Answer:
A) P_rad.abs = 2.4 × 10^(-8) Pa and P_rad.ref = 4.8 × 10^(-8) Pa
B) P_rad.abs = 2.369 × 10^(-13) atm and P_rad.ref = 4.738 × 10^(-13) atm
Explanation:
A) The formula for radiation pressure for absorbed light is given as;
P_rad = I/c
Where I is the intensity = 7.20 W/m² and c is the speed of light = 3 × 10^(8) m/s
Thus;
P_rad = 7.2/(3 × 10^(8))
P_rad.abs = 2.4 × 10^(-8) Pa
Now formula for radiation pressure for reflected light is given as;
P_rad = 2I/c
Thus;
P_rad = (2 × 7.2)/(3 × 10^(8))
P_rad.ref = 4.8 × 10^(-8) Pa
B) Now, 1.013 × 10^(5) Pa = 1 atm
Thus, for the absorbed surface, we have;
P_rad.abs = (2.4 × 10^(-8))/(1.013 × 10^(5))
P_rad.abs = 2.369 × 10^(-13) atm
For the reflecting surface, we have;
P_rad_ref = (4.8 × 10^(-8))/(1.013 × 10^(5))
P_rad.ref = 4.738 × 10^(-13) atm
What did the results of photoelectric-effect experiments establish?
Answer:
Option A
Electrons are emitted if low intensity, high-frequency light hits a metal surface.
Explanation:
From the experiments conducted to study the photoelectric effect, conclusions were made that the key factor that contributes to the emission of electrons from the surface of the metal is the frequency of the beam of light. This frequency has to be beyond a minimum threshold, if not, there will be no emission of electrons from the metal surface no matter the intensity of the beam of light or the length of time it is incident upon the metal surface.
This makes option A correct because it highlights the contributions made by the threshold frequency to the photoelectric effect.
Metal 1 has a larger work function than metal 2. Both are illuminated with the same short-wavelength ultraviolet light.
Do electrons from metal 1 have a higher speed, a lower speed, or the same speed as electrons from metal 2? Explain.
Answer:
a lower speed
Explanation:
Let us look closely at the Einstein's photoelectric equation;
KE= E-Wo
Where;
KE= kinetic energy of the emitted photoelectron
E= energy of the incident photon
Wo= work function of the metal
Hence,where Wo for metal 1 > Wo for metal 2, it follows that KE for metal 1 must also be less than KE for metal 2.
This is because the difference between E and Wo for metal 1 is smaller than the same difference for metal 2 hence the answer.
Suppose a 58-turn coil lies in the plane of the page in a uniform magnetic field that is directed into the page. The coil originally has an area of 0.150 m2. It is stretched to have no area in 0.100 s. What is the magnitude (in V) and direction (as seen from above) of the average induced emf if the uniform magnetic field has a strength of 1.10 T? magnitude V direction ---Select--- †\
Answer:
95.7v
Explanation
Using Faraday's law of electromagnetic induction we know that rate of change in magnetic flux will induce EMF in closed loop
So it is given as
E= Ndစ/dt
E= N BA-0/ deta t
Given that
N = 58turns
B = 1.10T
A = 0.150m^²
Deta t= 0.1s
now we have
E = 58(1.10x0.150)/0.1
= 95.7v
Magnetic flux is decreasing, so the direction of the current will be to aid the decreasing flux $decrease= CLOCKWISE
Explanation:
Light of wavelength 519 nm passes through two slits. In the interference pattern on a screen 4.6 m away, adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern. What is the separation of the two slits?
Answer:
The separation of the two slits is 0.456 mm.
Explanation:
Given the wavelength of light = 519 nm
The indifference pattern = 4.6 m
Adjacent bright fringes = 5.2 mm
In the interference, the equation required is Y = mLR/d
Here, d sin theta = mL
L = wavelgnth
For bright bands, m is the order = 1,2,3,4
For dark bands, m = 1.5, 2.5, 3.5, 4.5
R = Distance from slit to screen (The indifference pattern)
Y = Distance from central spot to the nth order fringe or fringe width
Thus, here d = mLR/Y
d = 1× 519nm × 4.6 / 5.2mm
d = 0.459 mm
Choose only one correct option. Explanation needed.
