The required flow work of the compressor is calculated using the flow work equation and the ideal gas law. First, we determine the final volume using the ideal gas law. Then we substitute these figures into the flow work equation, giving the result as 8.08 kJ/kg.
Explanation:The flow work required by the compressor is calculated using the equation
flow work = pressure * volume.
Given the initial pressure P1 = 120 kPa, volume V1 = 6 L and final pressure P2 = 1000 kPa,
we can substitute these values into the equation.
However, the volume at the end of compression is not given. To find this, we need to use the ideal gas law, P1V1/T1=P2V2/T2,
where T1 is the initial temperature and T2 is the final temperature.
Convert the temperatures to kelvins (T1 = 22 + 273 = 295 K, T2 = 400 + 273 = 673 K) and volume to m3 (V1 = 6 / 1000). Solving for V2 gives V2 = P1V1T2 / P2T1 = 0.00147 m3.
Now, substituting again into the flow work equation gives
flow work = (1000 kPa)(0.00147 m3) = 1.47 kJ.
This is the energy per unit volume, to find it per unit mass, divide it by the specific volume
v2 = V2 / m = R*T2 / P2 = 0.182 m3/kg,
Therefore, the required flow work = 1.47 kJ / 0.182 kg = 8.08 kJ/kg.
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The flow work required by the compressor is approximately 622.476 kJ/kg.
The flow work required by the compressor is given by the equation:
[tex]\[ w_{flow} = \int_{1}^{2} v \, dp \][/tex]
where [tex]\( v \)[/tex] is the specific volume of the air and [tex]\( dp \)[/tex] is the differential pressure change. For an ideal gas, the specific volume can be calculated using the ideal gas law:
[tex]\[ v = \frac{RT}{p} \][/tex]
where [tex]\( R \)[/tex] is the gas constant, [tex]\( T \)[/tex] is the absolute temperature in Kelvin, and [tex]\( p \)[/tex] is the pressure.
Given:
- Initial pressure [tex]\( p_1 = 120 \)[/tex] kPa
- Final pressure [tex]\( p_2 = 1000 \)[/tex] kPa
- Initial temperature [tex]\( T_1 = 22 + 273.15 = 295.15 \)[/tex] K (converting from Celsius to Kelvin)
- Final temperature [tex]\( T_2 = 400 + 273.15 = 673.15 \)[/tex] K
- Gas constant [tex]\( R = 0.287 \) kPa/m^3/kg/K[/tex]
Since the process is adiabatic and the air is being compressed, we can assume that the specific volume at the initial state [tex]\( v_1 \)[/tex] can be calculated using the initial conditions:
[tex]\[ v_1 = \frac{RT_1}{p_1} \][/tex]
Substituting the values:
[tex]\[ v_1 = \frac{0.287 \times 295.15}{120} \][/tex]
[tex]\[ v_1 = \frac{84.80}{120} \][/tex]
[tex]\[ v_1 = 0.7067 \text{ m}^3/\text{kg} \][/tex]
The flow work can be approximated if we assume the process to be isothermal at the inlet temperature (which is a common simplification for flow work calculation):
[tex]\[ w_{flow} = v_1(p_2 - p_1) \][/tex]
Substituting the values:
[tex]\[ w_{flow} = 0.7067(1000 - 120) \][/tex]
[tex]\[ w_{flow} = 0.7067 \times 880 \][/tex]
[tex]\[ w_{flow} = 622.476 \text{ kJ/kg} \][/tex]
A straight wire segment 2 m long makes an angle of 30degrees with a uniform magnetic field of 0.37 T. Find the magnitude of the force on the wire if it carries a current of 2.2 A.
Answer : 0.814 newton
Explanation:
force (magnetic) acting on the wire is given by
F= ? , I=2.2amp , B = 0.37 T
F = B i l sin (theta) = 0.37 x 2.2 x 2x 0.5 = 0.814N
A straight wire segment 2 m long makes an angle of 30degrees with a uniform magnetic field of 0.37 T. The magnitude of the force is
F= 0.814N
What is the magnitude of the force?Generally, the equation for the magnitude of the force is mathematically given as
[tex]F = B i l sin (\theta)[/tex]
Therefore
F= 0.37 x 2.2 x 2x 0.5
F= 0.814N
In conclusion, the magnitude of the force
F= 0.814N
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3) A person catches a ball with a mass of 145 g dropped from a height of 60.0 m above his glove. His hand stops the ball in 0.0100 s. What is the force exerted by his glove on the ball? Assume the ball slows down with constant acceleration.
Answer: 87KN
Explanation: F= m(v-u)/t
V= h/t
V= 60/0.0100
v= 6000m/s
M=145g/1000=0.145kg
F= 0.145*6000/0 .001
F= 87000N
F= 87KN
Total peripheral resistance is related to all of the following except the
Options to the question :
A- blood viscosity
B- osmolarity of interstitial fluid
C- turbulence
D-length of a blood vessel
E- blood vessel diameter
Answer:
Total peripheral resistance is NOT related to B ( osmolarity of interstitial fluid).
