The question is incomplete. The complete question is :
First Law of Thermodynamics - Sign Convention The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is AU = Q-W. Here AU is the change in internal energy U of the system. Q is the net heat transferred into the system that is, Q is the sum of all heat transfer into (positive) and out of (negative) the system. W is the net work done by the system—that is, W is the sum of all work done by (positive) and on (negative) the system. We use the following sign conventions: if Q is positive, then there is a net heat transfer into the system; if W is positive, then there is net work done by the system. So positive Q adds energy to the system and positive W takes energy from the system. Thus AU = Q-W. Note also that if more heat transfer into the system occurs than work done, the difference is stored as internal energy. The first law of thermodynamics AU = 9 - W Ur-U, Heat Work System AU--W system w Qin: talu Qout:
The first law of thermodynamics AU = Q - W U-U Heat Work System AUQ-W Qin: ta Qout: - Wout: + WK w Win: - - volume expands t volume decreases o All answers can be positive or negative. (a) Suppose there is heat transfer of 42 ) into a system, while the system does 6 ) of work. Later, there is heat transfer of 22 J out of the system while 6 ) of work is done on the system. What is the net heat transfer? 20 Correct (100.0%) Submit What is the total work? Enter a number Submit (5 attempts remaining) What is the net change in internal energy of the system? Enter a number
What is the net change in internal energy of the system? Enter a number Submit (5 attempts remaining) (b) What is the change in internal energy of a system when a total of 140 J of heat transfer occurs out of (from) the system and 165 ) of work is done on the system? Enter a number Submit (5 attempts remaining) (c) An athlete doing push-ups performs 645 kJ of work and loses 440 kJ of heat. What is the change in the internal energy (in kJ) of the athlete? Enter a number Submit (5 attempts remaining) kJ (d) An athlete doing push-ups performs 690 kJ of work and loses 450 kJ of heat. Then he takes in 830 kJ of energy from eating food, What is the total change in the internal energy (in kJ) of the athlete? Enter a number kJ.
Solution :
a). Given :
[tex]$Q_1 = 42 \ J$[/tex] , [tex]$Q_2 = -22 \ J , \ W_1 = 6 \ J, \ W_2 = -6 \ J $[/tex]
Net heat transfer
[tex]$Q= Q_1+Q_2$[/tex]
= 42 + (-22)
= 20 J
Total work
[tex]$W= W_1+W_2$[/tex]
= 6 + (-6)
= 0 J
∵ ΔU = Q - W
= 20 - 0
= 20 J
This is the net change in the internal energy of the system.
b). ΔU = Q + W
= (-140) + (-165)
= -305 J
c). ΔU = Q + W
= (-440) + (645)
= 205 J
d). ΔU = Q + W
= (-450) + (690)
= 240 J
______ is the total distance traveled divided by the total time of travel.
(5 Points)
speed
total speed
average speed
displacement
1 What is the resistance of the inductive coil
takes 5A current across 240V, 50Hz supply
at 0.8 power factor?
A 48
B 4250
C 38.40
D 26.60
Answer:
38.40
Explanation:
This is the answer for this question
A jet pilot puts an aircraft with a constant speed into a vertical loop. (a) Which is greater, the normal force exerted on the seat by the pilot at the bottom of the loop or that at the top of the loop
Answer:
A jet pilot puts an aircraft with a constant speed into a vertical loop is explained below in complete details.
Explanation:
Well, the difficulty does not provide the pilot's mass (or weight in regular gravity), but the difficulty can be resolved and declared in courses of m (the pilot's mass).
When the jet is at the foundation of the circuit, a free-body chart displays the centripetal energy working upward approaching the middle of the loop, and the sound force of the chair and the pilot also upward. The pilot's weight (mg) is earthward. From Newton's second law:
?F(c) = ma(c) = n - mg
n = mg + ma(c)
= m[g + a(c)]
Since centripetal acceleration equals v² / r, the equalization enhances:
n = m[g + (v² / r)]
When an object falls, it is reacting to the force of gravity. true or false
Answer:
The answer is true, as gravity affects everything.