Answer:
[tex]\large \boxed{\mathrm{C. \ \ \frac{500}{7 \times 15 \times 8} \ g/cm^3 }}[/tex]
Explanation:
[tex]\displaystyle \sf Density = \frac{mass}{volume}[/tex]
[tex]\displaystyle \rho = \frac{m}{V}[/tex]
[tex]\sf mass=500 \ g[/tex]
[tex]\sf volume \ of \ a \ cuboid=width \times length \times height=( 7 \times 15 \times 8) \ cm^3[/tex]
[tex]\displaystyle \rho = \frac{500}{7 \times 15 \times 8}[/tex]
An ac source of period T and maximum voltage V is connected to a single unknown ideal element that is either a resistor, and inductor, or a capacitor. At time t = 0 the voltage is zero and increasing toward a maximum. At time t = T/4 the current in the unknown element is equal to zero, and at time t = T/2 the current is I = -I max, where Imax is the current amplitude. What is the unknown element?
a. a resistor
b. an inductor or a capacitor
c. an inductor
d. a capacitor
A microwave oven operates at 2.4 GHz with an intensity inside the oven of 2300 W/m2 . Part A What is the amplitude of the oscillating electric field
Answer:
The amplitude of the oscillating electric field is 1316.96 N/C
Explanation:
Given;
frequency of the wave, f = 2.4 Hz
intensity of the wave, I = 2300 W/m²
Amplitude of oscillating magnetic field is given by;
[tex]B_o = \sqrt{\frac{2\mu_o I}{c} }[/tex]
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ m/A
I is intensity of wave
c is speed of light = 3 x 10⁸ m/s
[tex]B_o = \sqrt{\frac{2*4\pi *10^{-7}*2300}{3*10^8} } \\\\B_o = 4.3899 *10^{-6} \ T[/tex]
The amplitude of the oscillating electric field is given by;
E₀ = cB₀
E₀ = 3 x 10⁸ x 4.3899 x 10⁻⁶
E₀ = 1316.96 N/C
Therefore, the amplitude of the oscillating electric field is 1316.96 N/C
Consider two parallel wires where the magnitude of the left currentis 2 I0(io) and that of the right current is I0(io). Point A is midway between the wires,and B is an equal distance on the other side of the wires.
The ratio ofthe magnitude of the magnetic field at point A to that at point Bis________
Answer:
Explanation:
At the point midway between wires
magnetic field due to wire having current 2I₀
= 10⁻⁷ x 2 x2I₀ / r where 2r is the distance between wires .
magnetic field due to wire having current I₀
= 10⁻⁷ x 4 I₀ / r
magnetic field due to wire having current I₀
= 10⁻⁷ x 2I₀ / r
= 10⁻⁷ x 2 I₀ / r where 2r is the distance between wires .
these fields are in opposite direction as direction of current is same in both .
net magnetic field = (4 - 2 )x 10⁻⁷ x I₀ / r
= 2 x 10⁻⁷ x I₀ / r
At point A net magnetic field = 2 x 10⁻⁷ x I₀ / r
At point B , we shall calculate magnetic field
magnetic field due to nearer wire having current 2 I₀ = 10⁻⁷ x 4 I₀ / r
magnetic field due to wire far away = 10⁻⁷ x 2 I₀ / 3r
These magnetic fields act in the same direction so they will add up
net magnetic field = [ (4 I₀ / r) + (2 I₀ / 3r) ] x 10⁻⁷
= (14 I₀ / 3r ) x 10⁻⁷
Magnetic field at point B = (14 I₀ / 3r ) x 10⁻⁷
Ratio of field at A and B
= 3 / 7 . Ans
The ratio of the magnitude of the magnetic field at point A to point B is :
3 / 7
Given data :
Magnitude of the left current is 2I₀
Magnitude of the right current is I₀
First step : Determine the magnetic field at point A
The magnetic field due to the left current ( 2I₀ )
10⁻⁷ * 2 * 2I₀ / r ( 2r = distance between wires )
The magnetic field due to the right current ( I₀ )
10⁻⁷ * 2 I₀ / r
From the expressions above the magnetic fields are in opposite direction
∴ Net magnetic field = (4 - 2 )* 10⁻⁷ * I₀ / r = 2 * 10⁻⁷ * I₀ / r
Hence The magnetic field at point A = 2 * 10⁻⁷ * I₀ / r
Next step : determine the magnetic field at point B
Magnetic field due to the closest wire to point B ( i.e.2I₀ ) = 10⁻⁷ * 4 I₀ / r
Magnetic field due to the wire away from point A = 10⁻⁷ * 2 I₀ / 3r
Since the fields acts in the same directions
The net magnetic field = (4 I₀ / r) + (2 I₀ / 3r) ] * 10⁻⁷ = ( 14 I₀ / 3r ) * 10⁻⁷
Hence The magnetic field at point A = ( 14 I₀ / 3r ) * 10⁻⁷
Therefore the ratio of the magnitude of the magnetic field at point A to point B = 3/ 7
Hence we can conclude that the ratio of the magnitude of the magnetic field at point A to point B = 3 / 7
Learn more : https://brainly.com/question/22403676
A metal sample of mass M requires a power input P to just remain molten. When the heater is turned off, the metal solidifies in a time T. The heat of fusion of this metal is
Answer:
L = Pt/M
Explanation:
Power, P= Q/t = mL/t
we know that, (Q=m×l)
Now ⇒l= Pt/M
Thus l= Pt/M
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?
