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Virty [35]
3 years ago
8

Explain why the mean and standard deviation are not appropriate for particle size distributions

Physics
1 answer:
Eduardwww [97]3 years ago
8 0
The mean is the average of the sample data, while standard deviation is the average of the differences of each data to the mean. This could only be valid if there is a normal distribution. This distribution pertains to a bell-shaped frequency graph. However, in particle size distribution, the frequency of the sizes is rarely concentrated on the medium or middle size. It may be skewed to the left or skewed to the right. So, the mean and standard deviation is not a good measure for this type of distribution.
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The net force is 10 n because it has 2 different forces that are not the same you must subtract them
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A water rocket has a mass of 0.8kg and is launched in a school playground with an inital upwards force of 12newtons. what is the
Darina [25.2K]

Weight = (mass) x (gravity).
It always acts downward.

On Earth, the acceleration of gravity is 9.807 m/s².
On the Moon, the acceleration of gravity is 1.623 m/s².

On Earth, the rocket's weight is  (0.8kg) x (9.8 m/s²) = 7.84 newtons

On the Moon, the rocket's weight is  (0.8kg) x (1.62 m/s²) = 1.3 newtons

The force of the rocket engine acts upward.
Its magnitude is 12 newtons. (From the burning chemicals.
Doesn't depend on local gravity. Same force everywhere.)

Now we have all the data we need to mash together and calculate the
answers to the question.  You might choose a different method, but the
machine that I have selected to do the mashing with is Newton's 2nd law
of motion:

                           Net Force = (mass) x (acceleration).
 
Since the question is asking for acceleration, let's first solve Newton's law
for it.  Divide each side by (mass) and we have

                           Acceleration = (net force) / (mass) .

On Earth, the forces on the rocket are

        (weight of 7.84 N down) + (blast of 12 N up) =  4.16 newtons UP (net)

         Acceleration = (4.16 newtons UP) / (0.8 kg) = 5.2 m/s² UP .

On the moon, the forces on the rocket are

         (weight of 1.3 N down) + (blast of 12 N up) = 10.7 newtons UP (net)

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3 years ago
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Gravity is the cause of formation waves.
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3 years ago
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a large cargo trucks needs to cross a bridge the truck is 30m long, and 3.2 wide, the cargo exerts a force of 54,000 N the bridg
otez555 [7]

Answer:

It isn't safe for the truck to cross the bridge because the pressure exerted by the truck on the bridge is greater than the maximum tolerable pressure for the bridge, 562.5 Pa > 450 Pa.

Explanation:

Pressure is expressed in Force/Area.

So, for the truck, force exerted = 54000 N

Area covered = 30 × 3.2 = 96 m²

Pressure exerted by the truck = 54000/96 = 562.5 Pa

The pressure exerted by the truck on the bridge is greater than the maximum tolerable pressure for the bridge, 562.5 > 450, hence, it isn't safe for the truck to cross the bridge.

5 0
4 years ago
A student is experimenting with some insulated copper wire and a power supply. She winds a single layer of the wire on a tube wi
OverLord2011 [107]

Answer:

P=214.7187\,W

Explanation:

Given that:

Diameter of the solenoid, D=10\,cm=0.1\,m

length of the solenoid, L=90\,cm=0.9\,m

diameter of the wire, d=0.1\,cm=10^{-3}\,m

magnetic field at the center of the solenoid, B=7.4\times 10^{-3}\,T

<u>Now we need the no. of turns incorporated in the length of 90 cm:</u>

N=\frac{Length\,\,of\,\,solenoid}{diameter\,\,of\,\, wire}

N=\frac{L}{d}

N=\frac{0.9}{10^{-3}}

N=900\,\,turns

For solenoids we have:

B=\mu.n.I ...............................(1)

where:

\mu=permeability of the medium

n = no. of turns per unit length

I = current in the coil

So,

n=\frac{900}{0.9}

n=1000\,turns\,.\,m^{-1}

Now putting the respective values in the eq. (1)

7.4\times 10^{-3}=4\pi\times10^{-7}\times 1000\times I

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  • For copper we have resistivity:
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We know that resistance is given by:

R=\rho.\frac{l}{a} .....................................(2)

where:

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a = cross sectional area of the conducting wire

<u>Now we need the length (l) of the wire:</u>

Circumference of the solenoid,

C=\pi.D

C=0.1\pi\,m

\therefore l=C\times N

l=90\pi\,m

&

<u>Cross-sectional area of wire:</u>

a=\pi.\frac{d^2}{4}

a=\pi. \frac{(10^{-3})^2}{4}\,m^2

<u>Resistance from eq. (2):</u>

R=1.72\times 10^{-8}\times \frac{90\pi}{\pi. \frac{(10^{-3})^2}{4}}

R=6.192 \,\Omega

  • For power we have:

P=I^2.R

P=5.8887^2 \times 6.192

P=214.7187\,W

6 0
3 years ago
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