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A) Define the term mechanical advantage of a machine. b) The Figure below shows a pulley system being used to raise a load.

Physics

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Answer

c

−0.6 m/s2\mathbf{-0.6\, \text{m/s}^2}

Step-by-step solution

5 steps

Huu hapa ni ufumbuzi wa swali la 3:

Part a· 1 step

Eleza mwendo kati ya O na A

  1. Angalia grafu kati ya O na A.

    Grafu inaonyesha kasi ikiongezeka kutoka 0 m/s0\, \text{m/s} hadi 12 m/s12\, \text{m/s} kwa mstari ulionyooka.

    Mwendo kati ya O na A ni kasi inayoongezeka kwa usawa (uniform acceleration).

Part b· 1 step

Eleza mwendo kati ya A na B

  1. Angalia grafu kati ya A na B.

    Grafu inaonyesha kasi ikiwa 12 m/s12\, \text{m/s} na haibadiliki.

    Mwendo kati ya A na B ni kasi isiyobadilika (uniform velocity).

Part c

−0.6 m/s2\mathbf{-0.6\, \text{m/s}^2}

Bainisha mchapuko kati ya B na C

  1. Tambua pointi B na C kutoka kwenye grafu.

    Kwenye B, muda tB=100 st_B = 100\, \text{s} na kasi vB=12 m/sv_B = 12\, \text{m/s}.
    Kwenye C, muda tC=120 st_C = 120\, \text{s} na kasi vC=0 m/sv_C = 0\, \text{m/s}.

  2. Tumia fomula ya mchapuko.

    Mchapuko aa hutolewa na mabadiliko ya kasi kugawanya kwa mabadiliko ya muda:

    a=vC−vBtC−tBa = \frac{v_C - v_B}{t_C - t_B}
  3. Badilisha thamani na ukokotoe.

    a=0 m/s−12 m/s120 s−100 sa = \frac{0\, \text{m/s} - 12\, \text{m/s}}{120\, \text{s} - 100\, \text{s}} a=−12 m/s20 sa = \frac{-12\, \text{m/s}}{20\, \text{s}} a=−0.6 m/s2a = -0.6\, \text{m/s}^2

    Mchapuko kati ya B na C ni −0.6 m/s2\boxed{\mathbf{-0.6\, \text{m/s}^2}}.

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