A heat engine cannot have a thermal efficiency of 100% For all reversible processes, the second-law efficiency is 100%. The second-law efficiency of a heat engine cannot be greater than its thermal efficiency. The second-law efficiency of a process is 100% if no entropy is generated during that process
Imagine you have a ball tied to the end of a string. You hold the other end of the string and swing it around. Suppose the string breaks, what direction will the ball travel
Block 1, of mass m1, moves across a frictionless surface with speed ui. It collides elastically with block 2, of mass m2, which is at rest (vi=0). (Figure 1)After the collision, block 1 moves with speed uf, while block 2 moves with speed vf. Assume that m1>m2, so that after the collision, the two objects move off in the direction of the first object before the collision. What is the final speed vf of block 2?
The conservation of the momentum allows to find the velocity of the second body after the elastic collision is:
[tex]v_f = \frac{2u_o}{1- \frac{m_2}{m_1} }[/tex]
the momentum is defined by the product of the mass and the velocity of the body.
p = mv
The bold letters indicate vectors, p is the moment, m the mass and v the velocity of the body.
If the system is isolated, the forces during the collision are internal and the it is conserved. Let's find the momentum is two instants.
Initial instant. Before crash.
p₀ = m₁ u₀ + 0
Final moment. After crash.
[tex]p_f = m_1 u_f + m_2 v_f[/tex]
The momentum is preserved.
p₀ = [tex]p_f[/tex]
[tex]m_1 u_o = m_1 u_f + m_2 v_f[/tex]
Since the collision is elastic, the kinetic energy is conserved.
K₀ = [tex]K_f[/tex]
½ m₁ u₀² = ½ m₁ [tex]u_f^2[/tex] + ½ m₂ [tex]v_f^2[/tex]
Let's write our system of equations.
[tex]m_1 u_o = m_1 u_f + m_2 v_f \\m_1 u_o^2 = m_1 u_f^2 + m_2 v_f^2[/tex]
Let's solve
[tex]u_f = u_o - \frac{m_2}{m_2} \ v_f \\u_f^2 = u_o^2 - \frac{m_2}{m_1} \ v_f^2[/tex]
[tex]( u_o - \frac{m_2}{m_1} v_f)^2 = u_o - \frac{m_2}{m_1} \ v_f^2 \\u_o^2 - 2 \frac{m_2}{m_1} \ u_o v_f + (\frac{m_2}{m_1} )^2 v_f^2 = u_o^2 - \frac{m_2 }{m_1} \ v_f^2[/tex]
[tex]2 \frac{m_2}{m_1} \ u_o = \frac{m_2}{m_1} v_f \ ( 1 - \frac{m_2}{m_1}) \\v_f = \frac{2u_o}{1-\frac{m_2}{m_1} }[/tex]
In conclusion, using the conservation of momentum, we can find the velocity of the second body after the elastic collision is:
[tex]v_f = \frac{2u_o}{1-\frac{m_2}{m_1} }[/tex]
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change 176°F to the Celsius and Kelvin scales. pura batana kese hua fir answer dena plzz
Formulas change from F to degree C : C = (F – 32) / 1.8
so we have (176-32) /1.8= 80 oC
Formulas change from F to degree K : K = 5 / 9 x (F – 32) + 273.15
so we have 5/9 x (176-32) + 273.15 = 353 oK
ok done. Thank to me :>
The object in this diagram has a mass of 2 kg.
According to this diagram, what is the acceleration of this object?
A. 0 m/s2
B. 4 m/s2
C. 7 m/s2
D. 11 m/s2
Answer:
[tex]4 ms^{-2}[/tex]
Explanation:
Normal force - should be the reaction from the surface it's on - nullifies the weight of the object. At this point the only force is of [tex]22-14 = 8N[/tex] from whoever is pushing the square and the friction, towards the right. At this point we divide the force by the mass of the object to obtain an acceleration of [tex]8/2 = 4 ms^{-2}[/tex]
How much work must be done to stop a 925-kg car traveling at 95 km/h?
The amount of work done to stop the car is equal to 322,099.85 Joules.