Explanation:
Total peripheral resistance( also called systemic vascular resistance) is defined as the total opposition to the flow of blood in systemic circulation. Increase in total peripheral resistance leads to high blood pressure while it's decrease leads to low blood pressure. Factors that contributes to total peripheral resistance in systemic circulation includes:
- blood vessel diameter
- blood viscosity,
- lengthy of a blood vessel and
- turbulence.
Laboratory measurements show hydrogen produces a spectral line at a wavelength of 486.1 nanometers (nm). A particular star's spectrum shows the same hydrogen line at a wavelength of 486.0 nm. What can we conclude
Answer: we can conclude that the wavelength is decreasing. This means that the star is moving towards the observer on earth.
Explanation:
Since light has a constant speed of 3 x 10^8m/s, and this speed is a product of its wavelength and its frequency c = f¥
Where f is the frequency and ¥ is the wavelnght.
For a decreasing wavelength, it is seen that the frequency is increasing.
According to the doppler's effect, a moving body that is a source of a wave of frequency f, moving relatively towards an observer, the frequency will increase as they move closer to a frequency f' which is greater than f. This is known as the doppler shift of light wave.
A spring with a force constant of 5.0 N/m has a relaxed length of 2.59 m. When a mass is attached to the end of the spring and allowed to come to rest, the vertical length of the spring is 3.86 m. Calculate the elastic potential energy stored in the spring. Answer in units of J.
Answer:
Explanation:
Given
Spring force constant [tex]k=5\ N/m[/tex]
Relaxed length [tex]l_0=2.59\ m[/tex]
mass is attached to the end of spring and allowed to come at rest with relaxed length [tex]l_2=3.86\ m[/tex]
Potential energy stored in the spring
[tex]U=\frac{1}{2}k(\Delta x)^2[/tex]
[tex]\Delta x=l_2-l_0[/tex]
[tex]U=\frac{1}{2}\times 5\times (3.86-2.59)^2[/tex]
[tex]U=4.03\ J[/tex]
A 20.0 kg rock is sliding on a rough , horizontal surface at8.00 m/s and eventually stops due to friction .The coefficient ofkinetic fricction between the rock and the surface is 0.20
what average thermal power is produced as the rock stops?
Answer:
Therefore the average thermal power is = 19.61 W
Explanation:
Given that the mass of the rock = 20.0 kg
initial velocity (u) = 8.00m/s
Final velocity = 0
Kinetic fiction [tex](\mu_k)[/tex] = 0.20.
Friction force = Kinetic fiction× weight
=[tex]\mu_k mg[/tex]
Force = mass × acceleration
[tex]\Rightarrow acceleration =\frac{Force}{mass}[/tex]
[tex]=-\frac{\mu_kmg}{m}[/tex]
[tex]=-\mu_kg[/tex]
To find the time , we use the the following formula,
v=u+at
Here a [tex]=-\mu_kg[/tex]
[tex]\Rightarrow 0=8+(-\mu_kg)t[/tex]
[tex]\Rightarrow 0.20 \times 9.8\times t=8[/tex]
⇒t = 4.08 s
Now,
Thermal energy= work done by friction = change of kinetic energy
The change of kinetic energy
= [tex]\frac{1}{2} mu^2-\frac{1}{2} mv^2[/tex]
[tex]=\frac{1}{2}\times 20\times 8 -\frac{1}{2}\times 20\times 0[/tex]
=80 J
Thermal energy=80 J
Thermal power = Thermal energy per unit time
[tex]=\frac{80}{4.08}[/tex] w
=19.61 W
Therefore 19.61 W average thermal power is produced as the rock stops.
In a single-slit experiment, the slit width is 170 times the wavelength of the light. What is the width (in mm) of the central maximum on a screen 2.4 m behind the slit?
Answer:
[tex]w_c[/tex] = 28 mm
Explanation:
Displacement from the central maximum to minimum is as deterrmined as follows;
[tex]y_m[/tex] = [tex]\frac{m \lambda D}{a}[/tex]
If first minimum (m) = 1
single- Slit width is 170 times the wavelength of the light.
i.e
a = 170 λ
[tex]y_1[/tex] = [tex]\frac{(1) \lambda (2,.4m)}{170 \lambda}[/tex]
[tex]y_1[/tex] = 0.014 m
width of the central maximum can now be determined as:
[tex]w_c[/tex] = [tex]2y_1[/tex]
[tex]w_c[/tex] = 2(0.014 m)
[tex]w_c[/tex] = 0.028 m
[tex]w_c[/tex] = 28 mm
Hence, the width (in mm) of the central maximum on a screen 2.4 m behind the slit
= 28 mm
The width of the central maximum in a single-slit experiment is calculated by deriving the angle of the first minimum using light wavelength and slit width, and then applying it to the distance to the screen. The resulting value is an estimation.
Explanation:In a single-slit experiment, the width of the central maximum can be calculated with the use of physics and understanding of light's behavior. The primary formula that guides this phenomenon is given as sin θ = λ/a, where θ is the angle of the first minimum, λ is the wavelength of the light, and a is the width of the slit. Given that the slit width is 170 times the wavelength of the light, the angle of first minimum will be very small and approximated as θ = λ/a.
