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Zorluk: OrtaMeasurement of Mass and Weight

A chemical balance is used to determine the mass of a metallic sphere, giving a reading of 6.0 kg6.0\text{ kg}. The sphere is then suspended from a spring balance at a laboratory where the local acceleration due to gravity is 10 m s210\text{ m s}^{-2}. If the spring balance reads 3.0 N-3.0\text{ N} before any load is attached, what is the indicated reading on the spring balance when the sphere is suspended?

  1. 57.0 N57.0\text{ N}Cevap
  2. B
    63.0 N63.0\text{ N}
  3. C
    60.0 N60.0\text{ N}
  4. D
    6.0 N6.0\text{ N}

Cevap

The indicated reading on the spring balance is 57.0 N57.0\text{ N}.
The true weight of the metallic sphere is calculated as W=mg=6.0 kg×10 m s2=60.0 NW = mg = 6.0\text{ kg} \times 10\text{ m s}^{-2} = 60.0\text{ N}. Because the spring balance has an initial zero reading of 3.0 N-3.0\text{ N} before any load is attached, the indicated reading when the sphere is hung is 60.0 N3.0 N=57.0 N60.0\text{ N} - 3.0\text{ N} = 57.0\text{ N}.

Adım Adım Çözüm

1
Calculate the true weight of the metallic sphere
W=mg=6.0 kg×10 m s2=60.0 NW = mg = 6.0\text{ kg} \times 10\text{ m s}^{-2} = 60.0\text{ N}
Weight is the gravitational force acting on mass.
2
Set up the zero error relationship for instrument measurement
\text{Actual Weight} = \text{Indicated Reading} - \text{Zero Reading}
The actual physical value equals the observed pointer reading minus the zero reading of the unloaded instrument.
3
Substitute known values and solve for the indicated reading (RR)
60.0 N=R(3.0 N)    R=60.03.0=57.0 N60.0\text{ N} = R - (-3.0\text{ N}) \implies R = 60.0 - 3.0 = 57.0\text{ N}
Solving the linear relation yields an indicated pointer position of 57.0 N57.0\text{ N}.

Anahtar Kavram

Mass versus Weight Measurement and Zero Error Correction
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