Answer:
the temperature of the intermediate reservoir is 624.5 K
Explanation:
Given the data in the question
The two Carnot heat engines are operating in series;
[ T[tex]_H[/tex] ]
↓
((1)) ⇒ W[tex]_{out[/tex]
↓
[ T[tex]_M[/tex] ]
↓
((2)) ⇒ W[tex]_{out[/tex]
[ T[tex]_L[/tex] ]
The maximum possible efficiency for any heat engine is the Carnot efficiency;
η[tex]_{rev[/tex] = 1 - [tex]\frac{T_L}{T_H}[/tex]
the thermal efficiencies if both engines are the same will be;
η[tex]_A[/tex] = η[tex]_B[/tex]
1 - [tex]\frac{T_M}{T_H}[/tex] = 1 - [tex]\frac{T_L}{T_M}[/tex]
1 - 1 - [tex]\frac{T_M}{T_H}[/tex] = - [tex]\frac{T_L}{T_M}[/tex]
- [tex]\frac{T_M}{T_H}[/tex] = - [tex]\frac{T_L}{T_M}[/tex]
[tex]\frac{T_M}{T_H}[/tex] = [tex]\frac{T_L}{T_M}[/tex]
T[tex]_M[/tex]² = T[tex]_L[/tex] × T[tex]_H[/tex]
T[tex]_M[/tex] = √(T[tex]_L[/tex] × T[tex]_H[/tex])
source temperature of the first engine T[tex]_H[/tex] = 1300 K
sink temperature of the second engine T[tex]_L[/tex] = 300 K
we substitute
T[tex]_M[/tex] = √(300 × 1300)
T[tex]_M[/tex] = √390000
T[tex]_M[/tex] = 624.4998 K ≈ 624.5 K
Therefore, the temperature of the intermediate reservoir is 624.5 K
According to Newton's first law, an object at rest will _____.
never move
stay at rest forever
start moving
stay at rest unless moved by force
1. What types of natural phenomena could serve as time standards?
Answer:
The movement of Sun and moon
Explanation:
When the sun rise.it is am and when it sets .it is pm.
A class is learning about states of matter. The students set up the investigation in the diagram.
Which kinds of energy are needed in this investigation to change the state of matter of the owl made of wax?
At the base of a hill, a 90 kg cart drives at 13 m/s toward it then lifts off the accelerator pedal). If the cart just barely makes it to the top of this hill and stops, how high must the hill be?
Answer:
8.45 m
Explanation:
From the question given above, the following data were obtained:
Mass (m) = 90 Kg
Initial velocity (u) = 13 m/s
Final velocity (v) = 0 m/s
Height (h) =?
NOTE: Acceleration due to gravity (g) = 10 m/s²
The height of the hill can be obtained as follow:
v² = u² – 2gh (since the cart is going against gravity)
0² = 13² – (2 × 10 × h)
0 = 169 – 20h
Rearrange
20h = 169
Divide both side by 20
h = 169/20
h = 8.45 m
Therefore, the height of the hill is 8.45 m
Define Mechanical advantage
fe effort of 2125N is used to lift a Lead of 500N
through a Verticle high of 2.N using a buly System
if the distance Moved by the effort is 45m
Calculate 1. Work done on the load
2. work done by the effort
3. Efficiency of the System
Answer:
1) 1000Nm
2) 95,625Nm
3) 1.05%
Explanation:
Mechanical Advantage is the ratio of the load to the effort applied to an object.
MA = Load/Effort
1) Workdone on the load = Force(Load) * distance covered by the load
Workdone on the load = 500N * 2m
Workdone on the load = 1000Nm
2) work done by the effort = Effort * distance moves d by effort
work done by the effort = 2125 * 45
work done by the effort = 95,625Nm
3) Efficiency = Workdone on the load/ work done by the effort * 100
Efficiency = 1000/95625 * 100
Efficiency = 1.05%
Hence the efficiency of the system is 1.05%
The knee extensors insert on the tibia at an angle of 30 degrees (from the longitudinal axis of the tibia), at a distance of 3 cm from the axis of rotation at the knee. How much force must the knee extensors exert to produce an angular acceleration at the knee of 1 rad/s2 , given a mass of the lower leg and foot of 4.5 kg, and a radius of gyration of 23 cm
Answer:
the knee extensors must exert 15.87 N
Explanation:
Given the data in the question;
mass m = 4.5 kg
radius of gyration k = 23 cm = 0.23 m
angle ∅ = 30°
∝ = 1 rad/s²
distance of 3 cm from the axis of rotation at the knee r = 3 cm = 0.03 m
using the expression;
ζ = I∝
ζ = mk²∝
we substitute
ζ = 4.5 × (0.23)² × 1
ζ = 0.23805 N-m
so
from; ζ = rFsin∅
F = ζ / rsin∅
we substitute
F = 0.23805 / (0.03 × sin( 30 ° )
F = 0.23805 / (0.03 × 0.5)
