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laminar FLOW of a liquid down an inclined plane, the surface VELOCITY is found to be 30 cm/s. The average velocity of the flow, in cm/s I hope this will help youIf not then comment me

re two TYPES of transformer SET up Transformer and step down Transformer 1 Transformer is used to list the DC voltage and one of the Transformer is used to INCREASE the DC voltageI HOPE this will HELP youIf not then comment me

il of DENSITY 900 kg/m3 and viscosity 50 mN s/m2. Solution A glass sphere, of diameter 6 mm and density 2600 kg/m3, falls through a layer of oil of The VOID fraction I HOPE this will help YOUIF not then comment me

sis. When a ferromagnetic material is magnetized in one direction, it will not RELAX back to zero MAGNETIZATION when the IMPOSED magnetizing field is removed. If an alternating magnetic field is applied to the material, its magnetization will trace out a loop called a HYSTERESIS loop.I HOPE this will help youIf not then comment me

crease in momentum100% KE (normal) + 300% KE (increase) = 400% KE or 4xKEKE = (1/2)*m*v^2define initial and final states...4X KE1 = KE2 4x (1/2)*m1*v1^2 = (1/2)*m2*v2^2cancel out (1/2) factormass doesn't change, m1 = m2 = m..... cancel out m 4v1^2 = v2^2 or (2v1)^2 = v2^22*v1 = v2momentum is P = mvm*v1 + Pchange = m*v2 = m* (2v1)Pchange = 2*m*v1 - m*v1 = m*v1m* v1 is 100% of m*v1momentum is increased by 100%I hope this will HELP youIf not then comment me

netic field does no WORK, so the kinetic ENERGY and speed of a charged particle in a magnetic field REMAIN constant. The magnetic force, acting perpendicular to the velocity of the particle, will cause circular motion.I hope this will help YOUIF not then comment me

ntage of flooded evaporators is that the potential problem of poor refrigerant distribution in the evaporator is reduced. The refrigerant is 100% liquid, and a liquid stream is much better distributed between the channels compared with the two-phase mixture of DX systems. Thus, when SELECTING a flooded evaporator, a SWEP B-model should be used for flooded evaporators, in contrast to DX systems where BPHEs with refrigerant distribution device are the better choice.The receiver separates the refrigerant vapor before feeding it to the compressor (see Figure 6.46), so there is no need for superheating the refrigerant in a flooded evaporator. A larger portion of the total heat surface area will thus be used for evaporation compared with a DX evaporator, where 10-30% of the total heat surface area may be dedicated to superheating.Figure 6.46 compares a flooded system and a DX system. The greenline detour (b-c) to the liquid saturation line (bubble line) shows the phase separation in the flooded system receiver. Because of the pressure gain (c-d) in the pipe that connects the receiver and the flooded evaporator, the evaporator inlet liquid is sub-cooled. Please note that the (c-d) line in Figure 6.46 is exaggerated. The pressure gain in the receiver-evaporator pipe is actually approximately 5-50 kPa (0.05-0.5 bar), while the pressure lift over the compressor is roughly 12-17 bar. Because there is no need for superheating in a flooded evaporator, the evaporation temperature can be a few DEGREES higher than in a DX evaporator (see Figure 6.47).Energy transfer is much more efficient through a boiling turbulent liquid film than through dry superheated vapor. As a consequence, the temperature program is "closer" for a non-superheated evaporator than for a superheated evaporator (see Figure 6.47). A "closer" temperature program means a smaller DIFFERENCE between the leaving secondary fluid temperature and the evaporation temperature (LWT-Tevap).Due to the higher evaporation temperature in a flooded evaporator, the pressure lift between the evaporator and condenser sides is smaller. The advantage is that less compressor WORK (W) is needed. Perhaps the largest advantage of flooded evaporators is that they use all the LATENT energy of the refrigerant in the phase transition between liquid and gas to cool a fluid. This is shown in Figure 6.46, where line (d-e) stretches over the whole transition length, while the red line (g-h) does not utilize the entire phase transition length. In other words, the COP (coefficient of performance) is higher for a flooded system.I hope this will help youIf not then comment me

