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3 years ago

7

A cube with sides 12 cm is submerged in water to a depth of 30 cm. Given density of water is 1000 kg/m3. Calculate the pressure

at the bottom surface of the cube due to water.

1 answer:

yulyashka [42]3 years ago

8 0

Answer:

P=2940 Pa

Explanation:

Given that,

Side of a cube, a = 12 cm

It is submerged to depth of 30 cm

The density of water, d = 1000 kg/m^3

We need to find the pressure at the bottom surface of the cube. We know that the pressure exerted is given by :

$P=d gh\\\\P=1000\ kg/m^3\times 9.8\ m/s^2\times 0.3\ m\\\\P=2940\ Pa$

So, 2940 Pa of pressure is exerted at the bottom surface of the cube due to water.

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1/12 of 1/3 is what fraction?

Misha Larkins [42]

$\frac{1}{36}$

Explanation:

$Given problem; \frac{1}{12} of \frac{1}{3}$

The of word is used in mathematics determine how much of something is a part of another.

It simply translates to the mathematical operator of multiplication (x or *).

To solve this problem, we simply multiply both fractions to find how much of $\frac{1}{12}$ is $\frac{1}{3}$

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3 years ago

A baseball approaches home plate at a speed of 44.0 m/s, moving horizontally just before being hit by a bat. The batter hits a p

Luda [366]

Explanation:

It is given that,

Speed of the baseball, u = 44 m/s

Speed of the baseball, v = 53 m/s

Mass of the ball, m = 145 g = 0.145 kg

Time of contact between the ball and the bat, t = 2.2 ms = 0.0022 s

$F=ma$

$F=\dfrac{mv}{t}$

$F_1=\dfrac{0.145\ kg\times 44\ m/s}{0.0022\ s}$

F₁ = 2900 N...........(1)

$F=ma$

$F=\dfrac{mv}{t}$

$F_2=\dfrac{0.145\ kg\times 53\ m/s}{0.0022\ s}$

F₂ = 3493.18 N.........(2)

In average vector form force is given by :

$F=F_1+F_2$

$F=(2900i+(-3493.18)\ N$

$F=(2900i-3493.18j)\ N$

Hence, this is the required solution.

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4 years ago

A 75.0-kg person is riding in a car moving at 20.0 m/s when the car runs into a bridge abutment. (a) calculate the average force

iragen [17]

(a) $-1.5\cdot 10^6 N$

First of all, we need to calculate the acceleration of the person, by using the following SUVAT equation:

$v^2 -u ^2 = 2ad$

where

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d = 1.00 cm = 0.01 m is the displacement of the person

Solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.01)}=-20000 m/s^2$

And the average force on the person is given by

$F=ma$

with m = 75.0 kg being the mass of the person. Substituting,

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where the negative sign means the force is opposite to the direction of motion of the person.

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In this case,

v = 0 is the final velocity

u = 20.0 m/s is the initial velocity

a is the acceleration

d = 15.00 cm = 0.15 m is the displacement of the person with the air bag

So the acceleration is

$a=\frac{v^2-u^2}{2d}=\frac{0-20.0^2}{2(0.15)}=-1333 m/s^2$

So the average force on the person is

$F=ma=(75)(-1333)=-1.0\cdot 10^5 N$

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4 years ago

Cuál es el método que utiliza el factor de la gravedad y la difusión de un líquido sobre un sólido para separar las partes de un

Airida [17]

Answer:

chromatography

Explanation:

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In this technique the separation of the mixture by passing the mixture in some solution or some suspension through the medium where the components are made to move at some different rates.

The basic idea in this experiment is that the mixture sample is in mobile phase and it is forced either by pumping or by gravity or by capillary action through the stationary phase to separate the mixture.

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