Question

Please help me! will mark as brainliest! I will report absurd and incorrect answers!​

Answer 1

$\huge\fcolorbox{black}{aqua}{Solution:-}$

✏️ Answer:-

★ The possible shapes of the object must be a ring or hollow cylinder.

✏️ Explanation:-

★ Given:-

• Incline that rises by 3 in every 5 -- (will be) sin θ = 3/5

• Object gains a speed , v = √10 m/s

• Traverses a distance along the incline, s = 5/3

• Value of g = 10 m/s²

★ To find:-

• Probable shape/s of the object... i.e., to acquire the ratio of K²/R² which will

be further determining the possible rolling object.

★ Solution:-

• From the given scenario, we have,

Sin θ = 3/5

And the linear distance traversed by the object along the plane, s = h/sin θ.

⇒h = s

⇒sin θ = 5/3 × 3/5 = 1

Now, as we know,

The velocity of any rolling body is given by,

$v \: = \: \sqrt{ \frac{2gh}{1 + \frac{K^{2} }{R^{2} } } }$

Thus,

On plugging in the respective values and comparing the above expression, we get,

$\sqrt{10 \:} = \sqrt{ \frac{2gh}{1 + \frac{K^{2} }{R^{2} } } }$

$⇒ \sqrt{10 \: } = \: \frac{2 \times 10 \times 1}{1 + \frac{K ^{2} }{R^{2} } }$

$⇒ \: [ \: 1 + \frac{K^{2} }{R^{2} } ] = 2$

$⇒ \frac{K^{2} }{R^{2} } = 1$

Therefore,

We can conclude that the given object must either be a ring or a hollow cylinder.

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I hope this helps! :)

Answer 2

★ Given :-

• An object rolls down along an incline that rises by 3 on every 5(along it).
• The object gains a speed of √10m/s as it travels a distance of 5/3m along the incline.

★ Have to Find :-

What can be the possible shapes of object?

★ Solution :-

Now, from given we know that sin (theta) will be 3/5.

From given we know, the object covers linear distance which is equal to :-

s → $\frac{h}{sin\;\theta}$

Therefore, substituting the values to get the height.

s × sin θ = h

" s " = 5/3

" sinθ " = 3/5

$\frac{5}{3} \: \times \: \frac{3}{5} = \: 1$

Now, we are going to use the formula velocity of the object in order to find the k²/R²

So, we know the formula ,

Velocity → √2gh / ( 1 + k²/ r² )

$( \frac{2gh}{ \: 1 \: + \: \frac{ {k}^{2} }{ {r}^{2} } } )$

" g " → 10

" v " → √ 10m/s

" h " → 1

Simply substituting the values :-

√ 10 = √2gh / ( 1 + k²/ r² )

Root cancels root

➹ 10 = 2 * 10 * 1 / 1 + k²/R²

➹ 10/10 = 2 * 1 / 1 + k²/ R²

➹ 2 * 1 = 1 + k²/ R²

➹ 2 = 1 + k²/ R²

Transposing 1

2 - 1 =   1 + k²/R²

    1 =  k² /R²

Hence,

The object is either ring or hollow cylinder.

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