Answer:
The average induced emf in the coil is 0.0286 V
Explanation:
Given;
diameter of the wire, d = 11.2 cm = 0.112 m
initial magnetic field, B₁ = 0.53 T
final magnetic field, B₂ = 0.24 T
time of change in magnetic field, t = 0.1 s
The induced emf in the coil is calculated as;
E = A(dB)/dt
where;
A is area of the coil = πr²
r is the radius of the wire coil = 0.112m / 2 = 0.056 m
A = π(0.056)²
A = 0.00985 m²
E = -0.00985(B₂-B₁)/t
E = 0.00985(B₁-B₂)/t
E = 0.00985(0.53 - 0.24)/0.1
E = 0.00985 (0.29)/ 0.1
E = 0.0286 V
Therefore, the average induced emf in the coil is 0.0286 V
We have that for the Question, it can be said that the average induced emf in the coil is
E=0.028565V
From the question we are told
A fixed 11.2-cm-diameter wire coil is perpendicular to a magnetic field 0.53 T pointing up. In 0.10 s , the field is changed to 0.24 T pointing down. What is the average induced emf in the coil?
Generally the equation for the Average emf induced is mathematically given as
[tex]Emf_a=-NA\frac{dB}{dt}\\\\Where\\\\Area\\\\a=\pir^2\\\\a=\pi(0.056)^2\\\\a=0.00985\\\\[/tex]
Hence
[tex]dB=0.24-0.53\\\\dB=-0.29T[/tex]
Therefore
[tex]E=-\frac{1*0.00985*-0.29 }{0.10}[/tex]
E=0.028565V
For more information on this visit
https://brainly.com/question/23379286
Just wondering if I did this right
Yeah
All they are all correct
Vector has a magnitude of 6.0 m and points 30° north of east. Vector has a magnitude of 4.0 m and points 30° east of north. The resultant vector + is given by
Answer:
The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].
Explanation:
First, each vector is determined in terms of absolute coordinates:
6-meter vector with direction: 30º north of east.
[tex]\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)[/tex]
[tex]\vec A = 5.196\,i + 3\,j[/tex]
4-meter vector with direction: 30º east of north.
[tex]\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)[/tex]
[tex]\vec B = 2\,i + 3.464\,j[/tex]
The resultant vector is obtaining by sum of components:
[tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex]
The resultant vector is [tex]\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j[/tex].
Two coherent sources of radio waves, A and B, are 5.00 meters apart. Each source emits waves with wavelength 6.00 meters. Consider points along the line connecting the two sources.Required:a. At what distance from source A is there constructive interference between points A and B?b. At what distances from source A is there destructive interference between points A and B?
