Una sustancia de 75 g absorbió 3242 J. Calcular la variación de temperatura que sufrió. Calor
1 answer:
Answer:
dt = 138.55K
Explanation:
Dados los siguientes datos;
Masa = 75 g a kg = 75/1000 = 0.075 kg
Calor absorbido = 3242 J
Capacidad calorífica específica = 312 J/kg.K
Para encontrar el cambio de temperatura;
La capacidad calorífica viene dada por la fórmula;
Dónde;
Q representa la capacidad calorífica o la cantidad de calor.
m representa la masa de un objeto.
c representa la capacidad calorífica específica del agua.
dt representa el cambio de temperatura.
Sustituyendo en la fórmula, tenemos;
dt = 138.55K
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Answer:
(a) Elongation of rod = 0.19732 mm
(b) Change in diameter = 0.00651 mm
Explanation:
Circular area at end of steel rod = pi * diameter^2 / 4
Area = 3.801 * 10^(-4) meter squared
Stress = force / area
Stress = 75000 / (3.801 * 10^-4)
Stress = 197316495 Pa OR 0.197 GPa
Modulus of elasticity = stress / strain
200 = 0.19732 / Strain
Strain = 0.0009866 (longitudinal)
(a) Strain = change in length / initial length
0.0009866 = Elongation of rod / 200
Elongation of rod = 0.19732 mm
(b) Poisson ratio = lateral strain / longitudinal strain
0.3 = lateral strain / 0.0009866
Lateral strain = 0.000296
Lateral strain = Change in diameter / original diameter
0.000296 = Change in diameter / 22
Change in diameter = 0.00651 mm
Answer:
The initial velocity in the x-direction with which the branch was thrown is approximately 10.224 m/s
Explanation:
The given parameters of the motion of the branch are;
The height from which the branch is thrown = 3.00 m
The horizontal distance the branch lands from where it was thrown, x = 8.00 m
The direction in which the branch is thrown = Horizontally
Therefore, the initial vertical velocity of the branch, = 0 m/s
The time it takes an object in free fall (zero initial downward vertical velocity) to reach the ground is given as follows;
s = ·t + 1/2·g·t²
Where;
= 0 m/s
s = The initial height of the object = 3.00 m
g = The acceleration due to gravity = 9.8 m/s²
∴ s = 0·t + 1/2·g·t² = 0 × t + 1/2·g·t² = 1/2·g·t²
t = √(2·s/g) = √(2 × 3/9.8) = (√30)/7 ≈ 0.78246
The horizontal distance covered before the branch touches the ground, x = 8.00 m
Therefore, the initial velocity in the horizontal, x-direction with which the branch was thrown, 'uₓ', is given as follows;
uₓ = x/t = 8.00 m/((√30)/7 s)
Using a graphing calculator, we get;
uₓ = 8.00 m/((√30)/7 s) = (28/15)·√30 m/s ≈ 10.224 m/s
The initial velocity in the horizontal, x-direction with which the branch was thrown, uₓ ≈ 10.224 m/s.