Now, the angular width of the central maximum would be twice this angle in radians (θ), as the central maximum extends θ on either side of the central point. To find the actual width in mm on a screen 2.4 m away, you would multiply the total angular width (2*θ) by the distance of the screen (L) behind the slit, i.e. W = 2*θ*L. Please note that this is an approximation that widely works when the angle θ is small.
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A force of 4.9 N acts on a 14 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, and (c) the third seconds and (d) the instantaneous power due to the force at the end of the third second.
Explanation:
The work done on the body = force applied x displacement
in this case , the acceleration of body a = [tex]\frac{Force}{Mass}[/tex] = [tex]\frac{4.9}{14}[/tex] = 0.35 ms⁻²
The displacement in first second S₁ = u + [tex]\frac{1}{2}[/tex] a x t²
here u = 0 , because body was at rest
Thus S₁ = [tex]\frac{1}{2}[/tex] x 0.35 x ( 1 )² = = 0.175 m
The work done = 4.9 x 0.175 = 0.86 J
The displacement in 2 seconds = [tex]\frac{1}{2}[/tex] x 0.35 x ( 2 )² = 0.7 m
Work done in 2 seconds = 4.9 x 0.7 = 3.43 J
Work done in second second = 3.43 - 0.86 = 2.57 J
The displacement in three seconds = [tex]\frac{1}{2}[/tex] x 0.35 x ( 3 )² = 1.575 m
Work done in three seconds = 4.9 x 1.575 = 7.7 J
Work done in third second = 7.7 - 3.43 = 4.3 J
The power at the end of third second is 4.3 watt
Because 4.3 J of work is done in the last second .
Low-frequency vertical oscillations are one possible cause of motion sickness, with 0.30 Hz having the strongest effect. Your boat is bobbing in place at just the right frequency to cause you the maximum discomfort. The water wave that is bobbing the boat has crests that are 30 m apart. What will be the boat’s vertical oscillation frequency if you drive the boat at 5.0 m/s in the direction of the oncoming waves?.
Answer:
0.467 Hz
Explanation:
Wave properties are related thus
v = fλ
v = velocity of the wave = ?
f = 0.30 Hz
λ = wavelength = 30 m
v = 0.3×30 = 9.0 m/s
But if the boat is now moving at 5.0 m/s in the direction of the oncoming wave,
The speed of the wave relative to the boat = 9 - (-5) = 14.0 m/s
f = (v/λ) = (14/30) = 0.467 Hz
Hope this helps!
Answer:
The answer to the question is;
The boat’s vertical oscillation frequency if you drive the boat at 5.0 m/s in the direction of the oncoming waves moving at 9 m/s in the opposite direction is 0.467 Hz.
Explanation:
We note the frequency of the water wave = 0.3 Hz
Distance between crests or wavelength of water wave = 30 m
Speed v of a wave is given by Frequency f × Wavelength λ
Therefore the speed of the wave = f·λ = 0.3 × 30 = 9 m/s
(b) Distance between crests = 30 m = wavelength λ
Boat speed = 5.0 m/s v
Speed of the wave with respect to the boat = 5 + 9 = 14 m/s
From the wave speed relationship, we have
v = fλ f = [tex]\frac{v}{\lambda}[/tex] =[tex]\frac{14}{30} = \frac{7}{15}[/tex] = 0.467 Hz which is high.
while driving his sports car at 20.0 m/s down a four lane highway, eddie comes up behind a slow moving dump truck and decides to pass it in the left hand lane. of eddie can accelerate at 5.00 m/s^2, how long will it take for him to reach a speed of 30.0 m/s
Explanation:
Given:
u = 20 m/s
a = 5 m/s^2
v = 30 m/s
t = ?
Use the first kinematic equation of motion:
v = u + at
t = (v - u)/a = 10/5 = 2 seconds
Eddie will take 2 sec to reach a speed of 30 m/s
What is velocity ?
velocity is defined as rate of change of displacement of the object with respect to rate of change in time. In mathematics It is written as :
[tex]\begin{aligned}v&=\frac{\Delta d}{\Delta t}\end{aligned}[/tex]
Here it is given that :
initial speed of car (u) = 20 m/s
acceleration of car (a) = 5 m/s²
final speed of car (v) = 30 m/s²
it is to find the time t to achieve this speed which is calculated using the first equation of motion:
[tex]\begin{aligned}v&=u+at\\30&=20+5t\\&t=\frac{10}{5}\\&=2\text{\:sec}\end{aligned}[/tex]
Therefore, Eddie will take 2 sec to reach a speed of 30 m/s
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____ is a structural engineered wood that utilizes wood from species of wood that are not large or strong enough to be useful in solid lumber products, uses 12 in. long strands that are cut from logs, dried, and immersed in resin before being pressed into solid billets, and strands are aligned parallel to each other to take advantage of the natural strength of the wood.
