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)
No, because superconductivity cannot occur if there is resistance
In addition to explaining electrical resistance, equilibrium distance theory also foretells the existence of superconductivity. According to its postulates, electrical resistivity decreases with distance from the equilibrium. There is only superconductivity at zero distance, with no resistance
<h3>What is Superconductivity ?</h3>
The ability of some materials to transmit electric current with virtually little resistance is known as superconductivity.
- This ability has intriguing and maybe beneficial ramifications. Low temperatures are necessary for a material to exhibit superconductor behaviour. H. K. made the initial discovery of superconductivity in 1911.
- Aluminum, magnesium diboride, niobium, copper oxide, yttrium barium, and iron pnictides are a few well-known examples of superconductors.
Learn more about Superconductivity here:
brainly.com/question/17166152
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Answer:
F = 800N
the magnitude of the average force exerted on the wall by the ball is 800N
Explanation:
Applying the impulse-momentum equation;
Impulse = change in momentum
Ft = m∆v
F = (m∆v)/t
Where;
F = force
t = time
m = mass
∆v = v2 - v1 = change in velocity
Given;
m = 0.80 kg
t = 0.050 s
The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.
v2 = 25 m/s
v1 = -25 m/s
∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s
Substituting the values;
F = (m∆v)/t
F = (0.80×50)/0.05
F = 800N
the magnitude of the average force exerted on the wall by the ball is 800N
its B 0.225kPa using the formula p=f/A then change the pascals into kpa
