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3 years ago

The combined gas law does not account for changes in _____.

2 answers:

8 0

The combined gas law does not account for changes in power. The combined gas law has no official founder; it is simply the incorporation of the three laws that was discovered. The combined gas law is a gas law that combines Gay-Lussac’s Law, Boyle’s Law and Charle’s Law. Boyle’s law states that pressure is inversely proportional with volume at constant temperature. Charle’s law states that volume is directly proportional with temperature at constant pressure. And Gay-Lussac’s law shows that pressure is directly proportional with temperature at constant volume. The combination of these laws known now as combined gas law gives the ratio between the product of pressure-volume and the temperature of the system is constant. Which gives PV/T=k(constant). When comparing a substance under different conditions, the combined gas law becomes P1V1/T1 = P2V2/T2.

vitfil [10]3 years ago

8 0

d temperature is the correct answer

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A positron is an elementary particle identical to an electron except that its charge is . An electron and a positron can rotate

denis23 [38]

Explanation:

According to the energy conservation,

$F_{centripetal} = F_{electric}$

$\frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}$

$v^{2} = \frac{kq^{2}r}{d^{2}m}$

= $\frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}$

= $8.430 \times 10^{10} m^{2}/s^{2}$

v = $\sqrt{8.430 \times 10^{10} m^{2}/s^{2}}$

= $2.903 \times 10^{5} m/s$

Formula for distance from the orbit is as follows.

S = $2 \pi r$

= $2 \times 3.14 \times 0.75 \times 10^{-9} m$

= $4.71 \times 10^{-9} m$

Now, relation between time and distance is as follows.

T = $\frac{S}{v}$

$\frac{1}{f} = \frac{S}{v}$

or, f = $\frac{v}{S}$

= $\frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}$

= $6.164 \times 10^{13} Hz$

Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is $6.164 \times 10^{13} Hz$ .

7 0

3 years ago

This force on compass dials is an example of a force that _______.

romanna [79]

D. Acts at a distance, because magnetic forcefield is in two dots, Earth's north and south magnetic poles, and magnet is on some distance from them.

4 0

3 years ago

Read 2 more answers

Which glass is used in astrology? what is its focal length?

Maksim231197 [3]

Answer:

Convex lens is used in astrology

5 0

3 years ago

Read 2 more answers

Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid

Tanzania [10]

A) 18.4 m

a) mass of the load

b) mass of the truck

Explanation:

In order for the oad not to slide, its acceleration must be the same as the acceleration of the truck.

Since there is only one force acting on the load (the force of static friction), the acceleration of the load will be equal to the force of friction divided by the mass of the load (Newton's second law of motion):

$a=\frac{F_f}{m}=\frac{-\mu mg}{m}=-\mu g$ (1)

where

m is the mass of the load

$\mu=0.400$ is the coefficient of static friction

$g=9.8 m/s^2$ is the acceleration due to gravity

The acceleration of the truck (and the load) is also related to the stopping distance of the truck by the suvat equation:

$v^2-u^2=2as$ (2)

where

v = 0 is the final velocity of the car

u = 12.0 m/s is the initial velocity

a is the acceleration

s is the stopping distance

Since the acceleration must be the same, we can substitute (1) into (2), and solving for s we find:

$v^2-u^2=-2\mu g s\\s=\frac{v^2-u^2}{-2\mu g}=\frac{0^2-12.0^2}{-2(0.400)(9.8)}=18.4 m$

From part A, we see that the data that we have not used in the calculation are:

- The mass of the load

- The mass of the truck

Therefore, the two pieces of data unnecessary for the solution are

a) mass of the load

b) mass of the truck

8 0

3 years ago

The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?

Tju [1.3M]

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

$dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2$

$For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2$

The second distance, r₂, can be determined from sound intensity formula given as;

$I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m$

Therefore, the second distance of the sound from the source is 431.78 m.

7 0

3 years ago