F F = 0.23805 / 0.015
F = 15.87 N
Therefore, the knee extensors must exert 15.87 N
3. Two bullets have masses of 0.003 kg and 0.006 kg, respectively. Both are fired with a speed of 40.0 m/s.
A. Which bullet has more kinetic energy?
B. When you double the mass, what happens to the kinetic energy?
Answer:
A. The bullet with 0.006kg has more energy
B. When the mass is doubled the kinetic energy increases
Explanation:
Kinetic energy increases when mass increases
kinetic energy increases when velocity increases
३.रात में घूमने वाला write one word substitute
Explanation:
रात में घूमने वाला arthaarat निशाचर
If an observer on Earth sees a total lunar eclipse, Group of answer choices everyone on the nighttime side of Earth is seeing it. someone elsewhere on Earth must be seeing a partial lunar eclipse. someone elsewhere on Earth must be seeing a total solar eclipse
Answer:
everyone on the nighttime side of Earth is seeing it.
Explanation:
A lunar eclipse is a phenomenon that occurs when the Earth comes between the Moon and the Sun thereby causing it to cover the Moon with its shadow.
Simply stated, lunar eclipse takes place when the Moon passes or moves through the Earth's shadow thereby blocking any ray of sunlight from reaching the Moon. Thus, the full moon appears deep red (blood moon).
Also, a lunar eclipse would occur only when the Sun, Earth, and Moon are closely aligned to form a straight line known as the syzygy.
There are three (3) types of lunar eclipse and these are;
1. Total lunar eclipse.
2. Partial lunar eclipse.
3. Penumbra lunar eclipse.
Generally, if an observer on Earth sees a total lunar eclipse, everyone on the nighttime side of Earth is seeing it because it's quite easy to see a total lunar eclipse while the full moon passes through the innermost part of the shadow of the earth.
a sharp image is formed when light reflects from a
Answer:
Regular reflection
Explanation:
Regular reflection occurs when light reflects off a very smooth surface and forms a clear image.
i hope this helps a bit.
According to the context, a sharp image is formed when light reflects from a regular reflection.
What is regular reflection?It is reflection without diffusion that obeys the laws of geometrical optics, as in mirrors.
This reflection of light happens when the angles that the two rays determine with the surface are equal.
Therefore, we can conclude that according to the context, a sharp image is formed when light reflects from a regular reflection.
Learn more about regular reflection here: https://brainly.com/question/3778324
#SPJ2
During which phase is the moon not visible?
A) Full Moon
B) First quarter
C) New moon
D) Waxing crescent
Answer:
they are right it is a new moon
Explanation:
took the test
Mechanical energy is the most concentrated form of energy.
a. true
b. false
Is there a way to see moon and the sun at once?
A long, uninsulated steam line with a diameter of 100 mm and a surface emissivity of 0.8 transports steam at 150°C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20°C. (a) Calculate the rate of heat loss per unit length for a calm day. (b) Calculate the rate of heat loss on a breezy day when the wind speed is 8
Answer:
Heat loss per unit length = 642.358 W/m
The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m
Explanation:
From the information given:
Diameter D [tex]= 100 mm = 0.1 m[/tex]
Surface emissivity ε = 0.8
Temperature of steam [tex]T_s[/tex] = 150° C = 423K
Atmospheric air temperature [tex]T_{\infty} = 20^0 \ C = 293 \ K[/tex]
Velocity of wind V = 8 m/s
To calculate average film temperature:
[tex]T_f = \dfrac{T_s+T_{\infty}}{2}[/tex]
