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3 years ago

Which substance is an acid ?

Physics

2 answers:

Mekhanik [1.2K]3 years ago

8 0

Answer:

According to Arrhenius definition, an acid is a substance that releases hydrogen ion or hydronium ion.

there are also other definition like bronsted and lewis.

I HOPE IT WILL HELP

DENIUS [597]3 years ago

6 0

Answer:

if you like my answer markme barinlest:)

Explanation:

An acid is a substance or compound that releases hydrogen ions (H+) when in solution. In a strong acid, such as hydrochloric acid (HCl), all hydrogen ions (H+), and chloride ions (Cl-) dissociate (separate) when placed in water and these ions are no longer held together by ionic bonding.

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Veronica claims that she can throw a dart at a dartboard from a distance of 2 m and hit the 5 cm wide bullseye if she throws the

wariber [46]

First, find the amount of time for the dart to hit the board using this equation: t = d/v

t = 2 m/ 15 m/s = 0.133 s

Then, find the height the dart has fallen from its initial point using this equation: h = 0.5gt²

h = 0.5(9.81 m/s²)(0.133 s)² = 0.0872 m or 8.72 cm

Since the diameter of the bull's eye is only 5 cm, and you started at the same level of the top of the bull's eye, that means the maximum allowance would only be 5 cm. Since it exceeded to 8.72 cm, it means that <em>Veronica will not hit the bull's eye.</em>

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4 years ago

Which lighting direction is used to create silhouettes?

Ivahew [28]

Backlighting is used to create silhouettes

HOPED I HELPED

5 0

4 years ago

Read 2 more answers

¿Por qué si cargas a uno de tus compañeros por cierto tiempo no estás realizando un trabajo mecánico?

VladimirAG [237]

Answer:

I will answer this in English, we can translate it to:

Why if you charge a mate by an amount of time you are not doing work?

This happens because work is defined as the displacement done by a force:

W = d*F

where W is work, d is the distance, and F is the force.

This means that the amount of time that you are charging your mate does not affect the mechanical work, the only time that you are doing work is when you are lifting him.

4 0

4 years ago

What force (in N) must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2400 kg car (a large car

makvit [3.9K]

Answer:

Fm= 91.88 N

Explanation:

Pascal principle

The pressure acting on one side is transmitted to all the molecules of the liquid because the liquid is incompressible.

The pressure is definited like this:

P=F/A

Where:

P: Pressure in pascals (Pa)

F: Force acting in the area (N)

A : Area where the force acts (m²)

Pascal principle

Pm=Ps

Fm/ Am= Fs/ As Formula (1)

Where :

Pm : Pressure on the master piston

Ps : Pressure on the slave piston

Fm : Force on the master piston (N)

Fs: Force on the slave piston ((N)

Am: master piston area (m²)

As: slave piston area (m²)

Area Formula (A)

A= π*R²

R : piston radius

Calculation of the weight of the car (W)

W= m*g= 2400 kg*9.8m/s²= 23520 N

W = Fs

Data

Fs = 23520 N

Dm = 1.5 cm

Ds = 24 cm

Rm = 0.75 cm

Rs = 12 cm

Am = π*Rm² = π*(0.75)²

As = π*Rs² = π*(12)²

Force exerted on the master cylinder

We replace data in the formula (1)

$\frac{F_{m} }{A_{m} } = \frac{F_{s} }{A_{s} }$

$F_{m} = \frac{F_{s}*A_{m} }{A_{s}}$

$F_{m} = \frac{(23520 N)*(\pi *(0.75)^{2})(cm^{2})}{(\pi *(12)^{2})(cm^{2})}$

$F_{m} = (23520 N)*\frac{(0.75)^{2} }{(12)^{2} }$

Fm= 91.88 N

8 0

3 years ago

A student measures the diameter of a small cylindrical object and gets the following readings: 4.32, 4.35, 4.31, 4.36, 4.37, 4.3

Zinaida [17]

Answer:

a. $\bar{d}=4.34 cm$

b. $\sigma=0.023 cm$

c. $\rho=(0.0089\pm 0.00058) kg/cm^{3}$

Explanation:

a) The average of this values is the sum each number divided by the total number of values.

$\bar{d}=\frac{\Sigma_{i=1}^(N)x_{i}}{N}$

$x_{i}$ is values of each diameter
N is the total number of values. N=6

$\bar{d}=\frac{4.32+4.35+4.31+4.36+4.37+4.34}{6}$

$\bar{d}=4.34 cm$

b) The standard deviation equations is:

$\sigma=\sqrt{\frac{1}{N}\Sigma^{N}_{i=1}(x_{i}-\bar{d})^{2}}$

If we put all this values in that equation we will get:

$\sigma=0.023 cm$

Then the mean diameter will be:

$\bar{d}=(4.34\pm 0.023)cm$

c) We know that the density is the mass divided by the volume (ρ = m/V)

and we know that the volume of a cylinder is: $V=\pi R^{2}h$

Then:

$\rho=\frac{m}{\pi R^{2}h}$

Using the values that we have, we can calculate the value of density:

$\rho=\frac{1.66}{3.14*(4.34/2)^{2}*12.6}=0.0089 kg/cm^{3}$

We need to use propagation of error to find the error of the density.

$\delta\rho=\sqrt{\left(\frac{\partial\rho}{\partial m}\right)^{2}\delta m^{2}+\left(\frac{\partial\rho}{\partial d}\right)^{2}\delta d^{2}+\left(\frac{\partial\rho}{\partial h}\right)^{2}\delta h^{2}}$

δm is the error of the mass value.
δd is the error of the diameter value.
δh is the error of the length value.

Let's find each partial derivative:

1. $\frac{\partial\rho}{\partial m}=\frac{4m}{\pi d^{2}h}=\frac{4*1.66}{\pi 4.34^{2}*12.6}=0.0089$

2. $\frac{\partial\rho}{\partial d}=-\frac{8m}{\pi d^{3}h}=-\frac{8*1.66}{\pi 4.34^{3}*12.6}=-0.004$

3. $\frac{\partial\rho}{\partial h}=-\frac{4m}{\pi d^{2}h^{2}}=-\frac{4*1.66}{\pi 4.34^{2}*12.6^{2}}=-0.00071$

Therefore:

$\delta\rho=\sqrt{\left(0.0089)^{2}*0.05^{2}+\left(-0.004)^{2}*0.023^{2}+\left(-0.00071)^{2}*0.5^{2}}$

$\delta\rho=0.00058$

So the density is:

$\rho=(0.0089\pm 0.00058) kg/cm^{3}$

I hope it helps you!

3 0

3 years ago