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sleet_krkn [62]
3 years ago
14

A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t

he other is unknown. The center of mass of this system is on the rod a distance 1.60 m from the 1.00-kg mass object. What is the mass of the other object?A) 0.250kg B) 4.00kg C) 4.11kg D) 0.800kg E) 3.22kg
Physics
1 answer:
8090 [49]3 years ago
3 0

Answer:

<h2> 4kg</h2>

Explanation:

Step one:

given

length of rod=2m

mass of object 1 m1=1kg

let the unknown mass be x

center of mass<em> c.m</em>= 1.6m

hence 1kg is 1.6m from the <em>c.m</em>

and x is 0.4m from the <em>c.m</em>

Taking moment about the <em>c.m</em>

<em>clockwise moment equals anticlockwise moments</em>

1*1.6=x*0.4

1.6=0.4x

divide both sides by 0.4 we have

x=1.6/0.4

x=4kg

The mass of the other object is 4kg

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<h2>a</h2><h2> move with an increasing velocity</h2>
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John stirs the hot soup with a spoon the heat travels up the spoon to his hand by what process does the heat travel up the spoon
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Answer: is<u><em> Second Law of Thermodynamics (conduction)</em></u>.

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What are saturated and non saturated vapour? ​
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A saturated vapour is one that is in balance with its own liquid. Space is considered to be unsaturated if it contains fewer vapours than the maximum amount it can retain at a given temperature. Saturated vapour pressure is unaffected by volume, although it rises as temperature rises.

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8 0
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Given this graph plotting velocity versus time, estimate the acceleration of object A at points X and Y respectively.
swat32

Point X lies on a horizontal line. We can intuitively say that the slope of the graph at point X is 0, therefore the acceleration at point X is 0m/s²

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m = (y₂-y₁)/(x₂-x₁)

m = slope, (x₁, y₁) and (x₂, y₂) correspond to the coordinates of the line's endpoints.

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3 years ago
Read 2 more answers
Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m an
galben [10]

Answer:

\triangle P=1.95*10^{-4}

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Mass m=0.001

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Length l=10m

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Generally the equation for Friction factor is mathematically given by

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Generally the equation for Friction factor is mathematically given by

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Where

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\triangle P=\frac{19.9*10^{-9}}{10^3*(9.81)}

\triangle P=H*\rho g

\triangle P=1.95*10^{-4}

 

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