[tex]T_f = \dfrac{423+293}{2}[/tex]
[tex]T_f = \dfrac{716}{2}[/tex]
[tex]T_f = 358 \ K[/tex]
To calculate volume expansion coefficient
[tex]\beta= \dfrac{1}{T_f} \\ \\ \beta= \dfrac{1}{358} \\ \\ \beta= 2.79 \times 10^{-3} \ K^{-1}[/tex]
From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;
Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s
Thermal conductivity k = 30.608 × 10⁻³ W/m.K
Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s
Prandtl no. Pr = 0.698
Rayleigh No. for the steam line is determined as follows:
[tex]Ra_{D} = \dfrac{g \times \beta (T_s-T_{\infty}) \times D_b^3}{\alpha\times v}[/tex]
[tex]Ra_{D} = \dfrac{9.8 \times (2.79 *10^{-3})(150-20) \times (0.1)^3}{(31.244\times 10^{-6}) \times (21.7984\times 10^{-6})}[/tex]
[tex]Ra_{D} = 5.224 \times 10^6[/tex]
The average Nusselt number is:
[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(Ra_D)^{1/6}}{[ 1+ (0.559/Pr)^{9/16}]^{8/27}} \Big \}^2[/tex]
[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(5.224\times 10^6)^{1/6}}{[ 1+ (0.559/0.698)^{9/16}]^{8/27}} \Big \}^2[/tex]
[tex]Nu_D = \Big \{ 0.60 + \dfrac{5.0977}{[ 1.8826]^{8/27}}\Big \}^2[/tex]
[tex]Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2[/tex]
[tex]Nu_D = 23.29[/tex]
However, for the heat transfer coefficient; we have:
[tex]h_D = \dfrac{Nu_D\times k}{D_b} \\ \\ h_D = \dfrac{(23.29) \times (30.608 \times 10^{-3} )}{0.1}[/tex]
[tex]h_D = 7.129 \ Wm^2 .K[/tex]
Hence, Stefan-Boltzmann constant [tex]\sigma = 5.67 \times 10^{-8} \ W/m^2.K^4[/tex]
Now;
To determine the heat loss using the formula:
[tex]q'_b = q'_{ev} + q'_{rad} \\ \\ q'_b = h_D (\pi D_o) (T_t-T_{\infty})+\varepsilon(\pi D_b)\sigma (T_t^4-T_{\infty }^4)[/tex]
[tex]q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^{-8}) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}[/tex]
Now; here we need to determine the Reynold no and the average Nusselt number:
[tex]Re_D = \dfrac{VD_b}{v } \\ \\ Re_D = \dfrac{8 *0.1}{21.7984 \times 10^{-6}} \\ \\ Re_D = 3.6699 \times 10^4[/tex]
However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;
[tex]Nu_D = 0.3 + \dfrac{0.62 \times Re_D^{1/2}* Pr^{1/3}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\dfrac{Re_D}{282000})^{5/8}]^{4/5}[/tex]
[tex]Nu_D = 0.3 + \dfrac{0.62 \times (3.6699*10^4)^{1/2}* (0.698)^{1/3}}{[1+(0.4/0.698)^{2/3}]^{1/4}} [1+ (\dfrac{3.669*10^4}{282000})^{5/8}]^{4/5}[/tex]
[tex]Nu_D = (0.3 +\dfrac{105.359}{1.140}\times 1.218) \\ \\ Nu_D = 112.86[/tex]
SO, the heat transfer coefficient for forced convection is determined as follows afterward:
[tex]h_D = \dfrac{Nu_{D}* k}{D_b} \\ \ h_D = \dfrac{112.86*30.608 *10^{-3}}{0.1} \\ \\ h_D = 34.5 \ W/m^2 .K[/tex]
Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:
[tex]q'b = h_D (\pi D_b) (T_s-T_{\infty}) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^{-8}) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}[/tex]
Rhodium is in period 5 of the periodic table. What does this tell you about this element
Answer:
. It is an extraordinarily rare, silvery-white, hard, corrosion-resistant, and chemically inert transition metal. It is a noble metal and a member of the platinum group.
Explanation:
Highest density of electrostatic charges in a metal is found where
I don't know the answer but I just want points sorry
Calculate the kinetic energy of an 80,000 kg airplane that is flying with a velocity of 167 m/s.
Answer:
1115560000 J
Explanation:
1/2 * 80,000 * 167^2 m/s = 1115560000 J
A reaction occurs when a compound breaks down. This reaction has one reactant and two or more products. Energy, as from a battery, is usually needed to break the compound apart.
