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Ira Lisetskai [31]
3 years ago
15

If an object travels at a constant speed in a circular path, the acceleration of the object is

Physics
2 answers:
Sphinxa [80]3 years ago
6 0
There are two types of acceleration produced in a body if it moves in circular path !

first is ; tangential acceleration = dv/dt

and second is : radial acceleration = v^2/r

IRINA_888 [86]3 years ago
4 0
The acceleration of the object would be that <span>: 1- larger in magnitude the smaller the radius of the circle. 2- smaller in magnitude the smaller the radius of the circle. 3- in the same direction as the velocity of the </span>object<span>. 

Hope this helps!

</span>
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What is the IMA of the following pulley system?<br><br>34567
Lynna [10]

Answer:

    IMA of given system =   \frac{F_{r} }{F_{e} }

Explanation:

  • The "Ideal Mechanical advantage" (IMA) of given pulley is \frac{F_{r} }{F_{e} } .
  • Ideal Mechanical advantage of a system is defined by the ratio of achieved or output force to the implied force. In the pulley system above, output force is the resistant force denoted by F_{r}. The input force is analogous or equivalent to the effort applied i.e. F_{e} .
  • Hence by dividing these two forces we calculate the IMA of the above mentioned pulley system which is  \frac{F_{r} }{F_{e} } .
  • Its mathematical reference would be:

                                                IMA =   \frac{F_{r} }{F_{e} }

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A drop of oil of volume <br><img src="https://tex.z-dn.net/?f=%20%7B10%7D%5E%7B%20-%2010%7D%20" id="TexFormula1" title=" {10}^{
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Answer:

\frac{1}{10^{10}}\\

Explanation:

10^{-10}\\\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}\\\\10^{-10}=\\\frac{1}{10^{10}}\\\\\left(\mathrm{Decimal:\quad }\:0.0000000001\right)

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4 years ago
brick, dry soil, paper, and water. If all four substances were exposed to sunlight for the same amount of time, which substance
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4 years ago
A long thin rod of mass M and length L is situated along the y axis with one end at the origin. A small spherical mass m1 is pla
Yuri [45]

Answer:

= - 5.65\times 10^{-2} J

Explanation:

Given data:

L =2.00 *10^4 m

d = 18*10^4 m

M = 18  *10^6 kg

m_1 = 8*10^6 kg

Gravitational energy is given as

U =- G \frac{m_1 m_2}{r}

mass per unit length is given as

\sigma = \frac{M}{L} = \frac{18 \times 10^6}{2\times 10^4 m} = 900 kg/m

calculating potential energy

dU ==-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *dm}{r}

=-G\int_{16\times 10^4}^{18\times 10^4} \frac{m_1 *\sigma dr}{r}

=-G*m_1*\sigma\int_{16\times 10^4}^{18\times 10^4} \frac{dr}{r}

=-G*m_1*\sigma \left | ln r \right |_{16\times 10^4}^{18\times 10^4}

=-G*m_1*\sigma ln(1.125)

=-6.673 \times 10^{-11}*8*10^6*900*ln(1.125)

= - 5.65\times 10^{-2} J

5 0
3 years ago
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