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patriot [66]
3 years ago
8

You want to raise the temperature of 22kg of water from 60°C to 95°C. If electricity costs $0.08/kWh, how much would it cost you

to use an electric heater for this process? Use the specific heat of water is 4.186 kJ/kgoC and 1 kJ = 0.00028 kWh.
Physics
1 answer:
Sladkaya [172]3 years ago
3 0

Answer:

dgtgh

Explanation:

dgdfhdhfffff thdhfghjfj e

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Buco novia de 15n de edad
Zigmanuir [339]

Yo tengo quince y soy de cali y estoy viviendo en mexico

4 0
3 years ago
Why are the freezing points of water and the melting point the same
Sladkaya [172]
Because melting point<span> and </span>freezing point<span> describe the</span>same<span> transition of matter, in this case from liquid to solid (</span>freezing) or equivalently, from solid to liquid (melting<span>).</span>
5 0
3 years ago
Read 2 more answers
how do you find work when only given the angle a sled is pulled, the mass, the coefficent of kinetic friction and distance
Sergio039 [100]

Answer:

W = F * s    

Work done equals applied force * distance traveled

Apparent weight = M g (1 - sin θ)     since some of applied force will lighten sled

μ = coefficient of kinetic friction

F cos θ = force applied to motion of sled

s = distance traveled

[μ M g (1 - sin θ)] cos θ * s = work done in moving sled

Note that F = μ M g    if applied force is in the horizontal direction

8 0
2 years ago
A circular sign has a diameter of 40 cm and is subjected to normal winds up to 150 km/h at 10°C and 100 kPa. Determine the drag
Marat540 [252]

Answer:

147.7 N

221.55 Nm

Explanation:

P = Pressure = 100000 Pa

R_s = Mass-specific gas constant = 287.015 J/kg k

T = Temperature = 10+273 = 283 K

C = Drag coefficient = 1.1

A = Area

r = Radius = 0.2 m

v = Speed of wind = \frac{150}{3.6}\ m/s

L = Length of pole

Density

\rho=\frac{P}{R_sT}\\\Rightarrow \rho=\frac{100000}{287.058\times 283}\\\Rightarrow \rho=1.2309\ kg/m^3

Drag force

F=\frac{1}{2}\rho CAv^2\\\Rightarrow F=\frac{1}{2}\times 1.2309\times 1.1\times \left(\pi \times 0.2^2\right)\times \left(\frac{150}{3.6}\right)^2\\\Rightarrow F=147.7\ N

Force on the circular sign is 147.7 N

M=F\times L\\\Rightarrow M=147.7\times 1.5\\\Rightarrow M=221.55\ Nm

Bending moment at the bottom of the pole is 221.55 Nm

6 0
3 years ago
Galilee said that if you rolled a ball along a level surface it would be
kodGreya [7K]
It would <span>keep rolling without slowing down if no friction acted upon it.

</span>
4 0
3 years ago
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