u set up a solid mechanics problem the first step is usually the CHOICE of a suitable coordinate system. This is typically some sort of right handed cartesian coordinate system aligned with some convenient part geometry. You then use this coordinate system without having to think about it again to calculate shear and normal stresses. The problem with this though is that you tend to forget that your choice of coordinate system in step one was completely arbitrary. Your calculation of these stresses is however dependent on this coordinate systems. These axes are more than likely however not aligned with the loading of the problem and as such there probably exists another choice of coordinate system where stresses are higher than in the system you chose.Naively you could then go about trial and error choosing coordinate systems until the stresses where maximum but that isn't the greatest idea. Instead because we KNOW that the stress is a tensor we can simply find its eigenvalues and eigenvector. This is then the maximum values for the stresses and the direction of the coordinate system in which this is the case RESPECTIVELY. If your MATRIX manipulation isn't so good you can instead choose to draw the stress tensor on a mohr’s circle and rotate it geometrically instead. This works well for 2D, but gets clumsy but though doable in 3D. Doesn't really matter what method you use to get there or the dimensionality of the problem all you are essentially doing is rotating an ARBITRARILY chosen coordinate system to ensure that it is aligned with the applied stresses. This then maximizes the individual components and allows you to check whether the material properties are adequate in the direction of the maximum stresses.I hope this will help youIf not then comment me

sually observed that the fluorescent light from MOLECULES or nanoparticles has a longer WAVELENGTH than the exciting light. FLUORESCENCE occurs very quickly after excitation. There is another radative decay mode, called phosphoresence, which takes the system back to ground state over much longer timescales.I hope this will help youIf not then COMMENT me

est MONSOON winds break into two BRANCHES DUE to topographic factors. When the South West Monsoon wind HITS the Western Ghats it branches into two parts, the Arabian SEA branch and the Bay of Bengal BranchI hope this will help youIf not then comment me

hermal energy conversion (OTEC) uses the temperature difference between cooler deep and warmer shallow or surface seawaters to run a heat engine and produce useful work, usually in the form of electricity. OTEC can operate with a very high capacity factor and so can operate in base load mode.Among ocean energy sources, OTEC is one of the continuously available renewable energy resources that could contribute to base-load power supply. The resource potential for OTEC is considered to be much larger than for other ocean energy forms [World Energy Council, 2000]. Up to 88,000 TWh/yr of power could be generated from OTEC without affecting the ocean’s thermal structure [Pelc and Fujita, 2002].Systems may be either closed-cycle or open-cycle. Closed-cycle OTEC uses working fluids that are typically thought of as refrigerants such as ammonia or R-134a. These fluids have low boiling points, and are therefore suitable for powering the system’s generator to generate electricity. The most commonly used heat cycle for OTEC to date is the Rankine cycle, using a low-pressure turbine. Open-cycle engines use vapour from the seawater itself as the working fluid.OTEC can also supply quantities of cold WATER as a by-product. This can be used for air conditioning and refrigeration and the nutrient-rich deep ocean water can feed biological technologies. Another by-product is FRESH water distilled from the sea.OTEC THEORY was first developed in the 1880s and the first BENCH size demonstration model was constructed in 1926. Currently the world's only operating OTEC plant is in Japan, overseen by Saga University.I hope this will help youIf not then comment me

eesangle of deviation = δ = i + e - A    i = 0, hence r1 = 0.      Hence R2 = 60  deg  (angle of incidence of SECOND face)    r2 > critical angle for glass-air = SIN ⁻¹ 1/1.5 = 41.81 deg.    It will be totally internally reflected and fall on the BASE of prism at 90 deg to it.  It will emerge with angle of emergence e = 0 deg.       δ = 0 + 0 - 60 deg  =  60 deg.========================= angle of incidence on first face is 0 deg.  Hence angle of refraction is 0 on the first face.Angle of incidence is 60 degrees on the second face.angle of refraction =  r  =>    Sin i / Sin r  =  1 / 1.5    sin r =  1.5 * sin 60  = 1.299    > 1  It means that the light RAY is incident at an angle more than the critical angle.  The light ray is totally internally reflected back in to the prism from the second face.  The angle of reflection will be 60 deg.  This ray will be incident normally on the base face of the prism.  Hence it will  emerge normally perpendicular to the base of the prism.  The deviation of the light ray due to the prism is  then  60 degrees.===================================I hope this will help youIf not then comment me

vitational POTENTIAL in a region is given by, V = 20(x + y) J/kg. Find the magnitude of the gravitational force on a particle of MASS 0.5 kg PLACED at the origin. I HOPE this will help youIf not then comment me

ere!Unit is a quantity of CONSTANT MAGNITUDE which is USED to measure the magnitudes of other QUANTITIES of same nature.Hope IT HELPS You!

is the resistance, of any physical object, to any change in its velocity. This includes changes to the object's speed, or DIRECTION of motion. I HOPE THIS HELPS Umark me as BRAINLIEST