Given the following data:
Mass = 925 kilogramsVelocity = 95 km/hConversion:
Velocity = 95 km/h = [tex]\frac{95 \times 1000}{60 \times 60} = 26.39 \;m/s^2[/tex]
To determine the amount of work done to stop the car, we would apply the work-energy theorem:
Note: The amount of work done must balance the kinetic energy possessed by the car due to its motion.
From the work-energy theorem, we have:
[tex]Work\done = \Delta K.E\\\\Work\done = \frac{1}{2} MV^2[/tex]
Substituting the given parameters into the formula, we have;
[tex]Work\;done = \frac{1}{2} \times 925 \times (26.39)^2\\\\Work\;done =462.5 \times 696.4321[/tex]
Work done = 322,099.85 Joules
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how much work is done when a 30 kg mass is to be lifted through a height 6m?(1kg=9.8N
we know 1kg=9.8N so 30 kg= 30 x 9.8 = 294 N
work is done when a 30 kg mass is to be lifted through a height 6m :
A = 294 x 6 = 1764 J
ok done. Thank to me :>
A box of mass 20 kg has a base of area 4 m2
. Calculate the pressure exerted by it on the ground. (Use 1
kg-f = 10 N)
[tex]\huge \bf༆ Answer ༄[/tex]
First, find the force exerted by the box.
[tex] \sf \: F = ma [/tex]
The required terms are :
m = mass = 20 kga = Acceleration due to gravity = 10 m/s²Calculate the force : -
[tex] \sf \: F = 20 \times 10 [/tex][tex] \sf \: F = 200 \: N[/tex]_____________________________________
Pressure can be calculated using this formula : -
[tex] \sf \: P = \dfrac {F }{A }[/tex]The stated terms are :
P = pressure F = force = 200 NA = Area = 4 m²Calculate the Pressure now : -
[tex] \sf \: P = \dfrac{200}{4} [/tex][tex] \sf \: P = 50 \: pascals[/tex][tex]꧁ \: \large \frak{Eternal \: Being } \: ꧂[/tex]
How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
it is 1 / square root of the change in distance
Explanation:
since the equation of the gravitational force is
[tex]F=G{\frac{m_1m_2}{r^2}}[/tex]
then F and r are inversely related, by the equation
[tex]F \alpha \frac{1}{r^2}[/tex]
this means when F is multiplied by 2 then r^2 is multiplied by half
so when r is multiples by half we first square the number (0.5) ^ 2 = 0.25
so
[tex]F \alpha \frac{1}{(0.25)r^2}[/tex]
which means F is multiples by 1/0.25 or 4 when the radius is halfed
What time is the lunar eclipse tonight in california?
Find the work done while lifting a 115 kg sofa 1.5 m off the ground.
Find the weight first
P= 10m = 10 x 115 = 1150 N
work done while lifting a 115 kg sofa 1.5 m off the ground :
A= Ph = 1150 x 1.5 = 1725 J
P/s: I think the weight first is 115kg
ok done. Thank to me :>
2. Given what you know about the acceleration of Earth's gravity (g = 9.8 m/s2), is this number accurate?
accurate. If not explain why you think it is not accurate. Pleaseee help mee
Answer:
it is correct
Explanation:
Though no rounded numbers can be defined as accurate, if we were going by people's discovery, and research, we can define that the number, g = 9.8m/s^2, is accurate
A small sphere 0.70 times as dense as water is dropped from a height of 8 m above the surface of a smooth lake. Determine the maximum depth to which the sphere will sink. Neglect any energy transferred to the water during impact and sinking.
Answer:
18.66m
Explanation:
This was actually fun! Let set up the problem first: As you drop the sphere, it will accelerate till it hits the water with a given speed [tex]\dot z[/tex]. Once sinking the ball is subjected to two forces: its own weight [tex]m_sg = \rho_sVg[/tex] directed towards the bottom of the lake, and buoyancy (Archimedes law), ie a force upward equal to the volume of displaced water, [tex]m_wg=\rho_wVg[/tex]. At this point it's the classical "how high can I toss a ball before it falls down" with a water twist. Please note that I'm using z as the height of the sphere, and [tex]\dot z; \ddot z[/tex] represent the velocity (with one dot) and the acceleration (with two dots) in the z direction .Let's assume the sign of all quantities being positive if directed upwards, and negative if towards the bottom of the lake.