Answer:
decomposition
Explanation:
How much work is done when 100 N of force is applied to a rock to move it 20 m
Answer: 2000 J
Explanation: work W = F s
(d) Suppose you use a spring to launch a payload horizontally from the asteroid so that the payload ends up far from the asteroid, travelling at a speed of 3 m/s. The payload has a mass of 29 kg. If the spring is to be compressed initially an amount of 1.4 m, what stiffness ks must the spring be designed to have
Answer:
ks= 133.2 N/m
Explanation:
Assuming that we can neglect the gravitational potential energy of the mass, and that no other forces acting on the payload, total mechanical energy must be conserved.This energy, at any time, is part elastic potential energy (stored in the spring) and part kinetic energy.When the spring is initially compressed, the payload is at rest, so all energy is elastic potential.Once the spring has returned to its natural state, all this elastic potential energy must have been turned into kinetic energy.If the payload is launched horizontally, and no gravity is present,this means that its final speed will be horizontal only also, according to Newton's First Law.So, we can write the following equation:[tex]\Delta U + \Delta K = 0 (1)[/tex]
where ΔU = -1/2*k*(Δx)² (2)and ΔK = 1/2*m*v² (3)Replacing in (2) and (3) by the givens, and simplifying, we can find the stiffness ks as follows:[tex]k_{s} =\frac{m*v^{2}}{\Delta x^{2}} = \frac{29 kg*(3m/s)^{2}}{(1.4m)^{2}} = 133.2 N/M (4)[/tex]
On a 10 kg cart (shown below), the cart is brought up to speed with 50N of force for 7m, horizontally. At this point (A), the cart begins to experience an average frictional force of 15N throughout the ride.
Find:
a) The total energy at (A)
b) The velocity at (B)
c) The velocity at (C)
d) Can the cart make it to Point (D)? Why or why not?
Which time interval has the greatest speed?
Answer:
es la 2
Explanation:
epero que te curva
A projectile is fired with an initial velocity of 120.0 m/s at an angle, θ, above the horizontal. If the projectile’s initial horizontal speed is 55 meters per second, then angle θ measures approximately
Answer:
algm sabe tô precisando muito
1. A perspex box has a 10 cm square base and contains water to a height of 10 cm. A piece of rock of mass 600g is lowered into the water and the level rises to 12 cm.
(a) What is the volume of water displaced by the rock?
(b) What is the volume of the rock?
(c) Calculate the density of the rock
Answer:
(a) The volume of water is 100 cm³
(b) The volume of the rock is 20 cm³
(c) The density of the rock is 30 g/cm³
Explanation:
The given parameters of the perspex box are;
The area of the base of the box, A = 10 cm²
The initial level of water in the box, h₁ = 10 cm
The mass of the rock placed in the box, m = 600 g
The final level of water in the box, h₂ = 12 cm
(a) The volume of water in the box, 'V', is given as follows;
V = A × h₁
∴ The volume of water in the box, V = 10 cm² × 10 cm = 100 cm³
The volume of water in the box, V = 100 cm³
(b) When the rock is placed in the box the total volume, [tex]V_T[/tex], is given by the sum of the rock, [tex]V_r[/tex], and the water, V, is given as follows;
[tex]V_T[/tex] = [tex]V_r[/tex] + V
[tex]V_T[/tex] = A × h₂
∴ [tex]V_T[/tex] = 10 cm² × 12 cm = 120 cm³
The total volume, [tex]V_T[/tex] = 120 cm³
The volume of the rock, [tex]V_r[/tex] = [tex]V_T[/tex] - V
∴ [tex]V_r[/tex] = 120 cm³ - 100 cm³ = 20 cm³
The volume of the rock, [tex]V_r[/tex] = 20 cm³
(c) The density of the rock, ρ = (Mass of the rock, m)/(The volume of the rock)
∴ The density of the rock, ρ = 600 g/(20 cm³) = 30 g/cm³
A hollow sphere is attached to the end of a uniform rod. The sphere has a radius of 0.64 m and a mass of 0.48 kg. The rod has a length of 1.78 m and a mass of 0.50 kg. The rod is placed on a fulcrum (pivot) at X = 0.34 m from the left end of the rod.
(a) Calculate the moment of inertia (click for graphical table) of the contraption around the fulcrum. kg m2
(b) Calculate the torque about the fulcrum, using CCW as positive. N.m
(c) Calculate the angular acceleration of the contraption, using CCW as positive. rad/s2
(d) Calculate the linear acceleration of the right end of the rod, using up as positive. m/s2
The image of this hollow sphere and uniform rod is missing, so i have attached it.
