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4 years ago

20 kg object travels 28 meter and stops. coefficient friction= 0.085 how much work was done by friction?

1 answer:

7 0

Assuming it is on a horizontal surface:
friction = μR
R = 20g (g is gravity 9.81)
so Friction = 0.085 x 20g
Work done is force x distance
so Work done = 0.085 x 20g x 28
= 466.956 J

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The mass of a sample of sodium bicarbonate is 2. 1 kilograms (kg). There are 1,000 grams (g) in 1 kg, and 1 Times. 109 nanograms

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Mass is the quantity of the substance in the body or object. The SI unit of mass is Kilogram.

There are other units of measure,

Milligram: 1 g is equal to the $\bold {10^3 \ mg}$

Micro-gram: 1 g is equal to $\bold {10^{6} \ \mu g}$

Nano-gram: 1 g is is equal to $\bold {10^{9} \ ng}$

First convert kg to gram,

Since, 1 Kg = 1000 g

2.1 kg = grams of sample

So,

Do the cross multiplication,

$\rm mass\ of\ sample = \dfrac {2.1\ kg \times 1000\ g }{ 1 kg}\\\\\rm mass\ of\ sample =2100 g$

Now, convert 2100 g to nano-grams

Since, 1 g = 1 x 10⁹ ng

2100 g = ng of sample

So,

Do the cross multiplication,

$\rm mass\ of\ sample = \dfrac {2100 g \times 1 \times 10^9 ng }{1\ g}\\\\\rm mass\ of\ sample = 2.1 \times 10^1^2 ng$

Therefore, 2.1 kg of sodium bicarbonate is equal to the 2.1 x 10¹² ng of sample.

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4 years ago

A Rankine oval is formed by combining a source-sink pair, each having a strength of 36 ft2/s and separated by a distance of 15 f

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Distance between the sink of the source and origin= $a=\frac{15}{2}$ ft

Uniform velocity, U=12 ft/s

We have to find the length of the oval.

Formula to find the half length of the body

$\frac{l}{a}=(\frac{m}{\pi Ua}+1)^{\frac{1}{2}}$

Where a=Distance between sink of source and origin

U=Uniform velocity

m=Strength

l=Half length

Using the formula

$\frac{l}{\frac{15}{2}}=(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}$

$l=\frac{2}{15}(\frac{36}{\pi\times 12\times \frac{15}{2}}+1)^{\frac{1}{2}}$

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