Third-largest ocean?
2 answers:
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Before answering this question, first we have to understand the effect of ratio of surface area to volume on the rate of diffusion.
The rate of diffusion for a body having larger surface area as compared to the ratio of surface area to volume will be more than a body having less surface area. Mathematically it can written as-
V∝ R [ where v is the rate of diffusion and r is the ratio of surface area to volume]
As per the question,the ratio of surface area to volume for a sphere is given
The surface area to volume ratio for right circular cylinder is given
Hence, it is obvious that the ratio is more for right circular cylinder.As the rate diffusion is directly proportional to the surface area to volume ratio,hence rate of diffusion will be more for right circular cylinder.
Hence the correct option is B. The rate of diffusion would be faster for the right cylinder.
Answer:
t_{out} = t_{in}, t_{out} =
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D /
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D /
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D /
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =
Answer:
42m/s
6.06s
Explanation:
To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:
θ: angle = 45°
vo: initial velocity
g: gravitational constant = 9.8m/s^2
x_max: max distance = 180 m
t_max: max time
by replacing the values of the parameters and do vo the subject of the first formula you obtain:
with this value of vo you calculate the max time:
hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s
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TRANSLATION:
Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:
θ: ángulo = 45 °
vo: velocidad inicial
g: constante gravitacional = 9.8m / s ^ 2
x_max: distancia máxima = 180 m
t_max: tiempo máximo
reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:
con este valor de vo usted calcula el tiempo máximo:
por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s