Thus, the acceleration of the puck is determined from the change in velocity of the puck to change in time of motion of the pcuk.
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A fireworks rocket is launched vertically into the night sky with an initial speed of 44.2
m/s. The rocket coasts after being launched, then explodes and breaks into two pieces
of equal mass 2.50 s later. (a) If each piece follows a trajectory that is initially at 45.0° to
the vertical, what was their speed immediately after the explosion? (b) What is the
velocity of the rocket’s center of mass before and after the explosion? (c) What is the
acceleration of the rocket’s center of mass before and after the explosion?
Answer:
a)44.2 m/s
b)zero
c)zero
Explanation:
(a) The initial vertical component of velocity is given by v0y = 44.2 m/s. At an angle of 45 degrees to the vertical, the initial horizontal and vertical components of velocity are equal, so each piece has an initial speed of v0 = v0y / sin(45) = 62.4 m/s. When the rocket explodes, the momentum is conserved in both the x- and y-directions, so the horizontal component of velocity remains the same while the vertical component changes sign. Therefore, each piece has a velocity of v = v0 / sqrt(2) = 44.2 m/s after the explosion.
(b) The velocity of the rocket's center of mass before the explosion is zero, since it is at rest. After the explosion, the two pieces move in opposite directions with equal and opposite momentum. Since they have the same mass, their velocities are equal in magnitude, and the velocity of the center of mass is zero.
(c) Before the explosion, the rocket is at rest, so its acceleration is zero. After the explosion, the two pieces have equal and opposite acceleration vectors,
(a) The initial vertical component of velocity is given by v0y = 44.2 m/s. At an angle of 45 degrees to the vertical, the initial horizontal and vertical components of velocity are equal, so each piece has an initial speed of v0 = v0y / sin(45) = 62.4 m/s. When the rocket explodes, the momentum is conserved in both the x- and y-directions, so the horizontal component of velocity remains the same while the vertical component changes sign. Therefore, each piece has a velocity of v = v0 / sqrt(2) = 44.2 m/s after the explosion.
(b) The velocity of the rocket's center of mass before the explosion is zero, since it is at rest. After the explosion, the two pieces move in opposite directions with equal and opposite momentum. Since they have the same mass, their velocities are equal in magnitude, and the velocity of the center of mass is zero.
(c) Before the explosion, the rocket is at rest, so its acceleration is zero. After the explosion, the two pieces have equal and opposite acceleration vectors, but since they have the same mass, their acceleration vectors cancel out and the acceleration of the center of mass is zero.
You are designing a machine for a space exploration vehicle. It contains an enclosed column of oil that is 1.50 m tall, and you need the pressure difference between the top and the bottom of this column to be 0.125 atm. (a) What must be the density of the oil? (b) If the vehicle is taken to Mars, where the acceleration due to gravity is 0.379g, what will be the pressure difference (in earth atmospheres) between the top and bottom of the oil column?
Answer:
Explanation:
(a) To find the density of the oil, we can use the formula for pressure difference in a fluid column:
ΔP = ρgh
where ΔP is the pressure difference, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column.
Plugging in the given values, we have:
0.125 atm = ρgh = ρ(9.81 m/s^2)(1.50 m)
Solving for ρ, we get:
ρ = 0.125 atm / (9.81 m/s^2 x 1.50 m) ≈ 0.00847 g/cm^3
Therefore, the density of the oil must be approximately 0.00847 g/cm^3.
(b) On Mars, the acceleration due to gravity is 0.379 times that of Earth, or g_Mars = 0.379g_Earth. The pressure difference between the top and bottom of the oil column will be:
ΔP_Mars = ρgh_Mars = ρg_Earth(0.379)(1.50 m)
Using the density we found in part (a), we have:
ΔP_Mars = (0.00847 g/cm^3)(9.81 m/s^2)(0.379)(1.50 m) / (1 atm/101325 Pa)
ΔP_Mars ≈ 0.019 atm
So, the pressure difference between the top and bottom of the oil column on Mars will be approximately 0.019 atm, or about 0.15 times the pressure difference on Earth.
Answer:
The pressure difference (in Earth's atmosphere) between the top and bottom of the oil column on Mars is 0.045 atm.
Explanation:
(a) To find the density of the oil, we can use the formula for pressure difference in a fluid column: ΔP = ρgh, where ΔP is the pressure difference, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column.
We know that the pressure difference is 0.125 atm, the height of the column is 1.50 m, and the acceleration due to gravity on Earth is 9.81 m/s². Plugging in these values, we get:
0.125 atm = ρ(9.81 m/s²)(1.50 m)
Solving for ρ, we get:
ρ = 0.00803 g/cm³
Therefore, the density of the oil must be 0.00803 g/cm³.
(b) If the vehicle is taken to Mars, where the acceleration due to gravity is 0.379g, we can use the same formula to find the pressure difference:
ΔP = ρgh
We know that the height of the column is still 1.50 m, but the acceleration due to gravity is now 0.379g. Plugging in these values, we get:
ΔP = (0.00803 g/cm³)(9.81 m/s²)(0.379)(150 cm)
Solving for ΔP, we get:
ΔP = 0.045 atm
Therefore, the pressure difference (in Earth's atmosphere) between the top and bottom of the oil column on Mars is 0.045 atm.