Answer:
A) J = 0.7443 kg•m²
B) T = 1.9169 N•m CCW
C) α = 2.5754 rad/s²
D) a = 3.966 m/s²
Explanation:
A) The moment of inertia J of the contraption around the fulcrum is given by the formula;
J = Jℓ + Jr
Let's calculate Jℓ
Jℓ = [((0.34²/3) × 0.50 × 0.34)/1.78] + (0.48 × (0.34 + 0.64)²)
Jℓ = 0.4647 kg•m²
Now, let's Calculate Jr
Jr = ((1.78 - 0.34)²/3) × ((1.78 - 0.34)/1.78) × 0.50
Jr = 0.2796 kg•m²
Thus;
J = 0.4647 + 0.2796
J = 0.7443 kg•m²
(b) Using CCW as positive, Torque in Nm is calculated as;
T = Tℓ - Tr
Let's calculate Tℓ
Tℓ = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81
Tℓ = 4.7739 N•m CCW
Now, let's Calculate Tr;
Tr = [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81
Tr = 2.857 N•m CW
Thus;
T = 4.7739 - 2.857
T = 1.9169 N•m CCW
(c) The angular acceleration α of the contraption, using CCW is gotten from the formula;
α = T/J
α = 1.9169/0.7443
α = 2.5754 rad/s²
(d) The linear acceleration a of the right end of the rod, using up as positive is given by;
a = α*(1.78 - 0.34)
a = 2.5754 × 1.54
a = 3.966 m/s²
A) the moment of inertia of the contraption is 0.7443 kgm²
B) The torque about the fulcrum is 1.9169 Nm
C) Angular acceleration of the contraption is 2.5754 rad/s²
D) The linear acceleration of the contraption is 3.966 m/s²
Moment of inertia:(A) The moment of inertia I of the contraption around the fulcrum is given by :
[tex]I = [(0.34^2/3) \times 0.50 \times 0.34)/1.78 + (0.48 \times (0.34 + 0.64)^2)] + [(1.78 - 0.34)^2/3) \times (1.78 - 0.34)/1.78) \times 0.50][/tex]
I = 0.4647 + 0.2796
I = 0.7443 kgm²
(B) Using CCW as positive, Torque in Nm is given by;
T = [(0.48 × (0.64 + 0.34)) + (0.50 × 0.34/1.78) × 0.34/2)] × 9.81 - [(0.50 × (1.78 - 0.34)/1.78) × (1.78 - 0.34)/2)] × 9.81
T = 4.7739 - 2.857
T = 1.9169 Nm
(C) The angular acceleration (α) of the contraption is given by:
α = T/I
since, torque is defined as T = Iα
α = 1.9169/0.7443
α = 2.5754 rad/s²
(D) The linear acceleration (a) of the right end of the rod
a = αr
where r is the distance from the pivot
a = α × (1.78 - 0.34)
a = 2.5754 × 1.54
a = 3.966 m/s²
Learn more about moment of inertia:
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why doesn't a radio operating with two batteries function when one of the batteries is reversed?
Answer:
If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.
Explanation:
A student's backpack has a mass of 9.6 kg. The student applies a force of 94.08 N [up] while walking through 1.4 km [E] to get to school. Calculate the work done by the student on the backpack
The student does zero work on the backpack because the upward force applied by the student is acting perpendicular to the backpack's displacement parallel to the ground.
How much energy would be required to move the earth into a circular orbit with a radius 2.0 kmkm larger than its current radius
Answer:
[tex]3.52\times 10^{25}\ \text{J}[/tex]
Explanation:
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
M = Mass of Sun = [tex]1.989\times 10^{30}\ \text{kg}[/tex]
m = Mass of Earth = [tex]5.972\times 10^{24}\ \text{kg}[/tex]
[tex]r_i[/tex] = Initial radius of orbit = [tex]1.5\times 10^{11}\ \text{m}[/tex]
[tex]r_f[/tex] = Final radius of orbit = [tex]((1.5\times 10^{11})+2\times 10^3)\ \text{m}[/tex]
Energy required is given by
[tex]E=\dfrac{1}{2}\Delta U\\\Rightarrow E=\dfrac{GMm}{2}(\dfrac{1}{r_i}-\dfrac{1}{r_f})\\\Rightarrow E=\dfrac{6.674\times 10^{-11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{2}(\dfrac{1}{1.5\times 10^{11}}-\dfrac{1}{(1.5\times 10^{11})+2\times 10^3})\\\Rightarrow E=3.52\times 10^{25}\ \text{J}[/tex]
The energy required would be [tex]3.52\times 10^{25}\ \text{J}[/tex].
How fast were both runners traveling after 4 seconds?
40
Distance (in yards)
30
20
10
1
2.
3
0
Time in seconds
Answer:
they were fast ⛷⛷
In high air pressure the molecules are
A-Warm and moving fast
b-Close together and moving slowly
c-far apart and moving slowly
d-hot and moving rapidly
