[Homework Question] A room is 6 m long, 5 m wide and 3 m high . The four walls and ceiling are to be painted. If the windows and door is 14 square meters, what area is to be painted?
Given A room with 6 m= long 5m= wide3 m= high And window and door =14 square metersSo the panted area= area of cuboid-areas of the floor- area of windows and doors =2(6*5+6*3+5*3)-6*5-14= 82m^{2}Math Homework Help
A water tank is 3 m long, 2 m wide and 1 m high. How many litres can be stored in it?
First, let’s find the volume of the tank. This is a little tricky because it depends on what shape the water tank is. In this case, I will assume it’s cubic. We know the volume of a object is represented by the equation V=LWH, which tells that the volume is the product of the length, the width, and the height. In this case, the tank is 6 cubic meters.Although we found the volume, this is not the answer we’re looking for because we’re trying to find how many liters are in the tank, not cubic meters. The dictionary defines liter as equal to “1,000 cubic centimeters.” In that case, we convert the cubic meters to cubic centimeters, leaving us with 6,000,000 cubic centimeters. With our prior definition, we can divide 6,000,000 by 1,000 to get 6,000.Finally, assuming the tank is cubic, there can be at most 6,000 liters in a 3 x 2 x 1 tank.
A room with 3.0-m-high ceilings has a metal plate on the floor with V = 0 V ...?
a) Electric field strength between floor and ceiling .. E = V/d = 3.40^6V / 3.0m .. E = 1.13^6 N/C (≡V/m) Downward electric force on +charge .. F = Eq = 1.13^6N/C x 4.90^-9C .. F = 5.54^-3 N Grav force on charge = mg = 1.30^-3kg x 9.80N/kg .. mg = 1.27^-2 N Net force, F↓ = 5.54^-3 N + 1.27^-2 N .. F↓ = 1.82^-2 N Net ↓ acceleration = F↓ / m = 1.82^-2N / 1.30^-3kg .. .. a = - 14.0 m/s² At max height (h) final velocity v = 0, initial vel u = 4.90m/s, accel = -14.0m/s² v² = u² + 2ah 0 = 4.90² - (2 x 14 x h) .. .. h = 4.90² / 28 .. .. ►h = 0.86 m b) Repeat above method .. this time the electric force is ↑ and net F↓ = 5.54^-3N - mg
How thick are the floors of tall buildings?
Floor-to-floor heights on high-rise buildings — commercial and residential— are typically more than 11′ and under 12′. For a school project, you should use round numbers: 12′. The Ground and Second Floors are typically higher.For concrete buildings, the concrete slabs are typically 1′ thick. In reality, the thicknesses vary per section of the building, but 1′ should suffice for a school project (unless your actual assignment is figuring out the thickness).
Homework Question: An 8.0 kg toy is dropped from a height of 7.0 m. What is the kinetic energy of the toy just before it hits the ground?
Well, our toy has an energy E, which is the sum of the potential and kinetic energy.So, when the toy is at the top, 7 m, it has potential energy, because it is at rest (its speed is 0)Ep = m*g*hWhere m is the mass (8 kg), g is the Earth's gravity (9.81 N/kg) and h is the height (7m). Please note that sometimes teachers use g = 10, in order to simplify the calculations.For the sake of simplicity, I will use 10. If you want to be exact, use 9.81.So, plugging in our numbers, we get:Ep = 8 kg * 7 m * 10 N/kg = 560 JWhen the toy falls, its energy is transforming from potential to kinetic.So, when the height is 0, all the energy is kinetic.Ek = 1/2 * m * v^2So, we know that energy is conserved, which means that the initial energy is the same as the final energy. We had calculated the initial energy to be 560 J, which means that Ek is also 560JSubstituting, we get that:1/2 * 8 kg * v^2 = 560V^2 = 140So, v is the square root of 140, which is 11.83 m/sFor any questions, feel free to comment!
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
How High - Physics Potential?
a) W = F(net)x = 0 - 0.5mv^2, F(net) = F(coulomb) – mg Note: the final velocity must be zero. Therefore, E(final)=0 F = qE = qV/d = (5.1x10^-9)(3.4x10^6)/3.2 = 5.42x10^-3 N F(net) = F(coulomb:repulsion) – mg = (-5.42x10^-3) – (1.3x10^-3)(9.8) = (-18.2x10^-3) N W = Fh = 0.5mv^2 => h = (0.5mv^2)/F = [0.5(1.3x10^-3)(5.1)^2]/18.2.42x10^-3 = 0.93 m b) F = qE = qV/d = (5.1x10^-9)(2.8x10^6)/3.2 = 4.46x10^-3 N F(net) = F(coulomb:attraction) – mg = 4.46x10^-3 – (1.3x10^-3)9.8 = -8.28x10^-3 N h = 0.5mv^2/F = [0.5(1.3x10^-3)(5.1)^2]/8.28x10^-3 = 2 m