W = ∫ (x from 0.1 to +oo) F dx
= ∫ (x from 0.1 to +oo) A e^(-kx) dx
= A/k x [ - e^(-kx) ](between 0.1 and +oo)
= A/k x [ 0 + e^(-k * 0.1) ]
<span>
= A/k x e^(-k/10) </span>
<span>A. Comparison
</span>What type of organization is used in a paragraph that lists similarities between two objects? Comparison
NOT:
<span>B. Contrast
C. Chronological order
D. Cause and effect</span><span>
</span>
Answer:
i) 24.5 m/s
ii) 30,656 m
iii) 89,344 m
Explanation:
Desde una altura de 120 m se deja caer un cuerpo. Calcule a 2.5 s i) la velocidad que toma; ii) cuánto ha disminuido; iii) cuánto queda por hacer
i) Los parámetros dados son;
Altura inicial, s = 120 m
El tiempo en caída libre = 2.5 s
De la ecuación de caída libre, tenemos;
v = u + gt
Dónde:
u = Velocidad inicial = 0 m / s
g = Aceleración debida a la gravedad = 9.81 m / s²
t = Tiempo de caída libre = 2.5 s
Por lo tanto;
v = 0 + 9.8 × 2.5 = 24.5 m / s
ii) El nivel que el cuerpo ha alcanzado en 2.5 segundos está dado por la relación
s = u · t + 1/2 · g · t²
= 0 × 2.5 + 1/2 × 9.81 × 2.5² = 30.656 m
iii) La altura restante = 120 - 30.656 = 89.344 m.
As we know that when charge is released in electric field
It will have two forces on it
1. electrostatic force
2. gravitational force
now if the ball will accelerate upwards so we can say
net upward force = mass * acceleration
now we can find charge q on it by above equation
So above is the charge on the particle
P = density × gravity acceleration × height
P = 1200 × 9.81 × 15/100
P = 1765.8