Let's first determine how fast the ball hit the water. For me the easiest way is saying "at 8m it has a given potential energy, and 0 kinetic energy. When it hits the water it loses all potential energy and has only kinetic energy". In numbers:
[tex]m_sgz=\frac12m_s\dot z_0^2[/tex] Let's divide by the masses, g is a known value ([tex]9.81 ms^{-2}[/tex]), z is 8 meters, and we get that [tex]\dot z_0 = 4\sqrt g =12.52 m/s[/tex]
At this point, let's determine the force acting on the sphere, courtesy of Newton second law and the debate we had earlier:
[tex]m_s \ddot z = -m_sg + m_wg\\\rho_sV \ddot z = -\rho_sVg + \rho_wV g[/tex]
At this point we can divide by the volume of the sphere, and make use of the fact that [tex]\rho_s = 0.7 \rho_w[/tex]
[tex]0.7\rho_w \ddot z = -.7\rho_w g+ \rho_wg \rightarrow 0.7 \ddot z = 0.3 g\\\ddot z = \frac {0.3}{0.7} g= \frac 37g = 4.2 ms^-2[/tex]
We're almost there. At this point we can write the law of motion for the sphere.
[tex]\dot z = \dot z_0 + \ddot z t[/tex] that we will use to find the time for the ball to stop sinking - ie reaches 0 velocity.
[tex]0 = -12.52 +4.2t \rightarrow t= 12.52/4.2 \approx 3s[/tex]
At this point we can use the other law of motion to find out the distance traveled
[tex]z= z_0 + \dot z_ot + \frac12 \ddot z t^2\\z= 0 -12.52 (3) + \frac12 (4.2)(3)^2 = -37.56+18.90 = -18.66 m[/tex]
as usual, please double and triple check all the calculations, it's almost 1.30 am here and I am not the most confident number cruncher even when fully awake.
where would you look for the youngest stars in the milky way galaxy?
Answer:
at the center of the galaxy
Explanation:
as there r more new born stars found
Kiley went 5.7 km/h north and then went 5.8 km/h west. From start to finish, she went 8.1 km/h northwest.
Which best describes her numbers?
Answer:
5.7km/h north and 5.8km/h west are instantaneous speeds, while 8.1km/h Northwest is the average speed.
for what angle of projetion is the range is maximum
Answer:
46 degrees
Explanation:
How are fission and fusion similar?
A. A small fraction of the reactant mass is converted into small amounts of energy.
B. A large fraction of the reactant mass is converted into small amounts of energy.
C. A large fraction of the reactant mass is converted into large amounts of energy.
D. A small fraction of the reactant mass is converted to large amounts of energy.
Answer:
C
Explanation:
Fusion and fission are similar in that they both release large amounts of energy. Nuclear fusion is a process in which two nuclei join to form a larger nucleus. Nuclear fission is a process in which a nucleus splits into two smaller nuclei.
what is the main difference between a substance going through a physical change and one going through a chemical ?
Answer: Physical changes only change the appearance of a substance, not its chemical composition. Chemical changes cause a substance to change into an entirely substance with a new chemical formula. Chemical changes are also known as chemical reactions.
Answer:
just give the other person brainlyest
Explanation:
helpppp i am not smart
Answer:
it might be b maybe
Explanation:
How is the input energy carried to the light bulb's energy stores?
Answer:
the electric energy travels through the plug or wire or whatever is connecting the energy source to the lightbulb, which conducts electricity, and to the light bulb. the electric energy is transfered into light energy and thermal energy
Explanation:
i hope i understood this question and answered it correctly XD
You use a rope 2.00 m long to swing a 10 kg weight around your head. The tension in the rope is 20 N. In half a revolution how much work in J is done by the rope on the weight
Answer:
No work is done on the weight. :3
To solve this we must be knowing each and every concept related to work and its calculation. Therefore, 250J is the work done in joule by the rope on the weight.
What is work?Work in physics is the energy delivered to or out of an item by applying force across a displacement. It is frequently expressed in its most basic form as the combination of displacement and force.
When a force is applied, it is said to produce positive work if it has a proportion in the orientation of the movement of a site of application. Work done is positive when the direction of force acting on the object and displacement of the object both are in the same direction.
