Ask question

Ask question

All categories

Lubov Fominskaja [6]

3 years ago

5

Calculate the density of the stone given that its mass is 250g

1 answer:

stiks02 [169]3 years ago

8 0

Explanation:

density = mass divide volume

the volume must be there in order to answer the question.

You might be interested in

Joe is measuring the time it takes for a ball to roll down a ramp. In this experiment Joe takes the measurement 5 times and gets

PolarNik [594]

Step by step solution :

standard deviation is given by :

$\sigma = \sqrt\dfrac{{\sum (x-\bar{x})^2}}{n}$

where, $\sigma$ is standard deviation

$\bar{x}$ is mean of given data

n is number of observations

From the above data, $\bar{x}=24.88$

Now, if $x=24.8$ , then $(x-\bar{x})^2=0.0064$

If $x=23.9$ , then $(x-\bar{x})^2=0.9604$

if $x=26.1$ , then $(x-\bar{x})^2=1.4884$

If $x=25.1$ , then $(x-\bar{x})^2=0.0484$

If $x=24.5$ , then $(x-\bar{x})^2=0.1444$

so, $\sum (x-\bar{x})^2 =\frac{0.0064+0.9604+1.4884+0.0484+0.1444}{5}$

$\sum (x-\bar{x})^2 =2.648$

$\sqrt{\sum \frac{(x-\bar{x})^2}{n}}$

$\sigma =0.7277$

No, Joe's value does not agree with the accepted value of 25.9 seconds. This shows a lots of errors.

6 0

3 years ago

Which statement best describes the common features of sound waves and light waves

Viktor [21]

Sound waves or bounces off the wall and light waves are waves of light

7 0

4 years ago

Cesium-137 undergoes beta decay and has a half-life of 30.0 years. How many beta particles are emitted by a 14.0-g sample of ces

Mandarinka [93]

Answer: $0.81\times 10^{16}$ beta particles

Explanation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass = 14.0 g

Molar mass = 137 g/mol

$\text{Number of moles of cesium}=\frac{14.0g}{137g/mol}=0.102moles$

According to avogadro's law, 1 mole of every substance weighs equal to its molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

1 mole of cesium contains atoms = $6.023\times 10^{23}$

0.102 moles of cesium contains atoms = $\frac{6.023\times 10^{23}}{1}\times 0.102=0.614\times 10^{23}$

The relation of atoms with time for radioactivbe decay is:

$N_t=N_0\times \frac{1}{2}^{\frac{t}{t_{\frac{1}{2}}}}$

Where $N_t$ =atoms left undecayed

$N_0$ = initial atoms

t = time taken for decay = 3 minutes

${t_{\frac{1}{2}}}$ = half life = 30.0 years = $1.577\times 10^7$ minutes

The fraction that decays : $1-(\frac{1}{2})^{\frac{3}{1.577\times 10^7}}=1.32\times 10^{-7}$

Amount of particles that decay is = $0.614\times 10^{23}\times 1.32\times 10^{-7}=0.81\times 10^{16}$

Thus $0.81\times 10^{16}$ beta particles are emitted by a 14.0-g sample of cesium-137 in three minutes.

7 0

3 years ago

Đổi 20độC = ? độ F =? Độ k

san4es73 [151]

Answer:

IDC

Explanation:

I DON'T UNDERSTAND........

7 0

2 years ago

It is friction that provides the force for a car to accelerate, so for high-performance cars the factor that limits acceleration

bearhunter [10]

Answer:

About 12 seconds

Explanation:

6 0

3 years ago