Answer:
a
[tex]z= 2.5 \ m[/tex]
b
[tex]z = (1 \ m , 4 \ m )[/tex]
Explanation:
From the question we are told that
Their distance apart is [tex]d = 5.00 \ m[/tex]
The wavelength of each source wave [tex]\lambda = 6.0 \ m[/tex]
Let the distance from source A where the construct interference occurred be z
Generally the path difference for constructive interference is
[tex]z - (d-z) = m \lambda[/tex]
Now given that we are considering just the straight line (i.e points along the line connecting the two sources ) then the order of the maxima m = 0
so
[tex]z - (5-z) = 0[/tex]
=> [tex]2 z - 5 = 0[/tex]
=> [tex]z= 2.5 \ m[/tex]
Generally the path difference for destructive interference is
[tex]|z-(d-z)| = (2m + 1)\frac{\lambda}{2}[/tex]
=> [tex]|2z - d |= (0 + 1)\frac{\lambda}{2}[/tex]
=> [tex]|2z - d| =\frac{\lambda}{2}[/tex]
substituting values
[tex]|2z - 5| =\frac{6}{2}[/tex]
=> [tex]z = \frac{5 \pm 3}{2}[/tex]
So
[tex]z = \frac{5 + 3}{2}[/tex]
[tex]z = 4\ m[/tex]
and
[tex]z = \frac{ 5 -3 }{2}[/tex]
=> [tex]z = 1 \ m[/tex]
=> [tex]z = (1 \ m , 4 \ m )[/tex]
How wide is the central diffraction peak on a screen 2.20 mm behind a 0.0328-mmmm-wide slit illuminated by 588-nmnm light?
Answer:
[tex]y = 0.0394 \ m[/tex]
Explanation:
From the question we are told that
The distance of the screen is [tex]D = 2.20 \ m[/tex]
The distance of separation of the slit is [tex]d = 0.0328 \ mm = 0.0328*10^{-3} \ m[/tex]
The wavelength of light is [tex]\lambda = 588 \ nm = 588 *10^{-9} \ m[/tex]
Generally the condition for constructive interference is
[tex]dsin\theta = n * \lambda[/tex]
=> [tex]\theta = sin^{-1} [ \frac{ n * \lambda }{d } ][/tex]
here n = 1 because we are considering the central diffraction peak
=> [tex]\theta = sin^{-1} [ \frac{ 1 * 588*10^{-9} }{0.0328*10^{-3} } ][/tex]
=> [tex]\theta = 1.0274 ^o[/tex]
Generally the width of central diffraction peak on a screen is mathematically evaluated as
[tex]y = D tan (\theta )[/tex]
substituting values
[tex]y = 2.20 * tan (1.0274)[/tex]
[tex]y = 0.0394 \ m[/tex]
please help !!!!!!!!!!
Answer:
Lighthouse 1 during the day will be warmer, lighthouse 2 during the night will be warmer.
Explanation:
As the paragraph stated land absorbs heat and heats up faster than water. So during the day the lighthouse farthest away from the water will be hotter. But then the converse is true also land losses heat faster than water at night. So the water retains the heat from the day better making the lighthouse by the water warmer at night.
Kasek rides his bicycle down a 6.0° hill (incline is
6° with the horizontal) at a steady speed of 4.0
m/s. Assuming a total mass of 75 kg (bicycle and
Kasek), what must be Kasek's power output to
climb the same hill at the same speed?
Answer:
P = 2923.89 W
Explanation:
Power is
P = F v
for which we must calculate the force, let's use Newton's second law, let's set a coordinate system with a flat parallel axis and the other axis (y) perpendicular to the plane
X Axis
F - Wₓ = 0
F = Wₓ
Y Axis
N - [tex]W_{y}[/tex] = 0
let's use trigonometry for the components of the weight
sin 6 = Wₓ / W
cos 6 = W_{y} / W
Wₓ = W sin 6
W_{y} = W cos 6
F = mg cos 6
F = 75 9.8 cos 6
F = 730.97 N
let's calculate the power
P = F v
P = 730.97 4.0
P = 2923.89 W
In a LRC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect of this change on the impedance of the circuit
Answer:
Impedance increases for frequencies below resonance and decreases for the frequencies above resonance
Explanation:
See attached file
Explanation:
You are walking around your neighborhood and you see a child on top of a roof of a building kick a soccer ball. The soccer ball is kicked at 31° from the edge of the building with an initial velocity of 15 m/s and lands 63 meters away from the wall. How tall, in meters, is the building that the child is standing on?
Answer:
69.58 m tall
Explanation:
Pls see attached file
For a beam of light in air (n = 1) reflecting off glass (n = 1.5), what is Brewster's angle to the nearest degree?
