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Nady [450]
3 years ago
9

In which situation is maximum work considered to be done by a force

Physics
2 answers:
madam [21]3 years ago
7 0

Answer:

The angle between the force and displacement is 0°.

Explanation:

Katena32 [7]3 years ago
3 0

Answer:

  • Work done is maximum when the movement of object is in line and direction of force.  

OR

  • Work done is maximum, when displacement takes place along the direction of force.
  • Work done is given by the equation

                                W = F.S

                            <em>     W  = F. S cos Θ</em>

<em>When cos Θ = 0° ; cos 0 = 1</em>

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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)
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Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is U  =  \int\limits^T_0 {P(t)} \, dt

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Answer:

The value is  U  =  7.563 *10^{5} \  J

Explanation:

From the question we are told that

   The power rating of the bulb is P  =  100 \  W

   The resistance is   R =  143 \ \Omega

   The  voltage is  V  =  V_o  sin [2 \pi ft]

   The  energy expanded is U  =  \int\limits^T_0 {P(t)} \, dt

   The  voltage  V_o  =  110 \  V

   The frequency is  f =  60 \  Hz

    The  time considered is  t =  5 \  h  =  18000 \  s

Generally power is mathematically represented as

             P =  \frac{V^2}{ R}

=>          P =  \frac{( 110  sin [2 \pi * 60t])^2}{ 144}

=>           P =  \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}

So  

     U  =  \int\limits^T_0 { \frac{ 110^2*  [sin [120 \pi t])^2}{ 144}} \, dt

=>  U  =  \frac{110^2}{144} \int\limits^T_0 { (   sin^2 [120 \pi t]} \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | T} \atop {0}} \right.

=>  U =  \frac{110^2}{144} [\frac{t}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi t)}{240 \pi} ] ]\left  | 18000} \atop {0}} \right.

U =  \frac{110^2}{144} [\frac{18000}{2}  - [\frac{1}{2} *  \frac{sin(240 \pi (18000))}{240 \pi} ] ]

=>   U  =  7.563 *10^{5} \  J

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Answer:

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you are exerting force of 10 N on the ball.

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normal force acting on the ball will be

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F n = 10 - 1 × 9.8                        

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Wow !  I understand your shock.  I shook and vibrated a little
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The reason for our shock is all the extra junk in the question,
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That's all very picturesque, and it excites cosmic fantasies in
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It only gets in the way, and doesn't help a bit.

The real question is:

What is the acceleration of gravity 2000 m from
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      =  (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²

      =  (6.67 x 10⁻¹¹  ·  1.1 x 10³¹ / 4 x 10⁶)      (N) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴           (kg · m / s²) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴            m / s²      

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