Answer:
Explanation:
UR MOM SUCKQ
A large steel water storage tank with a diameter of 20 m is filled with water and is open to the atmosphere (1 atm = 101 kPa) at the top of the tank. If a small hole rusts through the side of the tank, 5.0 m below the surface of the water and 20.0 m above the ground, assuming wind resistance and friction between the water and steel are not significant factors, how far from the base of the tank will the water hit the ground?
Answer:
Explanation:
Their explanation is: We first need to determine the velocity of the water that comes out of the hole, using Bernoulli's equation.
P1+ρgy1 + ½ρv1² = P2+ρgy2+½ρv2²
The atmospheric pressure exerted on the surface of the water at the top of the tank and at the hole are essentially the same.
Then, P1-P2 = 0
. Additionally, since the opening at the top of the tank is so large compared to the hole on the side, the velocity of water at the top of the tank will be essentially zero. We can also set y1 as zero, simplifying the equation to:
0 = ρgy2 + ½ρv2²
Divide through by density
-gy2 = ½ v2²
y2= 5m and g =-9.81
-•-9.81×5 = ½v2²
9.81×5×2 = v2²
98.1 = v2²
v2 = √98.1
v2 = 9.9m/s
Using equation of motion to know the time of fall
We assume that the initial vertical velocity of the water is zero and the displacement of the water is -20 m
y-yo = ut + ½gt²
0-20 = 0•t -½ ×9.81 t²
-20 = -4.905t²
t² = -20÷-4.905
t² = 4.07747
t =√4.07747
t = 2.02secs
Using range formula
Then, R=Voxt
R= 9.9 × 2.02
R =19.99m
R ≈20m
Answer:
The distance from the base of the tank to the ground is 20 m
Explanation:
From Bernoulli equation where P is pressure, V is velocity, ρ is density, g is acceleration due to gravity and y is the height
P₁ + 1/2ρV₁² + ρgy₁ = P₂ + 1/2ρV₂² + ρgy₂
The pressure on the surface of the water on the top of the tank and that in the hole is the same (P₁ = P₂). Since the opening at the top of the tank is large compared to the hole at the side of the tank, the velocity at the top of the tank is 0 and y₁ is 0
Therefore: 0 = 1/2ρV₂² + ρgy₂
1/2ρV₂² = -ρgy₂ g = -10 m/s²
-10(5) = -1/2V₂²
V₂ = 10 m/s
The time it would take the water to fall on the ground is given as:
d = ut + 1/2gt²
u is the initial velocity = 0 , g = -10 m/s² and the displacement (d) = - 20 m
Therefore: -20 = 1/2(10)t²
t = 2 sec
The horizontal displacement (d) can be gotten from
d = V₂t
d = 10(2) = 20 m
The distance from the base of the tank to the ground is 20 m
A box is moved 10 m across a smooth floor by a force making a downward angle with the floor, so that there is effectively a 10 N force acting parallel to the floor in the direction of motion and a 5 N force acting perpendicular to the floor. The work done is
Answer:
Work done = 100 J
Explanation:
Given:
Distance moved by the box (d) = 10 m
Parallel force acting on it [tex](F_1)[/tex] = 10 N
Perpendicular force acting on it [tex](F_2)[/tex] = 5 N
Work done is the product of force and displacement along the line of action of force.
So, the perpendicular force doesn't do any work and it's only the parallel force that does work as the displacement caused is along the parallel force direction.
So, work done is given as:
[tex]Work=F_1\times d\\\\Work=10\ N \times 10\ m \\\\W=100\ J[/tex]
Therefore, work done is 100 J.
While you are in a bus that moves at 100 km/h you walk from the back to the front at 10 km/h. What is your speed relative to the road outside?
Answer: 110km/h
100+10=110km/h
Explanation:
Motion is defined as a change of position. The frame of reference is usually assumed to be at rest. If one is sitting on a bus, the road appears to be moving backwards relative to the observer. If he/she now walks to the front of the bus, he has a speed relative to the earth which is now greater than that of the bus hence the answer.
Your speed relative to the road, when walking from the back to the front of a bus moving at 100km/h, is 110km/h.
Explanation:Your speed relative to the road outside in this scenario will be the sum of the speed of the bus and your own speed walking within the bus. The bus is moving at 100 km/h, and you're moving toward the front of the bus at 10 km/h. Because you're moving in the same direction as the bus, you add your speeds together. Hence, your speed relative to the road outside would be 100 km/h (bus) + 10 km/h (you) = 110 km/h.
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A numerical scale of earthquake magnitude that takes into account the size of the fault rupture is the ____.
A) Richter scale
B) modifies Mercalli scale
C) moment magnitude scale
D) epicentral distance scale
Answer:
d. epicentral distance scale
Explanation:
The depth of focus from the epicenter, called as Focal Depth, is an important parameter in determining the damaging potential of an earthquake. Most of the damaging earthquakes have shallow focus with focal depths less than about 70km. Distance from epicenter to any point of interest is called epicentral distance
Approximately how fast is Jupiter orbiting the Sun? Approximately how fast is Jupiter orbiting the Sun? 10 km/skm/s a little less than 15 km/skm/s 20 km/skm/s cannot be determined from the information provided
Answer:
The correct answer is
a little less than 15 km/s.
Explanation:
The distance between the sun and Jupiter varies by about 75 million km between the perihelion and the aphelion with an average distance of 778 million km from the sun for which it takes Jupiter about 12 years to complete its orbit round the sun giving it an orbital speed of about 13.07 km/s
The size of Jupiter is more than the twice the combined size of all the other planets, which is about 1.300 times the size of earth.
Jupiter is orbiting the Sun from b. 15 km/s.
Jupiter orbits the Sun at an average distance (semi-major axis) of about 5.2 Astronomical Units (AU), where 1 AU is the average distance from the Earth to the Sun, approximately 149.6 million kilometers.
Using Kepler's third law of planetary motion and the fact that Jupiter takes approximately 11.86 Earth years to complete one orbit around the Sun, we can estimate its average orbital speed. The orbital speed (v) of a planet can be roughly calculated by the formula:
[tex]v = \frac{2 \pi r}{T}[/tex]
where:
[tex]r[/tex] is the average orbital radius (semi-major axis in meters), and[tex]T[/tex] is the orbital period in seconds.Converting the semi-major axis from AU to kilometers:-
[tex]r = 5.2 \times 149.6 \times 10^6 \text{ km} \\= 777.92 \times 10^6 \text{ km}[/tex]
Converting the orbital period from years to seconds:
[tex]T = 11.86 \text{ years} \times 3.154 \times 10^7 \text{ seconds/year} \\= 3.74 \times 10^8 \text{ seconds}[/tex]
Now, plugging in these values to our formula:
[tex]v = \frac{2 \pi \times 777.92 \times 10^6}{3.74 \times 10^8} \text{ km/s} \\\approx 13.07 \text{ km/s} \ {or} 15 km[/tex]
A 55.0 kg ice skater is moving at 4.07 m/s when she grabs the loose end of a rope, the opposite end of which is tied to a pole. She then moves in a circle of radius 0.808 m around the pole. (b) Compare this force with her weight by finding the ratio of the force to her weight.
Answer:
Explanation:
mass of ice skater, m = 55 kg
velocity, v = 4.07 m/s
radius, r = 0.808 m
The force is centripetal force, the formula for this force is
[tex]F_{c}=\frac{mv^{2}}{r}[/tex]
[tex]F_{c}=\frac{55\times 4.07\times 4.07}{0.808}[/tex]
Fc = 1127.56 N
Weight of the person, W = mg
W = 55 x 9.8 = 539 N
The ratio of force to he weight
[tex]\frac{F_{c}}{W}=\frac{1127.56}{539}[/tex]= 2.09
The term ____________________ refers to a single logical network composed of multiple physical networks, which may all be at a single physical location, or spread among multiple physical locations.
Answer: Internetwork
Explanation:
Internetworking is the process of connecting different networks together by the use of intermediary devices such as routers or gateway devices. Internetworking helps data communication among networks owned and operated by different bodies.
Answer:
Internetwork
Explanation:
Internetwork involves having a central network system which is shared into other systems and location through the help of intermediary devices such as routers. This usually helps to increase the orderliness and precision by which data is shared between the various connections.
The amount of medicine in micrograms, A, in a person's bloodstream t minutes after a certain quantity of it has been inhaled can be approximated by the equation Upper A equals negative 40 t squared plus 160 t. How long after inhalation will there be about 120 micrograms of medicine in the bloodstream?
Answer:
After 1 second and after 3 seconds.
Explanation:
the amount of medicine in time t is given by A = -40t2 + 160t.
To know the time when the amount of medicine will be 120 micrograms, we need to make the expression A equals 120, and then find the value of t:
120 = -40t2 + 160t
-40t2 + 160t - 120 = 0
dividing everything by -40, we have
t2 - 4t + 3 = 0
we have a second order equation, and to find its roots, we can use the baskhara formula:
D = b2 -4ac = 16 - 12 = 4.
The square root of D is +2 or -2
t_1 = (4 + 2)/2 = 3
t_2 = (4 - 2)/2 = 1
We have two values of t, both acceptable. So, after 1 second and after 3 seconds of the inhalation, there will be 120 micrograms of medicine in the bloodstream.
The time after inhalation when there will be about 120 micrograms of medicine in the bloodstream is approximately 2 minutes.
Explanation:The equation given is A = -40t^2 + 160t. To find the time, 't', when the amount of medicine in the bloodstream is 120 micrograms, we should set 'A' equal to 120 and solve for 't'. So, 120 = -40t^2 + 160t. Simplify this equation to 40t^2 - 160t + 120 = 0. Here, you can observe that all terms are divisible by 10, thus simplifying it further gives 4t^2 - 16t + 12 = 0. Splitting the middle term, we get 4t^2 - 8t - 8t + 12 = 0. On factorizing, we get 4t(t - 2) - 4(2t - 3) = 0. Hence, (4t-4)(t-2) = 0, implies that t = 1 or t = 2. Since 't' cannot be negative, t = 1 minute is invalid in this context (as the amount of medicine isn't enough), we have t = 2 minutes as the valid solution.
