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3 years ago

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Which of the following is a unit of mass?

2 answers:

Bogdan [553]3 years ago

7 0

the answer is c: kilogram

explanation: The standard International System of Units (SI) unit of mass is the kilogram (kg). The kilogram is 1000 grams (g), first defined in 1795 as one cubic decimeter of water at the melting point of ice.

larisa86 [58]3 years ago

5 0

Hi there!

Which of the following is a unit of mass?

Well, lets start with pound.

A pound will not be a unit of mass, it is a unit of weight because it holds a force of gravity on an object.

A millimeter is not a unit of mass, because it is a type of measure.

A newton is a unit of force, not a unit of mass.

Therefore, the only unit of mass will be Kilogram.

Hope this helped!~

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A car traveling at 60km/h undergoes uniform acceleration at a rate of 2/ms^2 until is velocity reached 120km/h determine the dis

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Given that,

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3 years ago

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3 years ago

Read 2 more answers

Why do things become hot or cold?<br> Relate to Thermal Energy and The Law of Conservation of Energy

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<u>Answer:</u>

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3 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

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$A=0.0198$

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$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

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From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

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From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

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