Ask question

Ask question

All categories

3 years ago

11

A closed container is filled with oxygen. the pressure in the container is 405 kpa . what is the pressure in millimeters of merc

ury? express the pressure numerically in millimeters of mercury.

1 answer:

kari74 [83]3 years ago

8 0

1 mm Hg pressure is the pressure due to 1 mm height of mercury

so it is given as

$P = \rho g h$

$P = 1 * 10^{-3} * 13700 * 9.8 = 134.26 Pa$

now we know that pressure is

P = 405 kPa

$P = 405 * 10^3 Pa$

so in mm Hg this pressure is

$P = \frac{405 * 10^3 }{134.26} = 3016.5 mm Hg$

so pressure in millimeter Hg will be 3016.5 mm Hg

You might be interested in

An astronaut exploring a distant solar system lands on an unnamed planet with a radius of 2530 km. When the astronaut jumps upwa

Natali [406]

Answer:

1.38*10^18 kg

Explanation:

According to the Newton's law of universal gravitation:

$F=G*\frac{m_a*m_p}{r^2}$

where:

G= Gravitational constant (6.674×10−11 N · (m/kg)2)

ma= mass of the astronaut

mp= mass of the planet

$F=m_a.a\\(v_f )^2=(v_o)^2+2.a.\Delta y\\\\a=\frac{(v_f)^2-(v_o)^2}{2.\Delta y}\\\\a=\frac{(0)^2-(4.29m/s)^2}{2.0.64m}=14.38m/s^2\\\\F=m_a*14.38m/s^2$

so:

$m_a*14.38m/s^2=(6.674*10^{-11}N.(m/kg)^2)*\frac{m_a.m_p}{(2.530*10^3m)^2}\\m_p=\frac{14.38m/s^2(2.530*10^3m)^2}{(6.674*10^{-11}N.(m/kg)^2)}\\\\m_p=1.38*10^{18}kg$

7 0

3 years ago

A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution abou

kupik [55]

Answer:

The acceleration of the satellite is $0.87 m/s^{2}$

Explanation:

The acceleration in a circular motion is defined as:

$a = \frac{v^{2}}{r}$ (1)

Where a is the centripetal acceleration, v the velocity and r is the radius.

The equation of the orbital velocity is defined as

$v = \frac{2 \pi r}{T}$ (2)

Where r is the radius and T is the period

For this particular case, the radius will be the sum of the high of the satellite ( $1.50x10^{7} m$ ) and the Earth radius ( $6.38x10^{6} m$ ) :

$r = 1.50x10^{7} m+6.38x10^{6}m$

$r = 21.38x10^{6}m$

Then, equation 2 can be used:

$T = 8.65 hrs \cdot \frac{3600 s}{1hrs}$ ⇒ $31140 s$

$v = \frac{2 \pi (21.38x10^{6}m)}{31140s}$

$v = 4313 m/s$

Finally equation 1 can be used:

$a = \frac{(4313m/s)^{2}}{21.38x10^{6}m}$

$a = 0.87 m/s^{2}$

Hence, the acceleration of the satellite is $0.87 m/s^{2}$

6 0

3 years ago

Which phenomenon occurs when light falls on a smooth mirror?

muminat

That would be reflection.

6 0

3 years ago

A wave amplitude 0.36m interferes with a second wave of amplitude 0.22m traveling in the same direction. What is the largest res

DerKrebs [107]

The largest resultant amplitude would be that created by constructive interference, basically when the two waves are of the same phase, so it would be 0.36m+0.22m= 0.58 m.

8 0

3 years ago

Which best describes a nonmagnetic material?

Natali [406]

Material that is not attracted to metal

3 0

3 years ago

Read 2 more answers