If a block slides down a 30 degree incline at a constant speed, what is the coefficient of dynamic friction between the block and the incline?
θ = 30 degreesvelocity is constant therefore acceleration a = 0Please refer to my analysis using the diagram below.The coefficient of dynamic friction is equal to 0.5773.
How much heat energy is required to change 2 kg of ice at 0°C to water at 20°C?
To make your concept clear about such type of questions, you need to understand the concept of latent heat and sensible heat.Latent Heat: The heat required to convert a solid into a liquid or vapour (latent heat of fusion), or a liquid into a vapour(latent heat of evaporation), without change of temperature.Sensible Heat: The heat required to change the temperature of a body without changing its phase.So, basically as your question suggests, you need to evaluate the heat required to convert 2 kg of ice at 0 °C to water at 20 °C.. the total heat content will comprise of:Total heat = Heat required to convert 2 kg of ice to 2 kg of water at 0 °C + Heat required to convert 2 kg of water at 0 °C to 2 kg of water at 20 °C.[math]Heat = m hfg + m Cp ΔT[/math]Here, m ( mass of ice) = 2 kghfg (latent heat of fusion of ice) = 334 KJCp of water (specific heat) = 4.187 KJ/Kg-KΔT(Temperature difference) = 20 °CTherefore, Heat required = 2 x 334 + 2 x 4.187 x (20 - 0 )Heat reqd= 835.48 KJTherefore, to melt 2 kg of ice 835.48 KJ of heat is required.Happy Reading!
How long does it take for ice to melt at average room temperature?
This depends on several factors:The temperature of the ice (it could be well below 0 degrees C)The mass, shape, and volume of the iceHow much room-temperature air is available to transfer heat into the iceThe rate at which the room-temperature air is able to transfer heat into the ice (depends on surface area of contact, ventilation)An extremely cold (perhaps -40 degrees C) spherical ball of solid ice about 1 meter across, if placed into an unventilated very small room at room temperature (about 20 degrees C), will take a very, very long time to melt. It could last several days, perhaps even up to a week. If the room is tiny enough, the ball of ice will chill the air in the room, so that it’s no longer at room temperature. (It will, effectively, turn the room into a large ice-box refrigerator.) If the room is very, very large (like a gymnasium with full-size basketball court and bleachers), then it won’t cool down noticeably, and the ball will melt faster, but may still last a few days to several days.If, however, the same mass of ice is already very near the melting point (perhaps -1 degrees C), and instead of being physically configured as a compact sphere, it’s configured as a large number of very thin sheets (each sheet being 10 meters by 10 meters by 2 millimeters thick), and ventilation fans are used to blow room-temperature gymnasium air over and between these sheets of ice, they will melt very, very quickly — perhaps being completely melted in less than an hour or as little as 20 or 30 minutes. The very high surface area, combined with the movement of air (for efficient heat transfer from the air to the ice), greatly accelerates the melting. Starting the ice at -1 degrees C instead of at -40 degrees C also gives it a head-start for melting.
How many joules of heat are needed to increase the temperature of 2 kg of copper by 10°C?
Ignoring losses and assuming that the temperature change is occurring near 20C.The specific heat of copper is 385 J/kg/C so 385 x 2 kg x 10C = 7.7 kJ.
A block of mass 3kg is placed on a rough surface. The coefficient of static friction between two surfaces is 0.2 then what is the minimum horizontal force required to move the block?
As stated, there is not sufficient information to answer.What is the acceleration due to gravity whereas the experiment is performed?If you can assume 1 Earth standard gravity, the the block is pressed down onto the surface by a force of 9.8 * 3 N. The coefficient of friction presents a retarding force of 0.2 * 29.4, or 5.88N
If a block of mass (m) is kept on a horizontal surface with friction coefficient [math]\mu[/math], then what would be the minimum force needed to move that block on the surface?
This is actually a tricky one since many people will end up gettingminimum force as [math]f = \mu.mg [/math]However Just read the question again , it says minimum force (There is no mention of minimum horizontal force)Now as we know Friction [math]force \sim Normal[/math]i.e. : if we minimize the normal reaction we can minimize the friction at the same time we also need to provide enough Horizontal force to ensure block gets a horizontal velocitySomething like shown in figure !Now the task is to find optimal angle “x” and Magnitude FApplying simple physics : (Assuming Normal reaction from surface is N)Balancing Vertical force[math]Fsinx+N = mg [/math] —>(1)[math]N = mg - Fsinx[/math]Balancing Horizontal force[math]Fcosx = Friction[/math]=> [math]Fcosx = \mu.N[/math]=> [math]Fcosx = \mu.(mg-Fsinx) [/math]=> [math]F(cosx+\mu.sinx) = \mu.mg [/math]=> [math]F = \frac{\mu.mg}{cosx+\mu.sinx} [/math] ==>(2)Now to minimize F(2) we need to maximize denominator of 2[math]max[/math]{[math]cosx+\mu.sinx[/math]}[math] = \sqrt{1+(\mu)^2} [/math] Basic TrigoSo minimum force needed would be[math]F = \frac{\mu.mg}{\sqrt{1+(\mu)^2}}[/math]
A 5kg block of iron with a specific heat capacity of 450 J/kg C receives 0.03MJ of energy. What is the temperature rise of the iron?
