All categories

3 years ago

Que unidad de medida se utiliza para expresar la capacidad o volumen de un objeto? Da un ejemplo

Physics

1 answer:

galina1969 [7]3 years ago

5 0

Answer:

Explicado a continuación

Explanation:

Hay una pequeña diferencia conceptual entre la capacidad y el volumen de un objeto, a saber:

El volumen hace referencia al espacio que ocupa un objeto, mientras que la capacidad hace referencia al espacio que este contiene. Calcular el volumen de un cuerpo es medir cuánto ocupa mientras que calcular su capacidad es medir cuánto cabe en él.

En la práctica, ambos conceptos son usados indistintamente, ya que tienen unidades equivalentes.

El volumen tiene unidades de longitud al cubo, como por ejemplo:

$m^3,\ cm^3\ ,dm^3,\ pie^3\ , pulg^3$

y la capacidad se suele expresar en litros o unidades derivadas: litro, mililitro, centilitro, etc.

Como mencionamos, hay equivalencia engre los dos grupos de unidades. Entre las más conocidas están:

$1\ lt=1\ dm^3,\ 1\ m^3=1000\ lt$

You might be interested in

Three vectors →a, →b, and →c each have a magnitude of 50 m and lie in an xy plane. Their directions relative to the positive dir

Damm [24]

Answer:

a) 38.27 b) 322.5°

c) 126.99 d) 1.17°

e) 62.27 e) 139.6°

Explanation:

First of all we have to convert the coordinates into rectangular coordinates, so:

a=( 43.3 , 25)

b=( -48.3 , -12.94)

c=( 35.36 , -35.36)

Now we can do the math easier (x coordinate with x coordinate, and y coordinate with y coordinate):

1.) a+b+c=( 30.36 , -23.3) = 38.27 < 322.5°

2.) a-b+c=( 126.96 , 2.6) = 126.99 < 1.17°

3.) (a+b) - (c+d)=0 Solving for d:

d=(a+b) - c = ( -40.36 , 47.42) = 62.27 < 139.6°

5 0

3 years ago

A ball is thrown straight up at 20 m. What is the balls velocity as it hits the ground?

deff fn [24]

Answer:

The velocity of the ball as it hit the ground = 19.799 m/s

Explanation:

Velocity: Velocity of a body can be defined as the rate of change of displacement of the body. The S.I unit of velocity is m/s. velocity is expressed in one of newtons equation of motion, and is given below.

v² = u² + 2gs.......................... Equation 1

Where v = the final velocity of the ball, g = acceleration due to gravity, s = the height of the ball

Given: s = 20 m, u = 0 m/s

Constant: g = 9.8 m/s²

Substituting these values into equation 1,

v² = 0 + 2×9.8×20

v² = 392

v = √392

v = 19.799 m/s.

Therefore the velocity of the ball as it hit the ground = 19.799 m/s

5 0

3 years ago

Help me please???!!!!

Mariulka [41]

I got you gurl so if your looking for speed kts speed =distance +time and when you do all your steps correctly its speed=320 m/s....... Hope this helped

6 0

3 years ago

Rory uses a force of 25 N to lift her grocery bag while doing 50 J of work. How far did she lift the

Dennis_Churaev [7]

Answer:

$W=Fd \\ 50 = 25d \\ d = \frac{50}{25} \\ \color{yellow} \boxed{d = 2m}$

8 0

3 years ago

The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2

Vlada [557]

Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]

Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the reaction with OH

Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]

Than, considering the concentration of Cl increases by a factor of 100 to about 4 × $10^{5}$ molecules $cm^{-3}$ , the result will be that the reaction with OH will be 535 + (100 to about 4 × $10^{5}$ molecules $cm^{-3}$ ) times faster than the reaction with Cl

Explanation:

4 0

3 years ago