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1 year ago

If the tension in the rope is 160 n, how much work does the rope do on the skier during a forward displacement of 270 m?

2 answers:

8 0

If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

Given,

Tension force in the rope is (T) = 160 N

Displacement of the skier (S) = 270 m

The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.

Therefore, work done by the rope on the skier is,

$W=T.S$

⇒ $W=270*160*cos\pi \\W=-43200 J$

Hence work done by the rope is - 43200 J.

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Mandarinka [93]1 year ago

6 0

The correct answer is - 43200 J.

If the tension in the rope is 160 n, - 43200 J work doen by the rope on the skier during a forward displacement of 270 m.

Given,

Tension force in the rope is (T) = 160 N

Displacement of the skier (S) = 270 m

The displacement takes place in forward direction while the direction of the tension in the rope is opposite to it.

Therefore, work done by the rope on the skier is,⇒

Hence work done by the rope is - 43200 J.

What is work ?

Work is the energy transferred to or from an object by applying force along a displacement in physics. It is frequently represented as the product of force and displacement in its most basic form. When a force is applied, it is said to do positive work if it has a component in the direction of the point of application's displacement. If a force has a component that is opposite the direction of displacement at the force's application point, it does negative work.

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We are given the equation:

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For (b)

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2 years ago

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When copper carbonate is heated, it decomposes to form a black residue. This black residue is actually copper(II) oxide. Along with this black residue, carbon-dioxide is released. The complete reaction is the following:

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A wooden block of dimensions of 6 cm ⨯ 6 cm ⨯ 15 cm floats with the long axis oriented horizontally and the of the square faces

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Explanation:

(a) Formula to calculate volume of the submerged wooden block is as follows.

$V_{sub} = l \times w \times d$

It is given data of the wooden block is as follows.

depth = 7.96 cm, length (l) = 6 cm

width (w) = 4 cm,

So, we will calculate the volume of the submerged wooden block as follows.

$V_{sub} = l \times w \times d$

= $6 \times 6 \times 7.96$

= 286.56 $cm^{3}$

Hence, the submerged volume of the block is 286.56 $cm^{3}$ .

(b) Expression for the buoyant force acting on the wooden block is as follows.

$F_{B} = \rho_{w} g V_{sub}$

And, expression for the force of gravity of the wooden block is as follows.

$F_{g} = m_{b}g$

As the wooden block is floating on the water hence, buoyant force is balanced by the weight of the block.

$F_{g} = F_{B}$

Hence, mass of the wooden block will be calculated as follows.

$F_{g} = F_{B}$

$m_{b}g = \rho_{w}gV_{sub}$

$m_{b} = \rho_{w}V_{sub}$

= $997 kg/m^{3} \times 286.56 cm^{3}$

= $997 kg/m^{3} \times 286.56 \times 10^{-6} m^{3}$

= 0.02857 kg

Therefore, mass of the given block is 0.02857 kg

(c) Expression for the density of the block is as follows.

$\rho_{b} = \frac{m_{b}}{V_{b}}$

Now, expression for the total volume of the wooden block is as follows.

$V_{b} = l \times w \times h$

Hence, density of the given block is as follows.

$\rho_{b} = \frac{m_{b}}{V_{b}}$

= $\frac{m_{b}}{lwh}$

= $\frac{0.02857 kg}{4 \times 4 \times 15}$

= $1.19 \times 10^{-4} kg/cm^{3}$

Therefore, density of the given block is $1.19 \times 10^{-4} kg/cm^{3}$ .

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