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liubo4ka [24]
3 years ago
14

Jessica has a mass of 55kg she sleds down a hill that has a slope of 32 degrees. what is the component of her weight that is alo

ng her direction of motion ?

Physics
2 answers:
Snowcat [4.5K]3 years ago
8 0

Answer:

286 rounded, - A P E X

Explanation:

Arturiano [62]3 years ago
7 0

Answer:

W = 285.62 N

Explanation:

It is given that,

Mass of Jessica is 55 kg

Slope of the hill is 32 degrees

We need to find the component of her weight that is along her direction of motion.

The component along her direction of motion is shown in attached figure. It means

W_y=mg\sin\theta\\\\W_y=55\times 9.8\times \sin(32)\\\\W_y=285.62\ N

So, the component of her weight that is along her direction of motion is 285.62 N.

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The statement shows a case of rotational motion, in which the disc <em>decelerates</em> at <em>constant</em> rate.

i) The angular acceleration of the disc (\alpha), in revolutions per square second, is found by the following kinematic formula:

\alpha = \frac{\omega_{f}-\omega_{o}}{t} (1)

Where:

  • \omega_{o} - Initial angular speed, in revolutions per second.
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  • t - Time, in seconds.

If we know that \omega_{o} = \frac{5}{12}\,\frac{rev}{s}, \omega_{f} = 0\,\frac{rev}{s} y t = 15\,s, then the angular acceleration of the disc is:

\alpha = \frac{0\,\frac{rev}{s}-\frac{5}{12}\,\frac{rev}{s}}{15\,s}

\alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}

The angular acceleration of the disc is \frac{1}{36} radians per square second.

ii) The number of rotations that the disk makes before it stops (\Delta \theta), in revolutions, is determined by the following formula:

\Delta \theta  = \frac{\omega_{f}^{2}-\omega_{o}^{2}}{2\cdot \alpha} (2)

If we know that \omega_{o} = \frac{5}{12}\,\frac{rev}{s}, \omega_{f} = 0\,\frac{rev}{s} y \alpha = -\frac{1}{36}\,\frac{rev}{s^{2}}, then the number of rotations done by the disc is:

\Delta \theta = 3.125\,rev

The disc makes 3.125 revolutions before it stops.

We kindly invite to check this question on rotational motion: brainly.com/question/23933120

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3 years ago
A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen
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Answer

given,

frequency from Police car= 1240 Hz

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Frequency received by the other car

  f_1 = \dfrac{f_0(u + v)}{u}..........(1)

u is the speed of sound = 340 m/s

v is the speed of the car

Frequency of the police car received

  f_2= \dfrac{f_1(u)}{u-v}

now, inserting the value of equation (1)

  f_2= f_0\dfrac{u+v}{u-v}

  1275=1240\times \dfrac{340+v}{340-v}

  1.02822(340 - v) = 340 + v

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The ocean’s tides are not only affected by the shape of the coastlines, or slope of the ocean floor, but also by the gravitation
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Answer:

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The gravitational force of the Moon on the Earth causes the Earth to be slightly bulged on the side directly facing the Moon

The gravitational force also pulls the water bodies on the Earth's surface towards the Moon in the same manner and the effect is more pronounced due to the ability of the liquid water to assume a shape based on the magnitude of the gravitational field attracting it

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Explanation:

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