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cupoosta [38]
3 years ago
12

10.0 ppm 10.4 ppm 10.2 ppm 10.8 ppm 10.1 ppm 10.5 ppm 10.1 ppm Using the new instrument on the first day the technicians got a v

alue of 9.0 ppm. Are the technicians right to exclude the measurement of the first day as an outlier at the 90% confidence level?
Physics
1 answer:
iren [92.7K]3 years ago
3 0

Answer:

The datapoint 9.0 ppm is outlier at the 90% confidence level.

Explanation:

The old data has following values

mean=10.5 mm

standard deviation 0.2 mm

Now the mean of new values is calculated as following

mean=\frac{10+10.4+10.2+10.8+10.1+10.5+10.1}{7}\\mean=10.3 ppm

So the value as 9.0 ppm can be considered easily as outlier in  this regard.

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A particle moves horizontally in uniform circular motion, over a horizontal xy plane. At one instant, it moves through the point
seropon [69]

since centripetal acceleration is always towards the center of the circle

so at the given position where speed and acceleration is given the center coordinate will be towards the center of circle

also we know that

a_c = \frac{v^2}{r}

11.6 = \frac{2.5^2}{R}

R = 0.54 m

so the coordinates of the center will be

(x, y) = (3.20 , (3.5 + 0.54))

(x, y) = (3.20, 4.04)

so the coordinate is (3.20 m, 4.04 m)

6 0
3 years ago
these are the types of attractions found between molecules. their strength determines the state of the substance at room tempera
Elina [12.6K]
<h2>Answer</h2>

The physical state of the elements depends upon the <u>attraction forces </u>and their <u>kinetic energy</u>.

<h2>Explanation</h2>

The elements or substances are fixed with each other with the help of different chemical forces including ionic bonding, covalent bonding, H- bonding etc. The strength of these forces is also one of the factors that affect their physical natures. For example, covalent or ionic bonds are the strongest bonds than all other bonds and metals that contain these forces are mostly in solid form. The kinetic motion of electrons in the element also affects the physical state of the element and potential of bonding.

7 0
3 years ago
Read 2 more answers
Does a light bulb with a greater wattage have a greater brightness
Tamiku [17]
Only within the same technology. / / / If both of the bulbs you're comparing are incandescent, or both fluorescent, or both CFL, or both LED, then the one that uses more power is brighter. But a CFL with the same brightness as an incandescent bulb uses less power, and an LED bulb with the same brightness as both of those uses less power than either of them.
8 0
2 years ago
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a piece of metal with a mass of 15.3 grams has a temperature of 50.0°C. When the metal is placed in 80.2 grams of water at 21.0°
AleksAgata [21]

Answer:

1.21

Explanation:

Heat rise in the body happens due to heat supplied by water to the body.

Heat rise in body = m₁ c₁ ΔT₁

Where m₁ is mass of body and c₁ is its specific heat of body

Heat lost from water to the body = m₂ c₂ ΔT₂

Where m₂ is mass of water and c₂ is its specific heat of water ( c₂ =1 (since water))

Equating both:

        15.3 x c₁ x 4.3 = 80.2 x 1 x 4.3

⇒   c₁ = 80.2 / (15.3 x 4.3) = 1.21

6 0
3 years ago
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Positive Charge is distributed along the entire x axis with a uniform density 12 nC/m. A proton is placed at a position of 1.00
lions [1.4K]

Answer:

b.  \Delta KE = 390 eV

Explanation:

As we know that the electric field due to infinite line charge is given as

E =\frac{\lambda}{2\pi \epsilon_0 r}

here we can find potential difference between two points using the relation

\Delta V = \int E.dr

now we have

\Delta V = \int(\frac{\lambda}{2\pi \epsilon_0 r}).dr

now we have

\Delta V = \frac{\lambda}{2\pi \epsilon_0}ln(\frac{r_2}{r_1})

now plug in all values in it

\Delta V = \frac{12\times 10^{-9}}{2\pi \epsilon_0}ln(\frac{1+5}{1})

\Delta V = 216ln6 = 387 V

now we know by energy conservation

\Delta KE = q\Delta V

\Delta KE = (e)(387V) = 387 eV

3 0
3 years ago
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