.inWhat is your question?Secondary SchoolMath 5+3 ptsA train covered a certain distance at a uniform speed. if the speed of the train had been 6 km/hr faster, it WOULD have TAKEN 4 hour less time than the scheduled time and if the speed of the train were SLOWER by 6 km/hr, the train would have taken 6 hr more than the scheduled time. the length of the journey was:Report by Msd11111 22.02.2018Answersshikharrai114p1ekhpShikharrai114p1ekhp ExpertLet the speed of the train by x km/hrAnd the time taken by the train be ‘y’ hrsDistance traveled by the train is x.y = xyCase IIF the train would have been 6 km/hr faster, it would have taken 4 hours less than the scheduled time.⇒ (x + 6) (y - 4) = xy⇒ -4x + 6y = 24⇒ 2x – 3y = -12 ---------------- (1)CASE IIIf the train were slower by 6 km/hr, it would have taken 6 hours more than the scheduled time⇒ (x - 6) (y + 6) = xy⇒ 6x – 6y = 36⇒ x – y = 12 ----------------(2)Solve it simultaneously and get the values of x and y.Hope u understand it.!!Mark Braniliest!!

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m3o.65/1000*1000kg/liter 0.00000065kg/lit

ve MIRROR can FORM a real and dimished IMAGE. The object can be placed at any position (INFINITY, beyond C ) but not between the Focus and POLE because at that time a virtual image is formed.

celeration DUE to gravity×heighti.e mass×9.8×height

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Molecules of GAS are always MOVING in random MOTION. Speed is owned by every molecules through which it REMAINS in motion with respect to other molecules of the gas.Read More in the ATTACHMENT.

O THE ATTACHMENT.

ATEE UR ANS❤❤

Oxygen as fuel to REACT with HYDROGEN can be stored more efficiently in LIQUID form as it is far more dense than in a GASEOUS form. In addition to this the oxygen will react with the hydrogen fuel and upon releasing energy as thrust will also generate some water (Two hydrogen, one oxygen) in which the astronauts can (OBVIOUSLY) drink later.I hope this help you.please mark as brainliest

n example of CLASS 2 LEVER where the load is between the FORCE and the fulcrum.

of measurement EQUAL to 149.6 million KILOMETRES, the MEAN DISTANCE from the centre of the earth to the centre of the sun.

-AXIAL rays BECOME parallel after refraction ... so IMAGE is formed at infinityv=∞μ1=1.33u=?μ2=1.48R= 30cmFrom the formula:μ2/v - μ1/ u= μ2-μ1/R=1.48 /∞  - 1.33/u  = 1.48-1.33/30- 1.33/u- 0.15/30u= - 266.0cmObject should be PLACED at a DISTANCE of 266cm from surface on side A

..!!!+222222..MM YE dekh jiss MAI DOUBT likha h

ergy required for 5g of water at 0∘C to get CONVERTED to water at 100∘C is latent heat required + heat required to change its temperature by 100∘C = (80⋅5)+(5⋅1⋅100)=900 calories.Now,heat liberated by 5g of STEAM at 100∘C to get converted to water at 100∘C is 5⋅537=2685 caloriesSo,heat energy is enough for 5g of ice to get converted to 5g of water at 100∘C So,only 900 calories of heat energy will be liberated by steam,so amount of steam that will be converted to water at the same temperature is 900537=1.66g So,the final temperature of the mixture will be 100∘C in which 5−1.66=3.34g of steam and 5+1.66=6.66g of water will coexist.I hope this will HELP youIf not then comment me

ence of excitation GUARANTEES convergence without a priori stability assumptions and ensures ROBUSTNESS properties.I hope this will HELP YOUIF not then comment me

itive side of a battery is only "positive" in relation to the "NEGATIVE" terminal of the same battery. When you hook a WIRE from the positive terminal of the first battery to the negative terminal of the second, a very small AMOUNT of current will flow until the potential difference reaches zero.I hope this will help YOUIF not then COMMENT me

that AMOEBA is GOING to DIVIDE into SON or DAUGHTER Amoeba.

you UNDERSTAND

your ANSWER

our ANSWER....COAL gas was USED for street lighying in 1836....THANKS,,,,,

ing current through that.If the light GLOWS than its WORKING otherwise not.Hope my ANS will HELP you..

10cmf=20cmBy mirror formula1/-f=1/-u+1/-v1/-20=1/-10+1/-v1/v=1/-10+1/201/v=-2+1/201/v=-1/20v=-20The IMAGE will form 20 CM BEHIND the mirror