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kumpel [21]
3 years ago
5

Explain whyusedwhy composite samples​

Physics
1 answer:
asambeis [7]3 years ago
4 0

Answer:

samples are combined, thoroughly homogenized, and treated as a single sample. Composite sampling can improve spatial or temporal coverage of an area without increasing sample number. Appropriateness of composite sampling is dependent upon the sampling objectives and the site characteristics.

Explanation:

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A solution is prepared by dissolving 49.3 g of KBr in enough water to form 473 mL of solution. Calculate the mass % of KBr in th
son4ous [18]

Answer:

9.31%

Explanation:

We are given that

Mass of KBr=49.3 g

Volume of solution=473 mL

Density of solution =1.12g/mL

We have to find the mass% of KBr.

Mass =volume\times density

Using the formula

Mass of solution=1.12\times 473=529.76 g

Mass % of KBr=\frac{mass\;of\;KBr}{Total\;mass\;of\;solution}\times 100

Mass % of KBr=\frac{49.3}{529.76}\times 100

Mass % of KBr=9.31%

Hence, the mass% of KBr=9.31%

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3 years ago
Suppose scientists discover two new moons.The average surface temperature of one of the moons is –180°C, but the temperature can
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The answer is a in the center of active volcanos
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A current carrying loop of wire lies flat on a table top. When viewed from above, the current moves around the loop in a counter
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Answer:

D &B

Explanation:

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5 0
3 years ago
On the sonometer shown below, a horizontal cord of length 5 m has a mass of 1.45 g. When the cord was plucked the wave produced
Korolek [52]

Answer:

(a) T = 0.015 N

(b) M = 1.53 x 10⁻³ kg = 1.53 g

Explanation:

(a) T = 0.015 N

First, we will find the speed of waves:

v =f\lambda

where,

v = speed of wave = ?

f = frequency = 120 Hz

λ = wavelength = 6 cm = 0.06 m

Therefore,

v = (120 Hz)(0.06 m)

v = 7.2 m/s

Now, we will find the linear mass density of the coil:

\mu = \frac{m}{l}

where,

μ = linear mass density = ?

m = mass = 1.45 g = 1.45 x 10⁻³ kg

l = length = 5 m

Thereforre,

\mu = \frac{1.45\ x\ 10^{-3}\ kg}{5\ m}\\\\\mu = 2.9\ x\ 10^{-4}\ kg/m

Now, for the tension we use the formula:

v = \sqrt{\frac{T}{\mu}}\\\\7.2\ m/s = \sqrt{\frac{T}{2.9\ x\ 10^{-4}\ kg/m}}\\\\(51.84\ m^2/s^2)(2.9\ x\ 10^{-4}\ kg/m) = T

<u>T = 0.015 N</u>

<u></u>

(b)

The mass to be hung is:

T = Mg\\\\M = \frac{T}{g}\\\\M = \frac{0.015\ N}{9.8\ m/s^2}\\\\

<u>M = 1.53 x 10⁻³ kg = 1.53 g</u>

4 0
3 years ago
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