Q. At what point in a waterfall do the drops of water contain the most kinetic energy ?
1.When they are halfway to the bottom
2.When they reach the bottom of the fall
Answer:
2.When they reach the bottom of the fall
Explanation:
The potential energy of the waterfall is maximum at the maximum height and decreases with decrease in height. Based on the law of conservation of mechanical energy, as the potential energy of the water fall is decreasing with decrease in height of the fall, its kinetic energy will be increasing and the kinetic energy will be maximum at zero height (bottom of the fall).
Thus, the correct option is "2" When they reach the bottom of the fall
A horizontal force F is used to pull a 5 kg block across a floor at constant speed of 3 m/s. The frictional force between the block and the floor is 10 N. The net work done on the block in 1 minutes is most nearly:
Answer:
5× 3 over 10 = 1.5
Explanation:
F = 1m/s × kg over n
The net work done on the block in 1 minutes is most nearly 1800 Joule.
What is work?
A force must be applied in order for work to be completed, and there must also be motion or displacement in the force's direction. The amount of force multiplied by the distance moved in the force's direction is known as the work done by a force acting on an item.
Work has no direction and only magnitude. Work is a scalar quantity as a result.
Given that: A horizontal force F is used to pull a 5 kg block across a floor at constant speed of 3 m/s.
As the block is moving with uniform velocity; this force is equal to the frictional force in magnitude.
So, The net work done on the block in 1 minutes is =
force× displacement
= force × velocity × time
= 10 × 3 × 60 Joule
= 1800 Joule.
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the maximum normal force a pilot can withstand is about eight times his weight. What is the maximum radius of curvature that a jet planes pilot, pulling out of a vertical dive
Complete Question
the maximum force a pilot can stand is about seven times his weight. what is the minimum radius of curvature that a jet plane's pilot, pulling out of a vertical dive, can tolerate at a speed of 250m/s?
Answer:
The value is [tex]r = \frac{250^2 }{6 * 9.8 }[/tex]
Explanation:
From the question we are told that
The weight of the pilot is [tex]W = mg[/tex]
The maximum force a pilot can withstand is [tex]F_{max} = 7 W = 7 (mg)[/tex]
The speed is [tex]v = 250 \ m/s[/tex]
Generally the centripetal force acting on the pilot is equal to the net force acting on the pilot i.e
[tex]F_c = F_{max} - mg[/tex]
Here N is the normal force acting on the pilot
Now
[tex]F_c = \frac{m v^2 }{r}[/tex]
So
[tex]\frac{m v^2 }{r} = 7(mg) - mg[/tex]
=> [tex]r = \frac{v^2 }{6g}[/tex]
=> [tex]r = \frac{250^2 }{6 * 9.8 }[/tex]
=> [tex]r = 1063 \ m[/tex]
1. What is the total distance traveled?
A 3.0m
B 4.0m
C 5.0m
D 6.0m
Answer:
c
Explanation:
a 60N at an angle of 30°from
horizontal
Explanation:
Force, F = 60 N
Angle, [tex]\theta=30^{\circ}[/tex]
We need to find horizontal and vertical component of the force.
Horizontal component,
[tex]F_x=F\cos\theta\\\\=60\times \cos(30)\\\\=51.96\ N[/tex]
Vertical component,
[tex]F_=F\sin\theta\\\\=60\times \sin(30)\\\\=30\ N[/tex]
So, the horizontal and vertical component are 51.96 N and 30 N respectively.
Question 1 of 25
Two asteroids with masses 3.71 x 10 kg and 1.88 x 104 kg are separated by
a distance of 1,300 m. What is the gravitational force between the asteroids?
Newton's law of gravitation is F gravity
Gm, 2 The gravitational
constant Gis 6.67 x 10-11 Nm²/kg?
A. 275 x 10"N
B. 4.13 x 10°N
C. 2.04 x 10°N
O D. 3.58 x 10-N
SUBMIT
Answer:
(A)
Explanation:
Answer:
275 x 10"N
Explanation:
A force of 22.7 N stretches an elastic band at room temperature. The rate at which its entropy changes as it stretches is about _____ J/Km. Round your answer to 3 decimal places.
Answer:
The value is [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]
Explanation:
From the question we are told that
The force is [tex]F = 22.7 \ N[/tex]
The value of room temperature is [tex]T = 298 \ K[/tex]
Generally the rate at which its entropy changes as it stretches is mathematically represented as
[tex]\frac{\Delta S }{ L} = - \frac{F}{T}[/tex]
=> [tex]\frac{\Delta S }{ L} = - \frac{21.5}{ 298 }[/tex]
=> [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]
what is force?answer in one line.
Can we use a hydrometer to
measure the density of milk?
Answer:
yes
Explanation:
i hope this helps not sure im right
stopping potential becomes more and more negative why
Answer:
stopping potential is the negative potential applied to the circuit to stop the moving electrons so as to stop the flow of current
for high current high negative potential is applied
which statement about force is incorrect
Answer:
What are the options?
Explanation:
A 2.00 kg copper pot is placed on a glass shelf and then one side of the shelf falls so that it becomes angled at 35.0° to the horizontal. The coefficient of static friction between copper and glass is 0.680 and the coefficient of kinetic friction is 0.530. Will the pot slide, and if so, what is its acceleration?
Answer:
Explanation: about 2 Mph when it falls and sorry if I’m wrong
when is a bandage best suited to be used?
Do clouds have respiration
Answer:
yes it also does have but not in the exact form but it does 30 percent of respiration to produce rain
If vector A= aj^ and vector B = bj^, then vector A×B is equal to
Answer:
0 because j×j =0
Explanation:
so a×b is 0
A television camera is positioned 4 km from the base of a rocket launching pad. In order to keep the rocket in focus as it takes off, the camera must be programmed with the distance to the rocket and how that distance is changing. If we assume that the rocket rises vertically at a speed of 200 km/hr, how fast is the distance from the camera to the rocket changing when the rocket has risen 3 km
Answer:
120 km/hr
Explanation:
Let D be the distance between the rocket and the camera as the rocket is moving upwards. Let d be the distance the rocket moves and L be the distance between the camera and the base of the rocket = 4 km.
Now, at any instant, D² = d² + L²
= d² + 4²
= d² + 16 since the three distances form a right-angled triangle with the distance between the rocket and the camera as the rocket is moving upwards as the hypotenuse side.
differentiating the expression to find the rate of change of D with respect to time, dD/dt ,we have
d(D²)/dt = d(d² + 16)/dt
2DdD/dt = 2d[d(d)/dt]
dD/dt = 2d[d(d)/dt] ÷ 2D
Now d(d)/dt = vertical speed of rocket = 200 km/hr
dD/dt = 200d/D [D = √(d² + 16)]
dD/dt = 200d/[√d² + 16]
Now substituting d = 3 km, the distance the rocket has risen into the equation, we have
dD/dt = 200(3)/[√(3² + 16)]
dD/dt = 600/[√(9 + 16)]
dD/dt = 600/√25
dD/dt = 600/5
dD/dt = 120 km/hr
So, the speed at which the distance from the camera to the rocket changing when the rocket has risen 3 km is 120 km/hr.
(c) The plates are moved farther apart with each plate maintaining the same net charge. In a coherent paragraph-length response, apply concepts of work and energy to explain how the electric potential difference between the plates changes, if at all, when the plates are moved farther apart.
Answer:
Energy is applied on the charge to do work.
Explanation:
Work is only be done when one charge moves against the electric field of anther charge that require huge amount of energy because both have same charge and the force of repulsion occurs between them. Work is defined as the product of force and displacement so from this equation we can conclude that work is done when a force is applied on an object and it moves in the direction in which force is applied. If the force or energy is removed from the charge which is present in the electric field of another charge so it moves away from that charge and the work is also be done..
A 52.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor covering. The heel on each shoe is circular and has a radius of 0.600 cm. If the woman balances on one heel, what pressure does she exert on the floor?
Answer:
4.5 × 10^6 Nm
Explanation:
Given that a 52.0-kg woman wearing high-heeled shoes is invited into a home in which the kitchen has vinyl floor covering. The heel on each shoe is circular and has a radius of 0.600 cm. If the woman balances on one heel, what pressure does she exert on the floor?
