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igor_vitrenko [27]
3 years ago
5

A 0.0502-kg pair of fuzzy dice is attached to the rearview mirror of a car by a short string. The car accelerates at constant ra

te, and the dice hang at an angle of 3.20° from the vertical because of the car's acceleration. What is the magnitude of the acceleration of the car?
Physics
1 answer:
Sauron [17]3 years ago
7 0

Answer:

a = 0.55 m/s/s

Explanation:

As the car accelerates in forward direction the string will make some angle with the vertical direction

So here horizontal component of the tension force will accelerates the dice in forward direction along with the car

Vertical component of the tension force will balance the weight of the dice

so we will have

Tcos\theta = mg

Tsin\theta = ma

here we know that

\theta = angle made with the vertical by string

now divide the two equations

\frac{sin\theta}{cos\theta} = \frac{ma}{mg}

so we have

a = g tan\theta

a = 9.81 tan3.20

a = 0.55 m/s^2

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Answer:

The right answer is:

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Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

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(a)

The mass of the string will be:

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or,

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On substituting the values, we get

⇒     =\frac{280\times (3)^2}{4 \pi^2}

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The speed of the string will be:

⇒  \frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2

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On substituting the values, we get

⇒                =\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }

⇒                =\sqrt{\frac{280(0.16-0.04)}{63.83} }

⇒                =\sqrt{\frac{280\times 0.12}{63.83} }

⇒                =0.725 \ m/s

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