The correct answer is Metals.
Generally, the specific heat of metals is low. Very high specific heat exists in water.A physical feature of matter known as heat capacity or thermal capacity is the quantity of heat that must be applied to an object in order to cause a unit change in temperature. Heat capacity is measured in joules per kelvin (J/K), the SI unit. A broad property is heat capacity. Use the following equation to determine heat capacity: heat capacity = E / T, where E is the quantity of delivered heat energy and T is the change in temperature. The formula would be as follows, for instance, if it takes 2,000 Joules of energy to raise a block's temperature by 5 degrees Celsius: 2,000 Joules per °C is the heat capacity.
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El factor mas importante para el exceso de peso es un exceso de energía creada por una alimentación excesiva
El peso de un cuerpo es definido por la relación entre la energía requerida para los procesos vitales del cuerpo, sus actividades físicas diarias y la energía suministra en forma de alimentos.
Cuando estos dos parámetros están en balance el peso es estable, pero cuando la cantidad de alimentos aumenta o el valor energético de los mismo aumenta se tiene un exceso de energía que el cuerpo almacena en forma de grasa corporal, este el el factor mas importante para el exceso de peso.
En conclusión el factor mas importante para el exceso de peso es un exceso de energía por una alimentación excesiva
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Answer:
ni = 2.04e19
Explanation:
we know that in semiconductor like intrinsic, when electron leave the band, it leave a hole in valence band so we have
n = p = ni
from intrinsic carrier concentration
1.7 = ni * 1.6*10^{-19} * (.35 + .17)
ni = 2.014 *10^{19} m^{-3}
ni = 2.04e19
Answer:
(A) 10052.2 m/s²
(B) 0.00678 seconds
Explanation:
From the question,
(A) Applying
V² = U²+2as..................... Equation 1
Where V = Final velocity, U = Initial velocity, a = acceleration due to gravity, s = distance.
make a the subject of the equation
a = (V²-U²)/2s........................ Equation 2
Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m
Substitute these values into equation 2
a = (68²-0²)/(2×0.230)
a = 10052.2 m/s²
(B) Using,
a = (V-U)/t......................... Equation 3
Where t= time.
make t the subject of the equation
t = (V-U)/a......................... Equation 4
Given: V = 68 m/s, U = 0 m/s, a = 10052.2
Substitute into equation 4
t = (68-0)/10052.2
t = 0.00678 seconds
Answer:
424.26 m/s
Explanation:
Given that Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direction along which Q is flying. Find the magnitude of velocity of the air craft P relative to air craft Q
The relative speed will be calculated by using pythagorean theorem
Relative speed = sqrt(300^2 + 300^2)
Relative speed = sqrt( 180000 )
Relative speed = 424.26 m/s
Therefore, the magnitude of velocity of the air craft P relative to air craft Q is 424.26 m/s
