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Talja [164]
3 years ago
5

The rock and meterstick balance at the 25-cm mark, as shown in the sketch. The meterstick has a mass of 1 kg. What must be the m

ass of the rock? (Show work).

Physics
1 answer:
liraira [26]3 years ago
3 0

Answer:

<h2>1 kg</h2>

Explanation:

Check the diagram attached below for the diagram.

Let the weight of the rock be W and the mass of the meter stick be M. Note that the mass of the meter stick will be placed at the middle of the meter stick i.e at the 50cm mark

Using the principle of moment to calculate the weight of the rock. It states that the sum of clockwise moments is equal to the sum of anti clockwise moment.

Moment = Force * perpendicular distance

The meterstick acts in the clockwise direction while the rock acys in the anti clockwise direction

Clockwise moment = 1kg * 25 = 25kg/cm

Anticlockwise moment = W * 25cm = 25W kg/cm

Equating both moments of forces

25W = 25

W = 25/23

W = 1 kg

The mass of the rock is also 1 kg

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» An electron is lighter than a proton.

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hence it's mass number is zero

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8 0
3 years ago
Tarzan swings on a 26.2 m long vine initially inclined at an angle of 28° from the vertical. (a) What is his speed at the bottom
Marianna [84]

Answer

given,

length of the swing = 26.2 m

inclined at an angle = 28°

let, the initial height of the Tarzan be h

h = L (1 - cos θ)

a) initial velocity v₁ = 0 m/s

   final velocity of Tarzan = v_f

law of conservation of energy

  PE_i + KE_i = PE_f + KE_f

mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2

       mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2

          mgh_i = \dfrac{1}{2}mv_f^2

             v_f = \sqrt{2gh_i}

                   = \sqrt{2gL(1- cos\theta)}

                   = \sqrt{2\times 9.8 \times 26.2(1- cos 28^0)}

                          = 7.75 m/s

the speed tarzan at the bottom of the swing

v_f = 7.75 m/s

b)initial speed of the  = 3 m/s

mgh_i + \dfrac{1}{2}mv_i^2= mgh_f + \dfrac{1}{2}mv_f^2

       mgh_i + 0 = 0 + \dfrac{1}{2}mv_f^2

          mgh_i+ \dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2

          gh_i+ \dfrac{1}{2}v_i^2 = \dfrac{1}{2}v_f^2

             v_f = \sqrt{v_1^2+2gh_i}

             v_f = \sqrt{3^2+2\times 9.8 \times (1- cos 28^0)}

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