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Nadusha1986 [10]
3 years ago
15

What do you mean by unit lb

Physics
2 answers:
Bingel [31]3 years ago
7 0

Unit lb mean pound. Which equals to about 0.45 kilograms(kg) /4536 grams,while 1kg equals to about 2.2lb.

Hope it helps!

andreyandreev [35.5K]3 years ago
4 0
Unit Pounds are often referenced to kilograms
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Which statement is true? A) The work done to lift an object 6 meters is greater than the gravitational potential energy it gains
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The work required to raise an object to a height is equal to the gravitational potential energy the object gains. <em>(C)</em>

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4 years ago
A car rolls backward from -2 to -10 m/s in 4 seconds. what is the acceleration. PLEASE HELP WILL GIVE BRAINLIEST
pashok25 [27]

\bf \underline{Given : }

  • Initial velocity,u = -2 m/s

  • Final velocity,v = -10 m/s

  • Time taken, t = 4 seconds

\bf \underline{To  \: find  \: out : }

Find the acceleration ( a ) .

\bf \underline{Solution : }

We know that,

\sf \: Acceleration =  \dfrac{v - u}{t}

Substituting the values in the above formula, we get

\implies \sf \: Acceleration =  \dfrac{ - 10 - ( - 2)}{4}

\implies\sf \: Acceleration =  \dfrac{ - 10 + 2}{4}

\implies \sf \: Acceleration = \cancel  \dfrac{ - 8}{4}

\implies\sf \: Acceleration =  - 2 \: ms {}^{ - 2}

Hence,the acceleration of a body is -2 m/s².

5 0
3 years ago
Atoms with the same atomic number but different atomic mass are called
Alexxx [7]
<span>Atoms with the same atomic number but different atomic mass are called:

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4 0
3 years ago
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An airplane is flying at a velocity of 122.4 m/s. It is getting ready to land so it slows down by accelerating at a rate of -2.8
PIT_PIT [208]

Answer:

115.12

Explanation:

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8 0
3 years ago
A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕
Slav-nsk [51]

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

6 0
3 years ago
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