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3 years ago

How much work is done when a hoist lifts a 290-kg rock to a height of 7 m?

1 answer:

5 0

Given: Mass m = 290 Kg; Height h = 7 m

Required: Work = ?

Formula: Work = Force x distance but, Force F = mg

W= fd

W = mgh

W= (290 Kg)(9.8 m/s²)(7 m)

W = 19,894 Kg.m²/s²

W = 19,894 J

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An RL circuit contains a resistor with R = 6800 Ω and an inductor with L = 2300 µH. If the impedance of this circuit is 160,000

Rainbow [258]

| Impedance | = √ [R² +(ωL)²]

R² = 6800² = 4.624 x 10⁷

(ωL)² = (2 · π · f · 2.3 · 10⁻³)²

= 2.0884 x 10⁻⁴ f²

| Z | = √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ] = 1.6 x 10⁵

   (1.6 x 10⁵)² = (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)

   (2.56 x 10¹⁰) - (4.624 x 10⁷) = 2.0884 x 10⁻⁴ f²

Frequency² = (2.56 x 10¹⁰ - 4.624 x 10⁷) / 2.0884 x 10⁻⁴

      = 2.555 x 10¹⁰ / 2.0884 x 10⁻⁴

=    1.224 x 10¹⁴

      =      122,400 GHz    <== my calculation

    11.1 MHz <== online impedance calculator

Obviously, I must have picked up some rounding errors
in the course of my calculation.

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3 years ago

Why is sun called a source of light??

mote1985 [20]

Answer:

because he give heat and energy

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3 years ago

Does anyone know this???

Gelneren [198K]

Answer:

Explanation:

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3 years ago

A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at

yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

Given:

Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

$\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)$

Therefore, the work done by the worker in lifting the bucket is given as:

$W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)$

Now, plug in the values given and solve for 'W'. This gives,

$W=(20\ kg)(9.8\ m/s^2)(25\ m)+\frac{1}{2}(20\ kg)(4^2-0^2)\ m^2/s^2\\\\W=4900\ J +160\ J\\\\W=5060\ J$

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

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3 years ago

On a distance-time graph, time is shown on the y-axis.<br> A) True<br> B) False

Artist 52 [7]

Answer:

false : In distance time graph,time is shown on the x -axis

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2 years ago