Let's first calculate the area covered by the heel shoe by using the area of a circle.
Convert cm to m
0.6 / 100 = 0.006
A = πr^2
A = 22/7 × 0.006^2
A = 1.131 × 10^-4 m^2
The force exerted by the woman = mg
Force exerted = 52 × 9.8
Force exerted = 509.6 N
Pressure = force / Area
Pressure = 509.6 / 1.131 × 10^-4
Pressure = 4505853.3 N/m
Therefore, she exerts pressure of 4.5 × 10^6 Nm approximately on the floor.
Two spherical balls are placed so that their centers are 3.61 m apart. The
force between them is 1.65 x 10-7 N. If the mass of the smaller ball is 81 kg,
what is the mass in kilograms of the other ball?
Answer:
The mass of the other ball is 397.775 kg.
Explanation:
Gravitation is the force of mutual attraction that bodies experience due to the fact that they have a certain mass.
The universal law of gravitation is a classical physical law that describes the gravitational interaction between different bodies with mass.
The law was formulated by Newton, who deduced that the force with which two bodies of different masses are attracted only depends on the value of their masses and the square of the distance that separates them.
In other words, the Law of Universal Gravitation predicts that the force exerted between two bodies of masses M1 and M2 separated by a distance "d" is proportional to the product of their masses and inversely proportional to the square of the distance, that is:
[tex]F=G\frac{M1*M2}{d^{2} }[/tex]
where:
F = It is the module of the force exerted between both bodies, and its direction is found on the axis that joins both bodies. G = It is the constant of Universal Gravitation, whose value is 6.67384*10⁻¹¹ [tex]\frac{N*m^{2} }{kg^{2} }[/tex]In this case:
F= 1.65*10⁻⁷ NG= 6.67384*10⁻¹¹ [tex]\frac{N*m^{2} }{kg^{2} }[/tex]M1= 81 kgM2= ?d= 3.61 mReplacing:
[tex]1.65*10^{-7} N=6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }\frac{81 kg*M2}{(3.61 m)^{2} }[/tex]
Solving for M2:
[tex]M2=\frac{1.65*10^{-7} N*(3.61 m)^{2}}{6.67384*10^{-11} \frac{N*m^{2} }{kg^{2} }*81 kg}[/tex]
M2= 397.775 kg
The mass of the other ball is 397.775 kg.
You want to move a spacecraft that is in 300 km circular orbit around Venus into another circular orbit around Venus of 600 km. Explain how you would accomplish this. Solve for the velocities that need to be applied.
Answer:
The final velocity will be half of the initial velocity of the spacecraft.
Explanation:
Angular momentum is conserved for the circular force motion and central force motion.
Considering
L = MVR = Constant
Where
M = Mass of the object
V = Velocity of the object
r = radius of circle
We know that
V = [tex]\frac{1}{R}[/tex]
So,
[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{R_{1} }{R_{2} }[/tex]
As per the given data
[tex]R_{1}[/tex] = Initial Radius = 300 km
[tex]R_{2}[/tex] = Final Radius = 600 km
[tex]V_{1}[/tex] = Initial Velocity =
[tex]V_{2}[/tex] = Final Velocity =
Placing values in the formula
[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{300 km}{600 km }[/tex]
[tex]\frac{V_{2} }{V_{1} }[/tex] = [tex]\frac{1}{2}[/tex]
[tex]{V_{2}[/tex] = [tex]\frac{1}{2} V_{1}[/tex]
Hence, The final velocity will half of the initial velocity of the spacecraft.
The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.
Answer:
The Starship Enterprise is powered by combining matter with antimatter. Suppose 1 kg of each are combined and ejected backward at the speed of light, what is the final speed of the Enterprise starting from rest? Assume that the mass of the Enterprise is 10,000 kg and the spaceship does not reach relativistic speed.
