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svp [43]
2 years ago
6

A child is sliding down a slide at the playground. is the mechanical energy conserved. why or why not.

Physics
2 answers:
aleksandrvk [35]2 years ago
6 0
C: the mechanical energy isn't conserved. Some energy was lost to friction. 
Nikolay [14]2 years ago
6 0

Answer: The correct answer is C.

Explanation:

According to the law conservation of the mechanical energy, the total mechanical energy in a closed system remains constant in the absence of friction.

In the given problem, a child is sliding down a slide at the playground.

When a child is on the top of the slide then the he has more potential energy. When he starts slides down the slide then the kinetic energy starts increasing.

At the bottom of the slide, he has more kinetic energy.

But some energy is lost here due to friction. Hence, the mechanical energy is not conserved.

Therefore, the correct option is C.

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At a given moment, a plane passes directly above a radar station at an altitude of 6 km. Let θ be the angle that the line throug
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- 187.237 km/hr fast is θ changing 12 min after the plane passes over the radar station

<u>Explanation:</u>

Let the distance x and angle θ be defined as in the figure below. Then

                  \tan \theta=\frac{6}{x}

Now, differentiate with respect to t, we get

                 \sec ^{2} \theta \frac{d \theta}{d t}=-\left(\frac{6}{x^{2}}\right) \frac{d x}{d t}

Now, calculate the travel distance from radar station to plane after 12min

Distance,  x=800 \times \frac{12}{60}=160

Substituting ‘x’ value, we get

                \tan \theta=\frac{6}{160}=\frac{3}{80}

Find the rate of change of theta after 12 min,

                \frac{d \theta}{d t}=-\frac{1}{\sec ^{2} \theta} \times \frac{6}{x^{2}} \times \frac{d x}{d t}

We know, the formula for,

                \sec ^{2} \theta=1+\tan ^{2} \theta=1+\frac{3^{2}}{80^{2}}

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               \frac{d \theta}{d t}=-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800

               =-\frac{1}{\left(1+\frac{3^{2}}{80^{2}}\right)} \times \frac{6}{160^{2}} \times 800

               =-\frac{1}{1.0014} \times \frac{6}{25600} \times 800

             =-\frac{4800}{25635.84}=-0.187237 \mathrm{rad} / \mathrm{hr}

When express the value in km/he, we get, the change in theta as

              =-187.237 \mathrm{km} / \mathrm{hr}

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