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IRISSAK [1]
3 years ago
15

A boat with a horizontal tow rope pulls a waterskier. She skis off to the side, the rope makes an angle of 15° with the forward

direction of motion. If the tension in the rope is 180 N, how much work does the rope two on the scare during the forward displacement of 300 m?
Physics
1 answer:
dsp733 years ago
4 0

Answer:

<h3>13,976.23Joules</h3>

Explanation:

Workdone by the rope is expressed using the formula;

W = Fd sin(theta)

F is the tension in the rope = 180

d is the displacement = 300m

theta is the angle of inclination = 15°

Substitute the given parameters into the formula;

W = 180(300)sin15

W = 54000sin 15

W = 13,976.23

Hence the workdone by the rope is 13,976.23Joules

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A pendulum has 711 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
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According to law of conservation of energy, 
<span>Energy can neither be constructed nor be destroyed but can be transformed from one form to another.
</span>
<span>At the highest point of the pendulum(point b), pendulum is associated with potential energy only and no kinetic energy.
</span><span>Therefore total energy at point b = potential energy = 711 J.... i
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<span>At the bottom most point(point a), pendulum is associated only with kinetic energy and no potential energy.
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Hence kinetic energy at the bottom most point is 711 J.
Hope this helps!!

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Explain how you think an asteroid impact could affect the tilt of Earth’s axis. Explain how this effect would change Earth’s sea
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3 years ago
If BHALA AHMAD KHAN applied the 20N force is applied on an object moving with the velocity 30 m/s. calculate the power in KW.
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Answer:0.6kw

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3 years ago
You hang a heavy ball with a mass of 10 kg from a gold wire 2.6 m long that is 1.6 mm in diameter. You measure the stretch of th
PolarNik [594]

<u>Answer:</u> The Young's modulus for the wire is 6.378\times 10^{10}N/m^2

<u>Explanation:</u>

Young's Modulus is defined as the ratio of stress acting on a substance to the amount of strain produced.

The equation representing Young's Modulus is:

Y=\frac{F/A}{\Delta l/l}=\frac{Fl}{A\Delta l}

where,

Y = Young's Modulus

F = force exerted by the weight  = m\times g

m = mass of the ball = 10 kg

g = acceleration due to gravity = 9.81m/s^2

l = length of wire  = 2.6 m

A = area of cross section  = \pi r^2

r = radius of the wire = \frac{d}{2}=\frac{1.6mm}{2}=0.8mm=8\times 10^{-4}m      (Conversion factor:  1 m = 1000 mm)

\Delta l = change in length  = 1.99 mm = 1.99\times 10^{-3}m

Putting values in above equation, we get:

Y=\frac{10\times 9.81\times 2.6}{(3.14\times (8\times 10^{-4})^2)\times 1.99\times 10^{-3}}\\\\Y=6.378\times 10^{10}N/m^2

Hence, the Young's modulus for the wire is 6.378\times 10^{10}N/m^2

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4 years ago
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