The generalised equation governing heat, internal energy and workdone in a system according to first law of thermodynamics isQ=∆U+WSince there is no information about point functions i.e pressure and volume of the system let us consider that the workdone by the system is Zero.Now, heat supplied is 30000 joulesI.Internal energy change is given by mc(∆T)∆U= 5*450*∆T.According to first law heat supplied is equal to change in internal energy of the system given here the workdone is zero30000=2250(∆T)Hence the change in temperature is 13.33 Celsius, since the specific heat is in Celsius scale.
What does it mean that the specific heat of water is 4,200 J per kg degree Celsius?
What does it mean that the specific heat of water is 4,200 J per kg degree Celsius?That arrangement is pretty much the way you speak it. But when you speak it, you don’t hear the parentheses. You have to know they are there. The units are J/(kg °C).When you have some water and you want to heat it up, you have to add heat to it. If you have more water, you need to add more heat. If you want to warm it up more degrees, you need to add more heat. the amount of heat needed is proportional to the mass and it’s proportional to the number of degrees you warm it up.If you have 1 kg of water and you want to heat it up by one degree C, then you have to add 4200 joules of heat. If you have 2 kg, then you have to add 8400 joules. That’s the per kilogram part. For each kilogram, you have to add 4200 J. It takes 4200 J per kg to heat up water by 1°C. Or 4200 J/kg for every degree C.Let’s go back to 1 kg. It takes 4200 J to warm it 1 degree. To warm it 2°C, it takes twice as much. It takes 8400 J. To raise it 3°, it will be 12,600 J. The amount of heat needed goes up in proportion to the number of degrees. That’s the per degree bit. So it’s 4200 J per degree C for each kg. Or 4200 J/°C per kg. Or 4200 J/°C/kg or 4200 J/(°C kg) or 4200 J/(kg °C).If you want to warm 5 kg of water by 7 degrees, you need to multiply that 4200 J/(kg°C) by both the number of kilograms and by the number of degrees.Heat added = 4200 J/(kg °C) [math]\times[/math] 5 kg [math]\times[/math] 7°C = 147,000 J.I remember when I was a kid struggling to understand why “per” was equivalent to divided by. Well if you read 4200 J/(kg °C) as 4200 per kg per °C and fold that into the above description, it begins to make sense. You have to multiply by the number of kilograms and multiply by the number of degrees to cancel those units that are already in the denominator.
If an object whose mass is 20 kg is dropped from height of 10m, what is the maximum kinetic energy it will attain?
Although Kinetic Energy = 1/2 * m * v^2, you don't have to calculate the velocity directly to solve this problem. The potential energy of an item is: m*g*h m is mass in kg. g is the acceleration due to gravity in m/s^2 (e.g. 9.81) h is the height in meters This yields 10*9.81*20 = 1962 kg*m^2/sec^2 (aka N*m or joules)immediately before impact, 100 % of the potential energy has been converted to kinetic energy. 1962 Joules is my final answer.
A ball of mass 8 kg is dropped from a height of 10 m. What is the velocity with which it strikes the ground?
First MethodJust Apply Third Equation Of Motionv[math]^2[/math]- u[math]^2[/math] = 2asa= accelerations= displacementu=initial velocityv=final velocityIf the body is freely falling under gravity then this equation can be modified asv[math]^2[/math]- u[math]^2[/math] = 2ghh= height from which body is droppedg=9.8 m/s[math]^2[/math]Now Come To Your Question According to you question body is dropped from height of 10m so we can consider its initial velicity as zeroh=10mu=0g=acceleration due to gravity on earth which is equal to 9.8 m/s[math]^2[/math]Now put the values of { height, initial velocity and acceleration due to gravity in third equation of motion }i.ev[math]^2[/math]- u[math]^2[/math] = 2ghv[math]^2[/math]- 0 = 2×9.8×10v[math]^2[/math] = 196v=14m/sHence the velocity with which it strikes the ground is 14m/sSecond MethodNow this question can also be solved by another waymgh = 1/2×mv[math]^2[/math]wher m is the mass of the body v is velocity and g is acceleration due to gravity and h is the height2gh=v[math]^2[/math]2×9.8×10 = v[math]^2[/math]v[math]^2[/math] = 196v= 14m/sBy any way you solve there is no role of mass here