An applied force is a force that is applied to an object by a person or another object.

Which of the following most accurately represents John Dalton’s model of the atom? A. a tiny, solid sphere with an unpredictable

aleksley [76]

A and c are the answersss

6 0

4 years ago

Note: Plagiarism will be reported

iren [92.7K]

Y no me han dicho que me ha dado un teléfono y me ha dicho la policía que han dado la cara para la que ha sido un policía que han dicho nada que no se han dado cuenta de

6 0

3 years ago

Read 2 more answers

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

What does the text mean when it says “naming an elements is like creating history

dolphi86 [110]

Answer: When you name an element, it's usually because the element is newly discovered. So by naming an element you are creating something that will last throughout history.

Explanation:

5 0

3 years ago

A car's acceleration is 3m/s2. If the car started at rest and it only took 10s for the car to reach this acceleration, what is t

Grace [21]

Answer:

30m/s

Explanation:

From law of motion equation

Vf= Vi + at

Where Vf= final velocity

Vi= initial velocity=0(the car started at rest)

a= acceleration= 3m/s2

t= time= 10s

Then substitute into the equation to get the final velocity.

Vf= 0+(10×3)

Vf= 30m/s

Hence, the car's final velocity is 30m/s

5 0

3 years ago