Answer:
You could move something across the Earth with a little push. It would make fuel really efficient on those pathways. You could make a floor that is impossible to walk on. Everybody would just fall without traction.
Explanation:
<em>12,25 km/h</em>
<em>≈ 3,4 m/s </em>
<em>v = d/t</em>
<em>= 12250m/h</em>
<em>= 12,25km/h</em>
<em>or</em>
<em>v = d/t</em>
<em>= 12250m/h</em>
<em>1h = 60m×60s = 3600s</em>
<em>= 12250m/3600s</em>
<em>≈ 3,4 m/s </em>
Answer:
Una Mezcla Homogénea es aquella mezcla en la que las sustancias que la forman poseen una combinación uniforme.Son ejemplos de Mezclas Homogéneas: Compuesta
Explanation:
Aire (es una mezcla de gases homogénea formada principalmente por de nitrógeno, oxígeno, vapor de agua, dióxido de carbono...)
Leche (mezcla de agua, carbohidratos, proteínas...)
Bebida alcohólica (mezcla de agua y alcohol etílico)
Acero (mezcla de elementos aleados como el hierro, el carbono y otras sustancias)
Petróleo (mezcla de hidrocarburos)
Agua de mar (mezcla de agua, cloruro sódico y otras sustancias)
Mezcla de agua y sal disuelta
Agua azucarada (mezcla de agua y azúcar)
Aleación metálica (las aleaciones metálicas son mezclas en las que se combinan diferentes metales de una manera homogénea y definida)
Perfume (mezcla de agua y otras sustancias olorosas cuya composición es uniforme)
Answer:
See below explanation
Explanation:
The correspondent chemical reaction for copper carbonate decomposed by heat is:
CuCO₃ (s) → CuO (s) + CO₂ (g)
Considering all molar mass (MM) for each element ( we consider rounded numbers) :
MM CuCO₃ = 123 g/mol
MM CuO = 79 g/mol
MM CO₂ = 44 g/mol
Statement mentions that scientis heated 123.6 g of CuCO₃ (almost a MM), until a black residue is obtained, which weights 79.6 g : this solid residue is formed by CuO, and the remaining mass (approximatelly 44 g) belongs to teh second product, this is, CO₂; as it is a gas compund, it is not certainly included on the solid residue.
So, law of conservation mass is true for this case, since: 123.6 g = 79.6 g + 44 g. As explained, on the solid residue, we don not include the 44 g, which "escaped" from our system, since it is a gas compound (CO₂)
Potential Energy (Initial one) = m * g * h
P.E. = 60 * 9.8 * 10
P.E. = 5880
Kinetic Energy (Final One) = 1/2 mv²
K.E. = 1/2 * 60 * (10)²
K.E. = 6000/2
K.E. = 3000
Lost Energy = 5880 - 3000 = 2880 J
In short, Your Answer would be 2880 Joules
Hope this helps!
