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2 years ago

If the gauge pressure of a gas is 114 kPa, what is the absolute pressure?

Physics

2 answers:

Anastasy [175]2 years ago

4 0

Answer:

D. 214 kPa

Explanation:

The absolute pressure is given by:

$p = p_a + p_g$

where

p is the absolute pressure

$p_a \sim 100 kPa$ is the atmospheric pressure

$p_g$ is the gauge pressure

In this problem, we have

$p_g = 114 kPa$

So, the atmospheric pressure is

$p = 100 kPa + 114 kPa = 214 kPa$

dalvyx [7]2 years ago

3 0

Answer:

your answer is 214kPa

Explanation:

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What happens to the gravitational force between two objects if:

tiny-mole [99]

Answer:

See the answers below

Explanation:

This problem and its respective questions can be easily solved using Newton's law of universal gravitation. Which can be calculated by means of the following expression.

$F =G*\frac{m_{1}*m_{2}}{r^{2} }$

where:

G = it is the universal gravitation constant. = 6.673 x 10⁻¹¹ [N*m²/kg²]

m1 = mass of the first body [kg]

m2 = mass of the second body [kg]

r = distance among the bodies [m]

a. the mass of one is doubled?

When this happens we see that the force is increased twice as well, since the mass is in the numerator of the expression.

b. The masses of both are doubled?

If both masses are doubled the force is increased to four times its original value since the terms of the masses are in the numerator of the expression.

c. The distance between them is doubled?

In this case the force is decreased to half of its original value, since the distance is in the denominator of the expression of universal gravitation.

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2 years ago

The moon Umbriel orbits Uranus (mass = 8.68 x 10^25 kg) at a distance of 2.66 x 10^8 m. What is Umbriel's orbital period (in hou

Whitepunk [10]

Answer:

T = 99.51 hour

Explanation:

Mass of Uranus, $M=8.68\times 10^{25}\ kg$

The moon Umbriel orbits Uranus at a distance of $2.66\times 10^8\ m$

We need to find Umbriel's orbital period. Let it is T. Using Kepler's third law of motion to find it.

$T^2\propto r^3\\\\T^2=\dfrac{4\pi^2r^3}{GM}\\\\T^2=\dfrac{4\pi^2\times (2.66\times 10^8)^3}{6.67\times 10^{-11}\times 8.68\times 10^{25}}\\\\T=358244.51\ s$

As 1 hour = 3600 s

358244.51 s = 99.51 hour

Hence, Umbriel's orbital period is 99.51 hour.

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2 years ago

Is walking just a constant state of falling?

atroni [7]

Answer:

Yes. Walking is controlled falling because you need to let go in order to move forward. If you never let your foot fall, your movements would be stilted and robotic. And according to medical engineers," When we walk normally we are constantly correcting tiny falls to keep ourselves stable."

Explanation:

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3 years ago

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Alex73 [517]

Answer:

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Explanation:

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2 years ago

Read 2 more answers

The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

$v=36.62 m/s$ This is the speed of the jogger

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3 years ago