Mathematically,
Work= force ×displacement
force =20 N
displacement=2.00 m
displacement=2×3.14×2=12.5m
substituting all the given values in the above equation, we get
Work= 20 ×12.5
Work =250J
Therefore, 250J is the work done in joule by the rope on the weight.
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LED lightbulbs are the most energy efficient way to light your home, as they use _____ of the energy of a comparable compact fluorescent bulb.
Light - emitting diode lightbulbs known as LED lightbulbs are energy saving lightbulbs. They consume very small amount of energy compare to other lightbulbs. Therefore, LED lightbulbs are the most energy efficient way to light your home, as they use about ten percent (10%) of the energy of a comparable compact fluorescent bulb.
There are varieties of different light bulbs. some of them are incandescent, compact fluorescent (CFL), halogen, and light-emitting diode (LED) to mention a few.
Before the advent of light - emitting diode (LED) bulbs, some of the bulbs consume a lot of energy. For instance, we have incandescent bulbs of 20 watts, 50 watts, 100 watts and even 200 watts.
But with the advent of light - emitting diode (LED) bulbs, consumers can save energy because light - emitting diode (LED) bulbs save a lot of energy and they are very bright.
We can therefore conclude that light - emitting diode lightbulbs known as LED lightbulbs are the most energy efficient way to light your home, as they use about ten percent (10%) of the energy of a comparable compact fluorescent bulb.
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How do you solve problems involving acceleration
Answer:
Velocity/time =acceleration
Acceleration x time = velocity
Velocity/acceleration=time
choose all the answers that apply.
Major areas of science include
____.
A.basic science
B.physical science
C.life science
D.social science
E. ethical science
Answer:
physical science
Explanation:
i guess??
The initial temperature of the mixture was +20 °C. The mixture froze at -1.5 °C.
A total of 165 kJ of internal energy was transferred from the mixture to cool and
freeze it.
specific heat capacity of the mixture = 3500 J/kg °C
specific latent heat of fusion of the mixture = 255 000 J/kg
Calculate the mass of the mixture.
Give your answer to 2 significant figures.
[6
M
Answe16000/3500(-1.5-2-)225000= 1.1
Explanation:
The mass of the mixture will be 2.19 ×10⁻³ kg. The unit of mass is kg and it is a scalar quantity.
What is mass?Mass is a numerical measure of inertia, which is a basic feature of all matter. It is, in effect, a body of matter's resistance to a change in speed or position caused by the application of a force.
In the International System of Units (SI), the kilogram is the unit of mass.
The given data in the problem is;
The initial temperature of the mixture was, [tex]\rm T_i = 20 ^0 \ C[/tex]
The mixture froze at the temperature of, [tex]\rm T_F = -1.5 ^0 \ C[/tex]
The internal energy is,[tex]\rm E = \rm 165 \ kJ[/tex]
The specific heat capacity of the mixture is,[tex]\rm C = 3500 J/kg ^0C[/tex]
The specific latent heat of fusion of the mixture is,[tex]\rm l = 255 000 J/kg[/tex]
m is the mass of the mixture,
The latent heat is equal to the heat transfer. Because the net work done is zero.
The formula for the heat transfer is found as;
[tex]\rm Q = mCdt \\\\ m = \frac{Q}{Cdt} \\\\ m = \frac{165 }{3500(20-(-1.5)} \\\\ m = 2.19 \times 10^{-3} \ kg[/tex]
Hence the mass of the mixture will be 2.19 ×10⁻³ kg.
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An element with a force of 0.00SN moves at a minimum: 0.2m / s². Calculate the mass of that element.
Please find attached photograph for your answer.
Hope it helps.
Do comment if you have any query.
A ball of mass 2.0 kg is travelling at a speed of 12 m/s. It moves towards an object of mass 3.0 kg which is at rest.
The ball hits the object and sticks to it.
Which row gives the total momentum, and the speed of both objects immediately after the collision?
total momentum (kg m/s)
A- 0
B- 0
С- 24
D- 24
AND..
speed (m/s)
A- 4.8
B- 8.0
C- 4.8
D- 8.0
Please find attached photograph for your answer.
Hope it helps.
