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3 years ago

1/12 of 1/3 is what fraction?

Physics

1 answer:

Misha Larkins [42]3 years ago

6 0

$\frac{1}{36}$

Explanation:

$Given problem; \frac{1}{12} of \frac{1}{3}$

The of word is used in mathematics determine how much of something is a part of another.

It simply translates to the mathematical operator of multiplication (x or *).

To solve this problem, we simply multiply both fractions to find how much of $\frac{1}{12}$ is $\frac{1}{3}$

Solving gives;

$\frac{1}{12} of \frac{1}{3} = \frac{1}{12} x \frac{1}{3} = \frac{1}{36}$

learn more:

Number properties brainly.com/question/5639299

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Answer:

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In a second order lever system the force ratio is 2.5, the load is at the distance of 0.5m from the fulcrum find distance of eff

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Answer:

1.25 m

Explanation:

From the question given above, the following data were obtained:

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3 years ago

Suppose astronomers discover a radio message from a civilization whose planet orbits a star 35 light-years away. Their message e

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Answer:

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Generally the time it would take for the message to get the the other civilization is mathematically represented as

$t = \frac{D}{c}$

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3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

For $m_{3}$ :

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

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3 years ago