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2 years ago

A metal ball has a mass of 2kg and a volume of 6 m3. What is its density

Physics

2 answers:

Lena [83]2 years ago

8 0

Answer: The answer is 333.3333 repeating

Explanation:

Divide the mass by the volume.

Whitepunk [10]2 years ago

7 0

Answer:

Density of the metal ball, $\rho=0.334\ kg/m^3$

Explanation:

It is given that,

Mass of the metal ball, m = 2 kg

Volume of the metal ball, $V=6\ m^3$

We need to find the density of the metal ball. The total mass divided by total volume of an substance is called its density. Its formula is given by :

$\rho=\dfrac{m}{V}$

$\rho=\dfrac{2\ kg}{6\ m^3}$

$\rho=0.334\ kg/m^3$

So, the density of the metal ball is $0.334\ kg/m^3$ . Hence, this is the required solution.

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Which data set has the largest standard deviation

AfilCa [17]

Answer:

<em>The data set marked as B has the largest standard deviation</em>

Explanation:

<u>Standard Deviation</u>

It's a number used to show how a set of measurements is spread out from the average value. A low standard deviation means that most of the values are close to the average. A high standard deviation means that the numbers are more spread out.

The formula for the standard deviation is

$\displaystyle \sigma=\sqrt{\frac{\sum (x_i-\mu)^2}{n}}$

Where $x_i$ is the value of each measurement, n is the number of elements in the set, and $\mu$ is the average or media of the values, defined as

$\displaystyle \mu=\frac{\sum x_i}{n}$

Let's analyze each set of data:

A.3,4,3,4,3,4,3

The average is

$\displaystyle \mu=\frac{3+4+3+4+3+4+3}{7}=3.43$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2+(4-3.43)^2+(3-3.43)^2}{7}}$

$\sigma=0.5$

B.1,6,3,15,4,12,8

The average is

$\displaystyle \mu=\frac{1+6+3+15+4+12+8}{7}=7$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(1-7)^2+(6-7)^2+(3-7)^2+(15-7)^2+(4-7)^2+(12-7)^2+(8-7)^2}{7}}$

$\sigma=4.7$

C. 20, 21,23,19,19,20,20

The average is

$\displaystyle \mu=\frac{20+21+23+19+19+20+20}{7}=20.29$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(20-20.29)^2+(21-20.29)^2+(23-20.29)^2+(19-20.29)^2+(19-20.29)^2+(20-20.29)^2+(20-20.29)^2}{7}}$

$\sigma=1.3$

D.12,14,13,14,12,13,12

The average is

$\displaystyle \mu=\frac{12+14+13+14+12+13+12}{7}=12.86$

Computing the stardard deviation:

$\sigma=\sqrt{\frac{(12-12.86)^2+(14-12.86)^2+(13-12.86)^2+(14-12.86)^2+(12-12.86)^2+(13-12.86)^2+(12-12.86)^2}{7}}$

$\sigma=0.8$

We can see the data set marked as B has the largest standard deviation

5 0

3 years ago

Explain why the mechanical advantage of a single fixed probably is always one

seropon [69]

Ideal mechanical advantage doesn't take energy lost to friction into account. Explain why the mechanical advantage of a single fixed pulley is always 1. A single fixed pulley changes only the direction of the effort force. ... Energy transforms from the object supplying the force to the object being moved.

4 0

2 years ago

Calculate the rate of heat conduction through a layer of still air that is 1 mm thick, with an area of 1 m, for a temperature of

max2010maxim [7]

Answer:

The rate of heat conduction through the layer of still air is 517.4 W

Explanation:

Given:

Thickness of the still air layer (L) = 1 mm

Area of the still air = 1 m

Temperature of the still air ( T) = 20°C

Thermal conductivity of still air (K) at 20°C = 25.87mW/mK

Rate of heat conduction (Q) = ?

To determine the rate of heat conduction through the still air, we apply the formula below.

$Q =\frac{KA(\delta T)}{L}$

$Q =\frac{25.87*1*20}{1}$

Q = 517.4 W

Therefore, the rate of heat conduction through the layer of still air is 517.4 W

6 0

3 years ago

Read 2 more answers

3. If I run 150m in 30 seconds, what speed will I have been running at?

Radda [10]

Answer:

speed = distance/time

Explanation:

speed = 150/30

speed =5m/s

you were running fast .....5m/s is a good speed

7 0

2 years ago

PLEASE HURRY AND HELP ME

laila [671]

Answer:

I believe the answer would be C. point z

5 0

3 years ago