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Dennis_Churaev [7]
3 years ago
7

20 N effort is used in the 20 cm long spanner to unscrew a nut, then calculate momentum to unscrew the nut. [ Ans=4 Nm]

Physics
1 answer:
irga5000 [103]3 years ago
7 0

Answer:

Refer to the attachment.

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A child walks 5m east, then 3m north, then 1m east. What is the magnitude of the child's displacement? What is the direction of
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A book is sitting on a table. The book is
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65 Newtons

Explanation:

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A still ball of mass 0.514kg is fastened to a cord 68.7cm long and is released when the cord is horizontal. At the bottom of its
Alex787 [66]

Answer:

1.21 m/s

Explanation:

From the law of conservation of energy,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

where U₁ = initial potential energy of system =initial potential energy of still ball = mgh where m = mass of still ball = 0.514 kg, g = acceleration due to gravity = 9.8 m/s² and h = height = length of cord = 68.7 cm = 0.687 m.

K₁ = initial kinetic energy of system = 0

E₁ = initial internal energy of system = unknown and

U₂ = final potential energy of system = 0

K₁ = final kinetic energy of system = final kinetic energy of ball + steel block = 1/2(m + M)v² where m = mass of still ball, M = mass of steel block = 2.63 kg and v = speed of still ball + steel block

E₁ = final internal energy of system = unknown

So,

U₁ + K₁ + E₁ = U₂ + K₂ + E₂

mgh + 0 + E₁ = 0 + 1/2(m + M)v² + E₂

mgh = 1/2(m + M)v² + (E₂ - E₁)

Given that (E₂ - E₁) = change in internal energy = ΔE = 1/2ΔK where ΔK = change in kinetic energy. So, ΔE = 1/2ΔK = 1/2(K₂ - K₁) = K₂/2 = 1/2(m + M)v²/2 = (m + M)v²/4

Thus, mgh = 1/2(m + M)v² + (E₂ - E₁)

mgh = 1/2(m + M)v² + (m + M)v²/4

mgh = 3(m + M)v²/4

So, making v subject of the formula, we have

v² = 4mgh/3(m + M)

taking square root of both sides, we have

v = √[4mgh/3(m + M)]

Substituting the values of the variables into the equation, we have

v = √[4 × 0.514 kg × 9.8 m/s² × 0.687 m/{3(0.514 kg + 2.63 kg)}]

v = √[13.8422/{3(3.144 kg)}]

v = √[13.8422 kgm/s²/{9.432 kg)}]

v = √(1.4676 m²/s²)

v = 1.21 m/s

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The answer would be 2.4m/s


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