Which one of the following is an example of an indicator?

Fluid pressure changes with depth are assumed to be linear. Which statement best explains why this does not hold true for atmosp

ankoles [38]

Answer:

Explanation:

Pressure due to fluid is directly proportional to the depth of fluid, density of the fluid and the value of acceleration due to gravity.

P = h d g

Where, h is the depth, d be the density and g be the acceleration due to gravity.

If we talk about teh atmospheric pressure, the density of air goes on decreasing as we go up and up. o we cannot say that it is directly depends only on the depth of air, it also depends on the changing density of air.

4 0

2 years ago

What are examples of a solution in solids, liquids, and gases

Nina [5.8K]

Answer:

Solid: metal alloy

Liquid: beer

Gas: Air

Explanation:

A solution is a type of mixture where the solvent and solute are homogeneously mixed. Homogeneous mixture means that the solute shouldn't be able to be seen with the naked eye, filtered and stable enough.

Metal alloy will be an example of a solution in solid-state. Beer is a solution made of liquid alcohol and liquid water. Air mostly composed of nitrogen, but it has oxygen, carbon dioxide, and many other substances in gaseous form.

3 0

2 years ago

The intensity at distance from a spherically symmetric sound source is 100 W/m2. What is the intensity at five times this distan

ss7ja [257]

To solve this problem it is necessary to apply the concepts related to intensity as a function of power and area.

Intensity is defined to be the power per unit area carried by a wave. Power is the rate at which energy is transferred by the wave. In equation form, intensity I is

$I = \frac{P}{A}$

The area of a sphere is given by

$A = 4\pi r^2$

So replacing we have to

$I = \frac{P}{4\pi r^2}$

Since the question tells us to find the proportion when

$r_1 = 5r_2 \rightarrow \frac{r_2}{r_1} = \frac{1}{5}$

So considering the two intensities we have to

$I_1 = \frac{P_1}{4\pi r_1^2}$

$I_2 = \frac{P_2}{4\pi r_2^2}$

The ratio between the two intensities would be

$\frac{I_1}{I_2} = \frac{ \frac{P_1}{4\pi r_1^2}}{\frac{P_2}{4\pi r_2^2}}$

The power does not change therefore it remains constant, which allows summarizing the expression to

$\frac{I_1}{I_2}=(\frac{r_2}{r_1})^2$

Re-arrange to find $I_2$

$I_2 = I_1 (\frac{r_1}{r_2})^2$

$I_2 = 100*(\frac{1}{5})^2$

$I_2 = 4W/m^2$

Therefore the intensity at five times this distance from the source is $4W/m^2$

3 0

3 years ago

Assume: The bullet penetrates into the block and stops due to its friction with the block. The compound system of the block plus

charle [14.2K]

Answer:

The total energy of the composite system is 7.8 J.

Explanation:

Given that,

Height = 0.15 m

Radius of circular arc = 0.27 m

Suppose, the entire track is friction less. a bullet with a m₁ = 30 g mass is fired horizontally into a block of wood with m₂ = 5.29 kg mass. the acceleration of gravity is 9.8 m/s.

Calculate the total energy of the composite system at any time after the collision.

We need to calculate the total energy of the composite system

Total energy of the system at any time = Potential energy of the system at the stopping point

$E=mgh+Mgh$

$E=(m+M)gh$

Put the value in to the formula

$E=(30\times10^{-3}+5.29)\times 9.8\times0.15$

$E=7.8\ J$

Hence, The total energy of the composite system is 7.8 J.

8 0

3 years ago

a teacher pushed a 10kg desk across a floor for a distance of 5m. she exerted a horizontal force of 20n. how much work was done?

serious [3.7K]

Work Done = Force x distance
Since she exerted a horizontal force of 20N over a distance of 5m, the work done is 20N x 5m which is equals to 100 joules

7 0

3 years ago