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Vladimir79 [104]
2 years ago
9

Force F1 acts on a particle and does work W1. Force F2 acts simultaneously on the particle and does work W2. The speed of the pa

rticle does not change. Which one of the following must be true?a. W1 is zero, and W2 is positiveb. W1 = - W2c. W1 is positive, and W2 is positived. W1 is positive, and W2 is zero
Physics
1 answer:
egoroff_w [7]2 years ago
7 0

Answer:

b.) W1 = -W2

Explanation:

According to Newton's third law of motion, action and reaction are equal and opposite. For the particle to maintain a constant speed, it means that the two workdone on the particle are of eqaul magnitude but act in opposite direction. The two weights, instead of adding up annul each other and has no effect on the speed of the particle.

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It would be B and D your welcome


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What is the value of work done on an object when a 0.1x102-newton force moves it 30 meters and the angle between the force and t
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Answer:306joules

Explanation:

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2 years ago
A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial
Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

               t=\frac{2usin\theta}{g}

           Here θ = 30° and t = 20 s

a) t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s

  Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

                  H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m

      Maximum altitude = 490.5 m

c) Range of projectile is given by

                              R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m

    Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

   Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

   We have equation of motion s = ut + 0.5 at²

   Horizontal motion

                         u = 169.91 m/s

                         a = 0 m/s²

                          t = 15 s

                Substituting

                          s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

      Vertical motion

                         u = 98.1 m/s

                         a = -9.81 m/s²

                          t = 15 s

                Substituting

                          s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

   \texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m

   Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0
2 years ago
A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it
makkiz [27]

Answer:D

Explanation:

Given

mass of object m=5 kg

Distance traveled h=10 m

velocity acquired v=12 m/s

conserving Energy at the moment when object start falling and when it gains 12 m/s velocity

Initial Energy=mgh=5\times 9.8\times 10=490 J

Final Energy=\frac{1}{2}mv^2+W_{f}

=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}

where W_{f} is friction work if any

490=360+W_{f}

W_{f}=130 J

Since Friction is Present therefore it is a case of Open system and net external Force is zero

An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .

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Experimentation has shown that the period of time that is person will test possible for gunshot residue is 5.27 days.

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