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3 years ago

If the pressure on an ideal gas is increased, what will happen to the volume of the gas?

1 answer:

iogann1982 [59]3 years ago

7 0

B). it will decrease.

But, you should keep the temperature constant, 'cause according to Boyle's law, pressure of the ideal gases is indirectly proportional to it's volume but at constant temperature. So, don't confuse in that.

Hope this helps!

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The mass of 2 cm3 of gold is 38.6 grams. find the density of the gold. Could you please help me??

Alex Ar [27]

Density = mass/ volume (so here this is how you would solve the problem)

D = 38.6 g/ 2 cm3 (first step)

D= 19.3 g/cm3 ( Do math and then you would get this)

Hope this helps!! :)

6 0

3 years ago

Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.160. If the patch i

Olegator [25]

Answer:

14.1 m/s

Explanation:

From the question,

μk = a/g...................... Equation 1

Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.

make a the subject of the equation

a = μk(g).................. Equation 2

Given: μk = 0.160, g = 9.8 m/s²

Substitute into equation 2

a = 0.16(9.8)

a = 1.568 m/s²

Using,

F = ma

Where F = force, m = mass.

Make m the subject of the equation

m = F/a................... Equation 3

m = 160/1.568

m = 102.04 kg.

Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.

Assuming the initial velocity of the skier to be zero

Fd+mgμ = 1/2mv²........................Equation 4

Where v = speed of the skier after crossing the patch, d = distance/width of the patch.

v = √2(Fd+mgμ)/m)................ Equation 5

Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16

Substitute these values into equation 5

v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]

v = √197.57

v = 14.1 m/s

v = 9.86 m/s

4 0

3 years ago

The smallest shift you can reliably measure on the screen is about 0.2 grid units. This shift corresponds to the precision of po

creativ13 [48]

Answer:

The distance is $d = 1.5 *10^{15} \ km$

Explanation:

From the question we are told that

The smallest shift is $d = 0.2 \ grid \ units$

Generally a grid unit is $\frac{1}{10}$ of an arcsec

This implies that 0.2 grid unit is $k = \frac{0.2}{10} = 0.02 \ arc sec$

The maximum distance at which a star can be located and still have a measurable parallax is mathematically represented as

$d = \frac{1}{k}$

substituting values

$d = \frac{1}{0.02}$

$d = 50 \ parsec$

Note $1 \ parsec \ \to 3.26 \ light \ year \ \to 3.086*10^{13} \ km$

So $d = 50 * 3.08 *10^{13}$

$d = 1.5 *10^{15} \ km$

3 0

3 years ago

Consider two antennas separated by 9.00 m that radiate in phase at 120 MHz, as described in Exercise 35.3. A receiver placed 150

alexgriva [62]

Answer:

$\phi=4.52 rad$

Explanation:

From the question we are told that

Distance b/e antenna's $d=9.00m$

Frequency of antenna Radiation $F_r=120 MHz \approx 120*10^6Hz$

Distance from receiver $d_r=150m$

Intensity of Receiver $i= 10$

Distance difference of the receiver b/w antenna's $(r^2-r^1)=1.8m$

Generally the equation for Phase difference $\phi$ is mathematically given by

$\phi=\frac{2\pi}{\frac{c}{f_r}} *(r^2-r^1)$

$\phi=\frac{2*\pi}{\frac{3*10^{8}}{120*10^6}} *1.8$

$\phi=\frac{4\pi}{5} *1.8$

Therefore phase difference f between the two radio waves produced by this path difference is given as

$\phi=4.52 rad$

7 0

2 years ago

A heavy boy and a lightweight girl are balanced on a massless seesaw. The boy moves backward, increasing his distance from the p

victus00 [196]

Answer:

The side the boy is sitting on will tilt downward.

Explanation:

According to the law of moments when the same force is applied at a greater distance from the pivot then the effect of moment is greater about that point.

Mathematically momentum is given as:

$M=F\times r$

where:

F is the applied force at a distance 'r' acting in a direction perpendicular to the line joining the point of application and the hinge.

Moment is the rotational effect of the applied force on the body.

When the boy of a heavier mass than the girl was sitting on a balanced see-saw then it is certain that he was closer to the hinge than the girl to balance the turning effect (in case of an unbiased see-saw). When the body moves farther his weight is same but the radial distance from the hinge increases which increases his moment of weight.

6 0

2 years ago