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The relationship between heat (q), work (w), and internal energy (U) can be described with which of the following?a. q = ΔU × wb. ΔU = qwc. w = ΔUqd. ΔU = q + w
Answer:
d. ΔU = q + w
Explanation:
Internal energy is the sum of kinetic energy (which comes from motion of molecules) and potential energy ( which comes from chemical bonds between atoms and other intermolecular forces that maybe present).
First law of thermodynamics states that: "the change in internal energy of a closed system is equal to the energy added to it in form of heat(q) and work (w) done on the system by the surroundings.
Mathematically written as:
[tex]E_{internal}[/tex] = q + w
Conventionally, the first law is based on the system doing work and the surrounding doing work.
[tex]E_{internal}[/tex] can also be written as ΔU.
Therefore [tex]E_{internal}[/tex] = ΔU = q + w
ΔU = q + w
A sealed tank contains 30 moles of an ideal gas at an initial temperature of 270 K. The pressure of the gas is increased until the final pressure equals 1.4 times the initial pressure. The heat capacity at constant pressure of the gas is 32.0 J(mol*K) What is the change in the internal energy of the gas? Let the ideal-gas constant R = 8.314 J/(mol • K).
130 kJ
77 kJ
-23 kJ
100 kJ
-50 kJ
The isochoric process involving a gas at constant volume undergoes a temperature change from 270 K to 378 K, resulting in a change in internal energy of approximately 40.4 kJ.
1. Identify the process:
This problem describes an isochoric process, meaning the volume of the gas remains constant throughout the pressure change. Since the volume doesn't change, we can utilize the pressure-temperature relationship for ideal gases.
2. Apply the pressure-temperature relationship:
The pressure-temperature relationship for an ideal gas at constant volume is:
P₁V₁ = P₂V₂
where:
P₁ is the initial pressure
V₁ is the initial volume (constant in this case)
P₂ is the final pressure
V₂ is the final volume (constant in this case)
Since the volume remains constant, we can rewrite the equation as:
P₁ = P₂ * (T₁ / T₂)
where:
T₁ is the initial temperature (270 K)
T₂ is the final temperature (unknown)
3. Solve for the final temperature:
We are given that the final pressure (P₂) is 1.4 times the initial pressure (P₁). Substitute this information into the equation:
P₁ = 1.4 * P₁ * (T₁ / T₂)
Simplify the equation:
1 = 1.4 * (T₁ / T₂)
Solve for T₂:
T₂ = 1.4 * T₁
T₂ = 1.4 * 270 K
T₂ = 378 K
4. Calculate the change in internal energy:
For an ideal gas at constant volume, the change in internal energy (ΔU) is given by:
ΔU = n * Cv * ΔT
where:
n is the number of moles (30 mol)
Cv is the heat capacity at constant volume (unknown)
ΔT is the change in temperature (T₂ - T₁)
5. Relate Cv and Cp:
We are given the heat capacity at constant pressure (Cp) as 32.0 J/(mol*K). However, we need Cv for the internal energy calculation. We can utilize the relationship between Cv and Cp for ideal gases:
Cv = Cp - R
where:
R is the gas constant (8.314 J/(mol*K))
Substitute the given value of Cp:
Cv = 32.0 J/(mol*K) - 8.314 J/(mol*K)
Cv = 23.686 J/(mol*K)
6. Calculate the change in internal energy:
Now, substitute all the known values into the equation for ΔU:
ΔU = 30 mol * 23.686 J/(mol*K) * (378 K - 270 K)
ΔU = 40429.8 J
7. Convert to kilojoules and round the answer:
Convert the answer to kilojoules:
ΔU = 40429.8 J / 1000 J/kJ
ΔU ≈ 40.4 kJ
Therefore, the change in internal energy of the gas is approximately 40.4 kJ.
Two small balls, A and B, attract each other gravitationally with a force of magnitude F. If we now double both masses and the separation of the balls, what will now be the magnitude of the attractive force on each one?A) 16F
B) 8F
C) 4F
D) F
E) F/4
The magnitude of the attractive force on each one will be F.
Explanation:
As two small balls are experiencing gravitational force between them ,then they will obey universal law of gravity. As per the universal law of gravity, the gravitational force acting between two objects will be directly proportional to the product of the masses of the objects and inversely proportional to the square of the distance between the two objects.
So if the mass of the two small balls A and B are considered as M and m, respectively, with the distance of separation be considered as r. Then the gravitational force of attraction acting between A and B will be
[tex]F = \frac{GMn}{r^{2} }[/tex] , This is the original or initial gravitational force between A and B.
Now, if the masses of A and B are doubled, then the new masses will be M' = 2 M and m' = 2m, respectively. Similarly, if the separation of the balls is also doubled then r' = 2r. So the new gravitational force exerting between A and B is
[tex]F' = \frac{GM'm'}{r'^{2} } = \frac{G*2M*2m}{(2r)^{2} } =\frac{4GMm}{4r^{2} } = \frac{GMm}{r^{2} }=F[/tex]
So after doubling the masses as well as the distance of separation, there will be no change in the gravitational force. So the magnitude of the attractive force on each one will be F.
In fireworks displays, light of a given wavelength indicates the presence of a particular element. What are the frequency and color of the light associated with each of the following?
Answer:
The four wavelengths of the problem are not given. Here they are:
a) [tex]Li^+,\lambda=671 nm[/tex]
b) [tex]Cs^+, \lambda=456 nm[/tex]
c) [tex]Ca^{2+}, \lambda=649 nm[/tex]
d) [tex]Na^+, \lambda=589 nm[/tex]
The relationship between wavelength and frequency of light wave is
[tex]f=\frac{c}{\lambda}[/tex]
where
f is the frequency
[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light
[tex]\lambda[/tex] is the wavelength
For case a), [tex]\lambda=671 nm = 6.71\cdot 10^{-7}m[/tex] (corresponds to red color), so its frequency is
[tex]f=\frac{3\cdot 10^8}{6.71\cdot 10^{-7}}=4.47\cdot 10^{14}Hz[/tex]
For case b), [tex]\lambda=456 nm = 4.56\cdot 10^{-7}m[/tex] (corresponds to blue color), so its frequency is
[tex]f=\frac{3\cdot 10^8}{4.56\cdot 10^{-7}}=6.58\cdot 10^{14}Hz[/tex]
For case c), [tex]\lambda=649 nm = 6.49\cdot 10^{-7}m[/tex] (corresponds to red color), so its frequency is
[tex]f=\frac{3\cdot 10^8}{6.49\cdot 10^{-7}}=4.62\cdot 10^{14}Hz[/tex]
For case d), [tex]\lambda=589 nm = 5.89\cdot 10^{-7}m[/tex] (corresponds to yellow color), so its frequency is
[tex]f=\frac{3\cdot 10^8}{5.89\cdot 10^{-7}}=5.09\cdot 10^{14}Hz[/tex]
In fireworks displays, the color of light is determined by its wavelength and frequency. The color red has the longest wavelength and the lowest frequency, while the color violet has the shortest wavelength and the highest frequency. Therefore, the order of the given colors from shortest wavelength to longest wavelength is blue, yellow, and red. Similarly, the order of the given colors from lowest frequency to highest frequency is also blue, yellow, and red.
Explanation:The colors of light in a fireworks display indicate the presence of different elements. The wavelength and frequency of the light determines its color. Within the visible range, our eyes perceive radiation of different wavelengths as light of different colors. The color red corresponds to the longest wavelength and the lowest frequency, while the color violet corresponds to the shortest wavelength and the highest frequency. Therefore, the order of the given colors from shortest wavelength to longest wavelength would be blue, yellow, and red. Similarly, the order of the given colors from lowest frequency to highest frequency would also be blue, yellow, and red.
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Air masses are identified on the basis of temperature and
Answer:
Moisture content
Explanation:
Air mass is a volume of air spread through a vast area may it cover hundred to thousand Kilometers and is identified on the basis of nearly equal temperature and water vapour content of the air thorough out the area.
The half-life of a radioactive isotope is the time it takes for a quantity of the isotope to be reduced to half its initial mass. Starting with grams of a radioactive isotope, how much will be left after half-lives
Answer:
Incomplete questions
This is the complete question
The half-life of a radioactive isotope is the time it takes for a quantity for the isotope to be reduced to half its initial mass. Starting with 150 grams of a radioactive isotope, how much will be left after 6 half-lives
Explanation:
Let analyse the question generally first,
The the mass of the radioactive element be M.
We want to know it mass after n half life
Then,
After first half life, it mass is
M1=M×½
After second half life, it mass is
M2= M×(½)²
After third half life, it mass is
M3= M×(½)³
But now we can see a pattern developing, because for each new half-life we are dividing the quantity by 2 to a power that increases as the number of half-lives.
Then we can take the original quantity and quickly compute for
nth half-lives:
So after nth half life will be
Mn= M × (½)ⁿ
Generally,
Now, let apply it to our questions
Give that the mass of the radioactive isotope is 150grams
It mass after 6th half life
Then, n=6
So applying the formula
Mn= M × (½)ⁿ
M6= 150 ×(½)^6
M6= 150×1/64
M6=2.34grams
The mass of the radioactive isotope after 6th half life is 2.34grams
A first-order reaction has a rate constant of 0.241/min. If the initial concentration of A is 0.859 M, what is the concentration of A after 10.0 minutes? 0.0772 M 0.00334 M 0.736 M 0.280 M
Answer:
option A.
Explanation:
given,
rate constant. k = 0.241/min
[A_0] = 0.859 M
[A_t] = ?
t = 10 minutes.
using first order reaction formula
[tex]k=\dfrac{2.303}{t}log(\dfrac{[A_0]}{[A_t]})[/tex]
[tex]0.241=\dfrac{2.303}{10}log(\dfrac{[0.859]}{[A_t]})[/tex]
[tex]log(\dfrac{[0.859]}{[A_t]}) = 1.0464[/tex]
[tex]\dfrac{[0.859]}{[A_t]} = 11.129[/tex]
[tex][A_t]=\dfrac{0.859}{11.129}[/tex]
[tex][A_t]=0.0772\ M[/tex]
the concentration of A after 10 minutes is equal to 0.0772 M.