Do comment if you have any query.
The velocity for the entire trip is 0.4 m/s as It takes her 500 seconds to make the round trip and 60 kg • m/s2 = (45 kg • v) + (20 kg • 3 m/s).
Path 1 = 400 m В B. A Path 2 = 200 m Path 3 = 300 m. Thus, option C is correct.
What is velocity?A particle's settling velocity known as the rate at which is travels through a still fluid. The specific gravity of the particles, their size, and their shape all have an impact on settling velocity.
A particle in still air will gravitationally settle and reach its terminal velocity fairly quickly. A particle's terminal velocity in a still fluid is referred to as the settling velocity (also known as the "sedimentation velocity").
Understanding variations in the hydraulic regime and interactions between sediment and fluid in the surf zone depends heavily on the particle settling velocity at the foreshore region. In contrast to sedimentation, which is the end product of the settling process, settling is the movement of suspended particles through the liquid.
Therefore, Thus, option C is correct.
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A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller diagram below.
1) Find the total energy for the roller coaster at the initial point.
2) Find the potential energy at point A using the PE formula.
3) Use the conservation of energy to find the kinetic energy (KE) at point B.
4) Find the potential energy at point C.
5) Use the conservation of energy to find the Kinetic Energy (KE) of the roller coaster at point C.
6) Use the Kinetic Energy from C, find velocity of the roller coaster at point C.
For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:
1) The total energy for the roller coaster at the initial point is 19820 J
2) The potential energy at point A is 19620 J
3) The kinetic energy at point B is 10010 J
4) The potential energy at point C is zero
5) The kinetic energy at point C is 19820 J
6) The velocity of the roller coaster at point C is 19.91 m/s
1) The total energy for the roller coaster at the initial point can be found as follows:
[tex] E_{t} = KE_{i} + PE_{i} [/tex]
Where:
KE: is the kinetic energy = (1/2)mv₀²
m: is the mass of the roller coaster = 100 kg
v₀: is the initial velocity = 2 m/s
PE: is the potential energy = mgh
g: is the acceleration due to gravity = 9.81 m/s²
h: is the height = 20 m
The total energy is:
[tex] E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J [/tex]
Hence, the total energy for the roller coaster at the initial point is 19820 J.
2) The potential energy at point A is:
[tex] PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J [/tex]
Then, the potential energy at point A is 19620 J.
3) The kinetic energy at point B is the following:
[tex] KE_{A} + PE_{A} = KE_{B} + PE_{B} [/tex]
[tex] KE_{B} = KE_{A} + PE_{A} - PE_{B} [/tex]
Since
[tex] KE_{A} + PE_{A} = KE_{i} + PE_{i} [/tex]
we have:
[tex] KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J [/tex]
Hence, the kinetic energy at point B is 10010 J.
4) The potential energy at point C is zero because h = 0 meters.
[tex] PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J [/tex]
5) The kinetic energy of the roller coaster at point C is:
[tex] KE_{i} + PE_{i} = KE_{C} + PE_{C} [/tex]
[tex] KE_{C} = KE_{i} + PE_{i} = 19820 J [/tex]
Therefore, the kinetic energy at point C is 19820 J.
6) The velocity of the roller coaster at point C is given by:
[tex] KE_{C} = \frac{1}{2}mv_{C}^{2} [/tex]
[tex] v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s [/tex]
Hence, the velocity of the roller coaster at point C is 19.91 m/s.
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2hrt 20 min +60 sec
Answer:
2hr 21 min
Explanation:
0 days 2 hours 20 minutes 0 seconds
+ 0 days 0 hours 0 minutes 60 seconds
= 0 days 2 hours 21 minutes 0 seconds
How much potential energy did the the milkshake have after it fell off of the counter, just before it hit the ground (at the bottom)? Make sure to write the correct units with
your answer!
Answer:
The formula for potential energy depends on the force acting on the two objects. For the gravitational force the formula is P.E. = mgh, where m is the mass in kilograms, g is the acceleration due to gravity (9.8 m / s2 at the surface of the earth) and h is the height in meters.
help me solve please
Answer:
please check the image
Explanation:
I hope this image helps you please follow the steps. Thank you
Answer:
1kg/```````2
Explanation:
solution