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3 years ago

7

A quantity of a gas is an absolute pressure of 400 K PA in the absolute temperature of 110° Kelvin when the temperature of the g

as is raised to 235° Kelvin what's the new pressure of the gas?

1 answer:

aliina [53]3 years ago

4 0

Given:

P1 = 400 kPa
T1 = 110 K

T2 = 235K

Required:

P2

Solution:

Apply Gay-Lussac’s law where P/T = constant

P1/T1 = P2/T2

P2 = T2P1/T1

P2 = (235K)(400kPa) / (110K)

P2 = 855 kPa

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Balboa Park in San Diego has an outdoor organ. When the air temperature increases, the fundamental frequency of one of the organ

musickatia [10]

The fundamental frequency of one of the organ pipes will go up or increase.

When pressured air is forced into an organ pipe, it echoes at a particular pitch, generating the sound of the pipe organ. Each pipe has been adjusted to a particular pitch on the musical scale.

A musical instrument called an outdoor pipe organ is used to perform music. It produces some calming tones and has a really serene sound. The organ pipe produces the sound of the outdoor organ. The wavelength of the sound is also dependent on the length of the pipe. The fundamental frequency of one of the organ pipes will grow as the speed of the sound increases as the ambient air temperature rises.

The correct option is (c).

Learn more about frequency here:

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6 0

1 year ago

A wire has a current density of 6.25 × 10 6 A / m 2 6.25×106 A/m2 . If the cross-sectional area of the wire is 1.79 mm 2 1.79 mm

joja [24]

Answer:17.44A

Explanation: Current density=I/Area

Area is given by 2.79mm^2=2.79×10^-6m^2

Current=I=current density ×Area=6.25×10^6 ×2.79×10^-6=17.44A

3 0

2 years ago

A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. Of a

Doss [256]

Answer:

Part(a): The frequency is $\bf{0.2~Hz}$ .

Part(b): The speed of the wave is $\bf{0.5~m/s}$ .

Explanation:

Given:

The distance between the crests of the wave, $d = 2.5~m$ .

The time required for the wave to laps against the pier, $t = 5.0~s$

The distance between any two crests of a wave is known as the wavelength of the wave. So the wavelength of the wave is $\lambda = 2.5~m$ .

Also, the time required for the wave for each laps is the time period of oscillation and it is given by $T = 5.0~s$ .

Part(a):

The relation between the frequency and time period is given by

$\nu = \dfrac{1}{T}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)$

Substituting the value of $T$ in equation (1), we have

$\nu &=& \dfrac{1}{5.0~s}\\~~~&=& 0.2~Hz$

Part(b):

The relation between the velocity of a wave to its frequency is given by

$v = \nu \lambda~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)$

Substituting the value of $\nu$ and $\lambda$ in equation (2), we have

$v &=& (0.2~Hz)(2.5~m)\\~~~&=& 0.5~m/s$

5 0

2 years ago

In severe head-on automobile accidents, a deceleration of 60 g’s or more (1 g 5 32.2 ft/s2) often results in a fatality. What fo

Musya8 [376]

Answer:

The resulting force on the child is 3000 lbf

Explanation:

To find the force that acts on a child of 50 lb with a deceleration of 60 g's, we can use the formula:

Force = mass * acceleration

To find the force in lbf, we need to use the mass in lb and the acceleration in g (standard unit of gravity).

So we have that:

Force = 50 * 60

Force = 3000 lbf

So the resulting force on the child is 3000 lbf.

3 0

2 years ago

How far from Earth must a space probe be along a line toward the Sun so that the Sun's gravitational pull on the probe balances

Tatiana [17]

Answer:

258774.9441 m

Explanation:

x = Distance of probe from Earth

y = Distance of probe from Sun

Distance between Earth and Sun = $x+y=149.6\times 10^6\ m$

G = Gravitational constant

$M_s$ = Mass of Sun = $1.989\times 10^{30}$

$M_e$ = Mass of Earth = $5.972\times 10^{24}\ kg$

According to the question

$\frac{GM_sm}{x^2}=\frac{GM_em}{y^2}\\\Rightarrow \frac{M_s}{x^2}=\frac{M_e}{y^2}\\\Rightarrow x=\sqrt{\frac{M_s\times y^2}{M_e}}\\\Rightarrow x=\sqrt{\frac{1.989\times 10^{30}\times y^2}{5.972\times 10^{24}}}\\\Rightarrow x=577.10852y$

$x+y=149.6\times 10^6\\\Rightarrow 577.10852y+y=149.6\times 10^6\\\Rightarrow 578.10852y=149.6\times 10^6\\\Rightarrow y=\frac{149.6\times 10^6}{578.10852}\\\Rightarrow y=258774.9441\ m$

The probe should be 258774.9441 m from Earth

7 0

2 years ago