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sammy [17]
3 years ago
12

Absolute differences between each observed mass and the accepted value,divided by the accepted value, multiplied by 100%

Physics
1 answer:
timama [110]3 years ago
3 0

Answer:

The percentage error    e% = Δm/m 100    

Explanation:

The absolute error is the assessment with which measurements are made, it is given by the minimum reading value of a given instrument.

The accepted value is the value measured in an instrument. Or the one calculated by an experiment.

The percentage error is the ratio of the absolute error between the value of the average per 100

      e% = Δm/m 100

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Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec
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Answer:

(a) F_{net} = 4.19 x 10^{-5} N, and its direction is towards m_{2}.

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Explanation:

Let, m_{1} = 165 kg, m_{2} = 465 kg, m_{3} = 60 kg, and the distance between m_{1} and m_{2} is 0.340 m.

(a) Since m_{3} is placed midway between m_{1} and m_{2}, then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = \frac{Gm_{1}m_{2}  }{r^{2} }

Where all variables have their usual meaning.

Then,

a. F_{net} = F_{23} - F_{13}

F_{13} = \frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }

     = 2.25 x 10^{-5} N

F_{23} = \frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }

     = 6.44 x 10^{-5} N

∴ F_{net} =  = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

The net force exerted by the two masses on the 60 kg object is 4.19 x 10^{-5}  N.

(ii) /F_{net}/ = /F_{23}/ - /F_{13}/

              = 6.44 x 10^{-5} - 2.25 x 10^{-5}

              = 4.19 x 10^{-5} N

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(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0
3 years ago
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Answer:

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