Hence, the correct answer is option A.
This is a change in the position of a body with respect to time relative to a reference point.
Answer: MOTION
Explanation:
motion is defined as the displacement of an object with respect to time relative to a stationary object (reference point). A good example of an object that can serve as a reference point includes: a tree or a building. The movement of a body at constant speed towards a particular direction at regular intervals of time can be determined and it's called uniform motion.
There are different types of motion, these includes: simple harmonic motion,
linear motion,
circular motion,
Brownian motion,
Rotatory motion
Answer:rest
Explanation:
A concrete highway is built of slabs 12 m long .How wide should be the expansion cracks between the slabs at 15 Celsius to prevent buckling if the range of temperature is -30 to 50 Celsius?
Answer:
[tex]x=4.2\ mm[/tex]
Explanation:
Given:
length of the concrete slab, [tex]l=12\ m[/tex]temperature of observation, [tex]T_o=15^{\circ}C[/tex]we've the coefficient of linear expansion for concrete, [tex]\alpha=10^{-5}\ ^{\circ}C^{-1}[/tex]lower temperature limit, [tex]T_l=-30^{\circ}C[/tex]upper temperature limit, [tex]T_u=50^{\circ}C[/tex]Change in length due to temperature can be given as:
[tex]\Delta l=l.\alpha.(T_u-T_l)[/tex]
[tex]\Delta l=12\times 10^{-5}\times (50-(-30))[/tex]
[tex]\Delta l=0.0096\ m=9.6\ mm[/tex]
Now at temperature 15°C:
[tex]\Delta l'=12\times 10^{-5}\times (15-(-30))[/tex]
[tex]\Delta l'=0.0054\ m=5.4\ mm[/tex]
Hence the expansion crack between the slabs at this temperature must be:
[tex]x=\Delta l-\Delta l'[/tex]
[tex]x=9.6-5.4[/tex]
[tex]x=4.2\ mm[/tex]
Mateo drew the field lines around the ends of two bar magnets but forgot to label the direction of the lines with arrows. At left a rectangular box with the long horizontal side labeled S. At right a rectangular box with the long horizontal side labeled N. There is a gap between the boxes. Curved lines labeled 1 emerge from the right end of S and reach the left end of N. In which direction should an arrow at position 1 point?
Answer:
left
Explanation:
The magnetic field lines always point outwards from the north pole and merges towards the south pole. Thus, the arrow, is from right to left.
What is bar magnet?A bar magnet is an object made of materials having permanent magnetism. They have two poles namely south poles and north poles. The like poles of magnets will repel and unlike pole attracts each other.
Conventionally, it is assumed that the magnet's field flows from its north pole inward to its south pole. Ferromagnetic materials can be used to create permanent magnets.
The magnetic field is strongest inside the magnetic substance, as seen by the magnetic field lines. Near the poles, there are the strongest external magnetic fields. A magnetic north pole will pull a magnet's south pole toward it while repelling another magnet's north pole. Hence, the arrow is pointing to left direction.
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When we breathe, we inhale oxygen and exhale carbon dioxide plus water vapor. Which likely has more mass, the air that we inhale or the same volume of air we exhale? Does breathing cause you to lose or gain weight?
Answer:
If the same volume of air is inhaled and exhaled, the air we breathe out normally weighs more than the air we breathe in.
Since the output from the body normally exceeds the input, breathing leads to weight loss.
Explanation:
If equal volumes of gas is inhaled and exhaled, the exhaled gas is heavier.
The inhaled gas contains Oxygen and majorly Nitrogen.
The exhaled gas contains CO₂, H₂O and a very large fraction of the unused inhaled air that goes into the lungs.
So, basically, the body exchanges O₂ with CO₂ and H₂O (and some other unwanted gases in the body) in a composition that CO₂, the heavier gas of the ones mentioned here, is prominent.
So, because the mass leaving the body is more than the mass entering, breathing leads to a loss of weight. This is one of the reasons why we need food for sustenance. Breathing alone will wear one out.
Final answer:
The air we exhale likely has less mass than the air we inhale due to the replacement of dense oxygen with less dense carbon dioxide and water vapor.
Explanation:
When we inhale, we take in air that is rich in oxygen, and when we exhale, we expel air with a higher concentration of carbon dioxide and water vapor. Given that a liter of oxygen weighs more than the same volume of water vapor, and we replace some of the oxygen with the less dense carbon dioxide, the air we exhale likely has less mass than the air we inhale.
The breathing process does result in weight loss, but only in small amount attributable to the exhaled carbon dioxide which comes from the metabolic processes in the body. As for the volume and density of our bodies, taking a deep breath increases the volume, but because the density of air is much smaller than that of the body, the overall density decreases.
During gas exchange, oxygen flows from the bloodstream into the body's cells, while carbon dioxide flows out of the cells into the bloodstream to be expelled. This process is driven by the differing partial pressures of oxygen and carbon dioxide in the blood and the cells, facilitating diffusion across the cellular membrane.