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Volgvan
2 years ago
5

A boat with a horizontal tow rope pulls a water skier. She skis off to the side, so the rope makes an angle of 12.0 ∘ with the f

orward direction of motion. For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Pushing a stalled car. Part A If the tension in the rope is 200 N, how much work does the rope do on the skier during a forward displacement of 320 m?
Physics
1 answer:
Sladkaya [172]2 years ago
4 0

Answer:

62.601kJ

Explanation:

We have a force with a horizontal component given by the angle of 15°. So wue can take this component to calculate the total work done, that is,

W_T = Fd(cos\theta)

Our values are,

d=320m

Angle = 12\°

Force=200N,

Substituting in Work equation,

W_T = 200(320)(cos12)

W_T = 62601.4J

W_T = 62.601kJ

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If the car speeds up at a steady 1.6 m/s2 , how long after starting is the magnitude of its centripetal acceleration equal to th
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8 0
3 years ago
A wheel has a radius of 5.9 m. How far (path length) does a point on the circumference travel if the wheel is rotated through an
posledela

Answer:

(a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

Explanation:

Given that,

Radius = 5.9 m

(a). Angle \theta=30°

We need to calculate the angle in radian

\theta=30\times\dfrac{\pi}{180}

\theta=0.523\ rad

We need to calculate the path length

Using formula of path length

Path\ length =angle\times radius

Path\ length=0.523\times5.9

Path\ length =3.09\ m

(b). Angle = 30 rad

We need to calculate the path length

Path\ length=30\times5.9

Path\ length=177\ m

(c). Angle = 30 rev

We need to calculate the angle in rad

\theta=30\times2\pi

\theta=188.4\ rad

We need to calculate the path length

Path\ length=188.4\times5.9

Path\ length =1111.56\ m

Hence, (a). The path length is 3.09 m at 30°.

(b). The path length is 188.4 m at 30 rad.

(c). The path length is 1111.5 m at 30 rev.

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