The intensity of sound is measured on the decibel scale, dB. The equation dB=10 log I represents the decibel level, where I is the ratio of the sound to the human hearing threshold. A noise is 150,000 times greater than the human hearing threshold. Which shows a valid step in the process of finding the decibel level of the noise?
a. 150,000 = 10 log I
b. 15,000 = log I
c. dB = 10 log 150,000
d. 10dB = log 150,000
e. 10/dB= log 150,000
Answer:
The correct option is c
Explanation:
From the question we are told that
The ratio of the noise to human hearing threshold is [tex]I = 150 000[/tex]
Generally from the equation given we have that
dB = 10 log I
So
dB = 10 log 150000
The expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000. Option C is correct
Given the equation for calculating the intensity of sound which is measured in decibel expressed as:
dB=10 log I
where;
I is the ratio of the sound to the human hearing threshold
Given that noise is 150,000 times greater than the human hearing threshold.
Substitute I = 150,000 into the expression above;
dB = 10 log 150,000
Hence the expression that shows a valid step in the process of finding the decibel level of the noise is dB = 10 log 150,000
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The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?
Answer:
The second distance of the sound from the source is 431.78 m..
Explanation:
Given;
first distance of the sound from the source, r₁ = 1.48 m
first sound intensity level, I₁ = 120 dB
second sound intensity level, I₂ = 70.7 dB
second distance of the sound from the source, r₂ = ?
The intensity of sound in W/m² is given as;
[tex]dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2[/tex]
[tex]For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2[/tex]
The second distance, r₂, can be determined from sound intensity formula given as;
[tex]I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m[/tex]
Therefore, the second distance of the sound from the source is 431.78 m.
A ball is thrown horizontally at 30 m/s from a height of 45m. How long is it in the air?
(i) How fast is it moving horizontally when it hits the ground?
Answer:
1.25s
1) 90.9m
Explanation:
From the question, we are given the following;
Speed v = 30m/s
Maximum height H = 45m
Required;
Time
Using the equation of motion S = ut + 1/2gt²
Substitute;
45= 30t + 1/2(9.8)t²
45 = 30t + 4.9t²
4.9t² + 30t - 45 = 0
Factorize;
t = -30 ±√30²-4(-45)(4.9)/2(4.9)
t = -30 ±√900+882/9.8
t = -30 ±42.21/9.8
t = -30 + 42.21/9.8
t = 12.21/9.8
t = 1.25secs
Hence the ball spent 1.25secs in air.
i) To get the horizontal distance, we will use the formula;
R = U√2H/g
R = 30√2(45)/9.8
R = 30√90/9.8
R = 30√9.18
R = 30(3.03)
R = 90.9m
Hence the horizontal distance is 90.9m
5. 3 women push a stalled car. Each woman pushes with a 400N force. What is the mass of the car if the car accelerates at 0.85 m/s??
Answer:
470.59 kgExplanation:
The the mass of the car can be found by using the formula
[tex]m = \frac{f}{a} \\ [/tex]
f is the force
a is the acceleration
From the question we have
[tex]m = \frac{400}{0.85} \\ = 470.58823...[/tex]
We have the final answer as
470.59 kgHope this helps you
A football quarterback throws a football for a long pass. While in the motion of throwing, the quarterback moves the ball , starting from rest, and completes the motion in . Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass
This question is incomplete, the complete question is;
A football quarterback throws a 0.408 kg football for a long pass. While in the motion of throwing, the quarterback moves the ball 1.909 m, starting from rest, and completes the motion in 0.439 s. Assuming the acceleration is constant, what force does the quarterback apply to the ball during the pass ;
a) F_throw = 8.083 N
b) F_throw = 9.181 N
c) F_throw = 2.284 N
d) F_throw = 16.014 N
e) None of these is correct
Answer:
the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer
Explanation:
Given that;
m = 0.408 kg
d = 1.909 m
u = 0 { from rest}
t = 0.439 s
Now using Kinetic equation
d = ut + 1/2 at²
we substitute
1.909 = (0 × 0.439) + 1/2 a(0.439)²
1.909 = 0 + 0.09636a
1.909 = 0.09636a
a = 1.909 / 0.09636
a = 19.8111 m/s²
Now force applied will be;
F = ma
we substitute
F = 0.408 × 19.8111
F = 8.0828 ≈ 8.083 N
Therefore the quarterback applied a force of 8.083 N to the ball during the pass
so Option a) F_throw = 8.083 N is the correct answer