Answer: 56°
Explanation:
Brewster's angle refers to the angle at the point where light of a certain polarization passes through a transparent dielectric surface and is transmitted perfectly such that no reflection is made.
The formula is;
[tex]= Tan^{-1} (\frac{n_{2} }{n_{1}} )[/tex]
[tex]= Tan^{-1} (\frac{1.5 }{1} )[/tex]
= 56.30993247
= 56°
You have three resistors: R1 = 1.00 Ω, R2 = 2.00 Ω, and R3 = 4.00 Ω in parallel. Find the equivalent resistance for the combination
Answer:
4 / 7
Explanation:
1/total resistance = 1/1 + 1/2 + 1/4
= 1¾
total resistance = 1 ÷ 1¾
= 4/7
Can a person break a wall with a punch?(I mean technically,maybe with inhuman speed)
Answer:
yes, shaolin monks can do. you can find tutorials on Youtb
Which unbalanced force accounts for the direction of the net force of the rocket?
a. Air resistance
b. Friction
c. Gravity
d. Thrust of rocket engine
It depends on what stage of the mission you're talking about.
==> While it's sitting on the pad before launch, the forces on the rocket are balanced, so there's no net force on it.
==> When the engines ignite, their thrust (d) is greater than the force of gravity. So the net force on the rocket is upward, and the spacecraft accelerates upward.
==> After the engines shut down, the net force acting on the rocket is due to Gravity (c).
. . . If the rocket has enough vertical speed, it escapes the Earth completely, and just keeps going.
. . . If it has enough horizontal speed, it enters Earth orbit.
. . . If it doesn't have enough vertical or horizontal speed, it falls back to Earth.
A rocket will preserve to speed up so long as there's a resultant pressure upwards resulting from the thrust of the rocket engine.
What unbalanced force bills for the course of the internet pressure of the rocket?A rocket launches whilst the pressure of thrust pushing it upwards is greater than the burden force because of gravity downwards. This unbalanced pressure reasons a rocket to accelerate upwards. A rocket will maintain to hurry up so long as there's a resultant force upwards resulting from the thrust of the rocket engine.
What's the net pressure of unbalanced?
If the forces on an item are balanced, the net pressure is zero. If the forces are unbalanced forces, the results do not cancel each difference. Any time the forces acting on an object are unbalanced, the net pressure is not 0, and the movement of the item modifications.
Learn more about the thrust of the rocket engine. here: https://brainly.com/question/10716695
#SPJ2
Simple harmonic oscillations can be modeled by the projection of circular motion at constant angular velocity onto the diameter of a circle. When this is done, the analog along the diameter of the acceleration of the particle executing simple harmonic motion is
Answer:
the analog along the diameter of the acceleration of the particle executing simple harmonic motion is the projection along the diameter of the centripetal acceleration of the particle in the circle
A light beam is traveling through an unknown substance. When it strikes a boundary between that substance and the air (nair≈1), the angle of reflection is 29.0∘ and the angle of refraction is 39.0∘. What is the index of refraction n of the substance?
Answer:
0.7707
Explanation:
From Snell's law,
n(1) * sin θ1 = n(2) * sinθ2
Where n(1) = refractive index of air = 1.0003
θ1 = angle of incidence
n(2) = refractive index of second substance
θ2 = angle of refraction
The angle of reflection through the unknown substance is the same as the angle of incidence of air. Thus this means that θ1 = 29°
=> 1.0003 * sin29 = n(2) * sin39
n(2) = (1.0003 * sin29) / sin39
n(2) = 0.7707
Explanation:
The index of refraction n of the substance is 0.7707
Snell law:Here we know that
n(1) * sin θ1 = n(2) * sinθ2
here
n(1) = refractive index of air = 1.0003
θ1 = angle of incidence
n(2) = refractive index of second substance
θ2 = angle of refraction
The angle of reflection should be via the unknown substance that represent the same as the angle of incidence of air.
So,
θ1 = 29°
1.0003 * sin29 = n(2) * sin39
n(2) = (1.0003 * sin29) / sin39
n(2) = 0.7707
learn more about refraction here: https://brainly.com/question/21312906
From a hot air balloon 2 km high, a person looks east and sees one town with angle of depression of 16 degrees. He then looks west to see another town with angle of depression of 84 degrees. What is the distance between the two towns?
Answer:
7km
Explanation:
The angle of depression is congruent to the angle of elevation and can be explained as angle below horizontal in which the person observing an object must view for him/her to view object's that are lower than him/her.
In angle of depression, there is assumption that object is closer to the person observing it, so there is parallel horizontal for both observing and object been observed.
hot air balloon 2 km high,
there exist two triangles
From trigonometry
Tanx= opposite/adjacent
Opp= 2km
Adj= X1
first triangle have base length of
Tan(16)=2/X1
X1=2/ tan(16)
X1=6.97
For Second triangle
Tanx= opposite/adjacent
Opp= 2km
Adj= X2
the other with a base length of
X2=2/tan(84)
X2=0.21
Therefore,, the total distance between the two towns is
x1+x2=6.97+0.21=7.18km
a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring force constant with
Complete question:
a 2.0 kg block slides on the horizontal, frictionless surface until it counters a spring with force constant of 955 N/m. The block comes to rest after compressing the spring a distance of 4.6 cm. Find the initial speed (in m/s) of the block.
Answer:
The initial speed of the block is 1.422 m/s
Explanation:
Given;
mass of the block, m = 2.0 kg
force constant of the spring, K = 955 N/m
compression of the spring, x = 4.6 cm = 0.046 m
Apply Hook's law to determine applied force on the spring;
F = Kx
F = (955 N/m)(0.046 m)
F = 43.93 N
Apply Newton's 2nd law to determine the magnitude of deceleration of the block when it encounters the spring;
F = ma
a = F / m
a = 43.93 / 2
a = 21.965 m/s²
Apply kinematic equation to determine the initial speed of the block;
v² = u² + 2ax
where;
v is the final speed of the block = 0
u is the initial speed of the block
x is the distance traveled by the block = compression of the spring
a is the block deceleration = -21.965 m/s²
0 = u² + 2(-21.965 )(0.046)
0 = u² - 2.021
u² = 2.021
u = √2.021
u = 1.422 m/s
Therefore, the initial speed of the block is 1.422 m/s
Which one of the following actions would make the maxima in the interference pattern from a grating move closer together?
A. Increasing the number of lines per length.
B. Decreasing the number of lines per length.
C. Increasing the distance to the screen.
D. Increasing the wavelength of the laser.
Answer:
Answer:
A. Increasing the number of lines per length.
A plastic balloon that has been rubbed with wool will stick to a wall.
a. Can you conclude that the wall is charged? If not, why not? If so, where does the charge come from?
b. Draw a series of charge diagrams showing how the balloon is held to the wall.
Answer:
Explanation:
When plastic balloon is rubbed with wool , charges are created on both balloon and silk in equal amount . Rubber balloon will acquire negative charge and silk will acquire positive charge .
Now when balloon is brought near a wall , there is induction of charge on the wall due to charge on the balloon . On the near surface of wall positive charge is produced and on the surface deep inside the wall negative charge is produced . The charge deep inside goes inside the earth but the positive charge near the surface of wall can not escape . It remains trapped by negative charge on the balloon .
hence there is mutual attraction between balloon and surface of wall is just like attraction between opposite charges . But once the ballon due to mutual attraction comes in contact with the wall , the charge on balloon and on wall neutralises each other and hence after some time the balloon falls off from the wall on the ground . It does not remain attracted to wall for ever . It happens due to neutralisation of charges on balloon and wall .
within which type of system is the total mass conserved but not the total energy
In a closed system the mass is conserved, but the energy is not conserved.
To find the answer, we have to study about different systems in thermodynamics.
What is thermodynamic system?A system, which can be expressed in terms of thermodynamic coordinates is called Thermodynamic system.Open system: System can exchange both energy and matter, thus, both energy and matter is not conserved here.Closed system can exchange energy with its surroundings (as heat or work), but not matter.Isolated system: A system that is open to the environment can interchange energy and matter, but a system that is insulated from it cannot.Thus, we can conclude that, in closed system the mass is conserved, but the energy is not conserved.
Learn more about Thermodynamic system here:
https://brainly.com/question/26035962
#SPJ1
48. A patient presents with a thrombosis in
the popliteal vein. This thrombosis most likely
causes reduction of blood flow in which of the
following veins?
Answer:
the interation blood veins
Explanation:
