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3 years ago

Which expression is equivalent to g−m ÷ gn?

Physics

1 answer:

Vinvika [58]3 years ago

5 0

Answer

(g²n - m)/(gm)

Explanation

g - m ÷ gn = g - m/gn

Make the equation have the same denominator

g - m ÷ gn = g - m/gn = (ggn)/gn - m/gn

= (g²n)/gn - m/gm

Since they have the same denominator, we can carry out the subtraction on the numerator and then put them under one denominator.

(g²n)/gn - m/gm = (g²n - m)/(gm)

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A piano tuner strikes a tuning fork at the same time he strikes a piano key with a note of a similar pitch. If he hears 3 beats

DaniilM [7]

The frequency produced by the string could be 437 Hz or it could be 443 Hz.

The frequency of the beats ... 3 Hz ... tells the piano tuner that
the difference between the fork and string frequencies is 3 Hz,
but it doesn't tell her which one is higher or lower.

7 0

3 years ago

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Paladinen [302]

Answer:

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Explanation:

7 0

3 years ago

Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh

Alinara [238K]

Answer:

a) $k_{avg}=6.22\times 10^{-21}$

b) $k_{avg}=8.61\times 10^{-21}$

c) $k_{mol}=3.74\times 10^{3}J/mol$

d) $k_{mol}=5.1\times 10^{3}J/mol$

Explanation:

Average translation kinetic energy ( $k_{avg}$ ) is given as

$k_{avg}=\frac{3}{2}\times kT$ ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

T = Temperature in kelvin

a) at T = 27.8° C

T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8$

$k_{avg}=6.22\times 10^{-21}J$

b) at T = 143° C

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416$

$k_{avg}=8.61\times 10^{-21}J$

c ) The translational kinetic energy per mole of an ideal gas is given as:

$k_{mol}=A_{v}\times k_{avg}$

here $A_{v}$ = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

$k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}$

$k_{mol}=3.74\times 10^{3}J/mol$

d) now at T = 143° C

$k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}$

$k_{mol}=5.1\times 10^{3}J/mol$

8 0

3 years ago

Conductivities are often measured by comparing the resistance of a cell filled with the sample to its resistance when filled wit

Bezzdna [24]

Answer:

1200 Sm^2mol^-1

Explanation:

Given data :

conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1

conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1

Kkcl = 1.1639 - 0.076 = 1.0879 Sm^-1

Resistance = 33.21 Ω

where conductivity can be expressed as = $\frac{Cell constant}{Resistance }$

hence cell constant = conductivity * Resistance

= 1.0879 * 33.21 = 36.13m^-1

conductivity of CH3COOH ( kCH3COOH ) = 36.13 / 300

= 0.120 Sm^-1

Determine the molar conductivity of acetic acid

= ( kCH3COOH * 1000 ) / C

C = 0.1 mol dm

= (0.120 * 1000) / 0.1 = 1200 Sm^2mol^-1

3 0

3 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago