3 years ago

A 2.0 kg box is placed on an inclined plane that forms an angle of 21 ° with the horizontal

Physics

1 answer:

Dmitrij [34]3 years ago

5 0

Answer:

.-2.0

Explanation:

as at 80 friction it's a 360?

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Find the 24th term of the arithmetic sequence 11, 14, 17, … . Express the 24 terms of the series of this sequence using sigma no

Anna71 [15]

Answer:

The 24th term is 80 and the sum of 24 terms is 1092.

Explanation:

Given that,

The arithmetic series is

11,14,17,........24

First term a = 11

Difference d = 14-11=3

We need to calculate the 24th term of the arithmetic sequence

Using formula of number of terms

$t_{n}=a+(n-1)d$

Put the value into the formula

$t_{24}=11+(24-1)\times3$

$t_{24}=80$

$t_{24}=u_{24}=80$

We need to calculate the sum of the first 24 terms of the series

Using formula of sum,

$S_{n}=\dfrac{n}{2}(a+u_{24})$

Put the value into the formula

$S_{n}=\dfrac{24}{2}\times(11+80)$

$S_{n}=1092$

Hence, The 24th term is 80 and the sum of 24 terms is 1092.

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4 years ago

Consider the following four objects: a hoop, a flat disk, a solid sphere, and a hollow sphere. Each of the objects has mass M an

Mumz [18]

Answer:

The hoop

Explanation:

We need to define the moment of inertia of the different objects, that is,

DISK:

$I_{disk} = \frac{1}{2} mR^2$

HOOP:

$I_{hoop} = mR^2$

SOLID SPHERE:

$I_{ss} = \frac{2}{5}mR^2$

HOLLOW SPHERE

$I_{hs} = \frac{2}{3}mR^2$

If we have the same acceleration for a Torque applied, then

$mR^2>\frac{2}{3}mR^2>\frac{1}{2} mR^2>\frac{2}{5}mR^2$

$I_{hoop}>I_{hs} >I_{disk}>I_{ss}$

The greatest momement of inertia is for the hoop, therefore will require the largest torque to give the same acceleration

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Explanation:

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Answer:

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