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Anastasy [175]
2 years ago
7

A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end

of the wire, a pulse travels down the wire and returns 7.84 ms later, having reflected off the lower end. The stone is heavy enough to prevent the lower end of the wire from moving. What is the mass m of the stone
Physics
1 answer:
Triss [41]2 years ago
3 0

Answer: m= 35.6 kg

Explanation:

For finding the mass of the stone we have the formula

v= \sqrt{\frac{Tension}{Linear. Mass. density} }

Here, Tension= m*g = m*9.81

and linear mass density= \frac{8.25 g}{65 cm}

Linear mass density= \frac{8.25*10^-3}{65*10^-2}

Linear mass density= 0.0127 kg/m

Velocity= 2*\frac{l}{t}

Velocity= 2 * \frac{65*10^-2}{7.84}

Velocity= 165.8 m/s

So putting all these values in equation we get

v= \sqrt{\frac{Tension}{Linear. Mass. density} }

165.8= \sqrt{\frac{m*9.81}{0.0127} }

Solving we get

m= 35.58 kg

or m= 35.6 kg

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In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li
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Total energy saving will be 0.8 KWH

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6 0
2 years ago
On the earth, when an astronaut throws a 0.250-kg stone vertically upward, it returns to his hand a time T later. On planet X he
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Answer:

correct is d) a ’= g / 2

Explanation:

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     a = (v₀- v) / T

On planet X

    v = v₀ - a' t’

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1 year ago
The coefficient of friction between the 2-lb block and the surface is μ=0.2. The block has an initial speed of Vβ =6 ft/s and is
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Answer:

x = 0.0685 m

Explanation:

In this exercise we can use the relationship between work and energy conservation

            W = ΔEm

Where the work is

             W = F x

The energy can be found in two points

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           Em₀ = K = ½ m v²

Final. When the system is at rest

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            ∑ F = F - fr

Axis y

           N- W = 0

           N = W

The friction force has the equation

          fr = μ N

          fr = μ W

  The job

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            (F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)

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We substitute values ​​and solve

           ½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0

         x² 30 + 14.4 x - 1,125 = 0

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We solve the second degree equation

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        x₂ = -0.549 m

The first result results from compression of the spring and the second torque elongation.

The result of the problem is x = 0.0685 m

4 0
2 years ago
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