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3 years ago

What is the mechanical advantage of the wheel and axle shown below?

Physics

2 answers:

Salsk061 [2.6K]3 years ago

7 0

Answer:

Mechanical advantage of the wheel and axle is 32.5

Explanation:

Given that,

The radius of the axle, $r_a=0.8\ cm$

The radius of wheel , $r_w=26\ cm$

We need to find the mechanical advantage of the wheel and axle. It can be calculated as :

$MA=\dfrac{r_w}{r_a}$

$MA=\dfrac{26}{0.8}$

MA = 32.5

So, the mechanical advantage of the wheel and axle shown below is 32.5 . Hence, this is the required solution.

BigorU [14]3 years ago

6 0

The correct answer is A. 32.5

Mechanical advantage is the ratio of force that is input into a machine to the force output.

Mechanical advantage of a wheel and axle is calculated by dividing the radius of the wheel by that of the axle.

MA=R/r where R is the radius of the wheel and r is the radius of the axle.

Substituting for the values in the question gives:

MA=26cm/0.8cm

=32.5

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3 years ago

Read 2 more answers

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 $\frac{F_{r} }{F_{e} }$ .

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3 years ago

Two pipes of equal length are each open at one end. Each has a fundamental frequency of 470 Hz at 310 K. In one pipe the air tem

Tresset [83]

Answer:

Given:

Fundamental frequency: 470Hz

T1:310k,T2:315k

Calculating velocity

Recall v=(331m/s)✓[T1/273k)

V=331✓(310/273)

V1=331*(1.0656)=352.72m/s

V2=331✓(315/273)=355.5m/s

Fundamental frequency=4L

F2=F1(V2/V1)

F2=470(355.5/352.72)=474.4Hz

Beat=[F2-F1]=474.4-470=4.4Hz

Explanation:

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3 years ago

Las condiciones iniciales de un gas son 3000 cm3

slava [35]

Answer:

T'=92.70°C

Explanation:

To find the temperature of the gas you use the equation for ideal gases:

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P: pressure = 1250mmHg; 1 mmHg = 0.001315 atm

n: number of moles

R: ideal gas constant = 0.082 atm.L/mol.K

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For the given values you firs calculate the number n of moles:

$n=\frac{PV}{RT}=\frac{(1520[0.001315atm])(3L)}{(0.082\frac{atm.L}{mol.K})(300.15K)}=0.200moles$

this values of moles must conserve when the other parameter change. Hence, you have V'=2L and P'=3atm. The new temperature is given by:

$T'=\frac{P'V'}{nR}=\frac{(3atm)(2L)}{(0.200\ moles)(0.082\frac{atm.L}{mol.K})}=365.85K=92.70\°C$

hence, T'=92.70°C

8 0

3 years ago