28g of copper pellets are removed from a 300∘C oven and immediately dropped into 90 mL of water at 25∘C in an insulated cup.
The final temperature of the water and copper pellets is 43.8°C.
We can use the equation:
[tex]mc\Delta T = mwCw\Delta T[/tex]
Substituting the values given in the problem:
m c = 28 g
m w = 90 mL = 90 g
c = 0.385 J/g°C
Cw = 4.18 J/g°C
ΔT = T f - 25°C
where T f is the final temperature of the water and copper pellets.
Simplifying the equation:
28 g x 0.385 J/g°C x (T f - 300°C) = 90 g x 4.18 J/g°C x (T f - 25°C)
Solving for T f:
T f = (90 g x 4.18 J/g°C x 25°C + 28 g x 0.385 J/g°C x 300°C) / (90 g x 4.18 J/g°C + 28 g x 0.385 J/g°C)
T f = 43.8°C
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--The complete Question is, What are the final temperature of the water and the copper pellets after thermal equilibrium is reached, assuming no heat is lost to the surroundings? --
I NEEDDD HELP LIKE RIGHT NOW ITS SIMPLE
What is the best description of originality?
nonobvious
special or interesting
convergent
having a low probability, unique
Answer: a
Explanation:
At 3:00 P.M., a bank robber is spotted driving north on I-15 at milepost 129. His speed is 135.0 mi/h. At 3:37 P.M., he is spotted at milepost 182 doing 103.0 mi/h. (Assume a straight highway). Assume north to be positive. What is the bank robber’s displacement during this time interval? Enter a positive value if the displacement is toward north and enter a negative value if the displacement is toward south.
To find the displacement of the bank robber during this time interval, we need to calculate the distance he traveled in each direction and then find the vector difference between his initial and final positions.
From 3:00 P.M. to 3:37 P.M., the bank robber traveled for 37 minutes = 0.617 hours at a speed of 135.0 mi/h. Therefore, he traveled a distance of:
d1 = v1 x t1 = 135.0 mi/h x 0.617 h = 83.3 miles north
From 3:00 P.M. to 3:37 P.M., the bank robber traveled for 37 minutes = 0.617 hours at an average speed of 103.0 mi/h. Therefore, he traveled a distance of:
d2 = v2 x t2 = 103.0 mi/h x 0.617 h = 63.5 miles north
To find the displacement, we need to subtract the initial position from the final position. The initial position is at milepost 129 and the final position is at milepost 182. Therefore, the displacement is:
displacement = final position - initial position
displacement = 182 - 129 = 53 miles north
Since the displacement is in the positive direction, we enter a positive value. Therefore, the bank robber's displacement during this time interval is 53 miles north.
Gravitational acceleration (g) and gravitation constant (G)
What evidence did Wegener NOT use to support his idea of continental drift?
A. Mountain ranges on different continents lined up when coastlines were matched up.
B. Fossils of plants and animals in climates where their survival would have been impossible. C. The thickness of layers of ice in the Antarctic.
D. Rock strata on different continents lined up when coastlines were matched up.
The evidence that Wegener did NOT use to support his idea of continental drift is "the thickness of layers of ice in the Antarctic.
option C
What is Wegener's primary evidence for continental drift?Wegener's primary evidence for continental drift included the fit of the coastlines of different continents, the distribution of fossils across different continents, and the alignment of rock strata on different continents.
So the thickness of layers of ice in the Antarctic, was not used by Wegener to support his idea of continental drift. While this evidence is important for supporting the theory of glaciation, it is not relevant to the theory of continental drift.
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A system underwent an adiabatic process during which the internal energy decreased 613 J, the temperature decreased 461 K, the pressure decreased to a quarter of its original value, & the number of moles remained constant. Determine the total heat flow of the system. And complete table
The total heat flow of the system is -2413 J.
How to calculate the heat flowV1 = nRT1/P1
V2 = nRT2/P2 = 4nRT1/P1
Since the number of moles is constant, the work done by the system is:
W = -PΔV = -P(V2 - V1) = -P(3V1) = -(3nRT1)
Substituting the given values, we get:
W = -1800 J
Finally, we can use the first law of thermodynamics to calculate the heat flow:
Q = ΔU + W = -613 J - 1800 J = -2413 J
Therefore, the total heat flow of the system is -2413 J. Note that this is a negative value, indicating that the system lost heat during the process.
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the magnitude of kilogram is same all over the world give reason
The SI unit is a unit of measurement that is a globally recognized system that is used for technical and scientific activities.
The kilogram, which corresponds to the mass of the international prototype of the kilogram, is the SI base unit of mass.
The weight of a specific international prototype constructed of platinum-iridium and stored at the International Bureau of Weights and Measures is referred to as one kilogram.
Initially, it was specified as being equal to the mass of one litre (10-3 cubic metres) of pure water. A mass of 1 kg weighs around 2.20 pounds (lb) at the surface of the Earth.
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