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3 years ago

PLZ HELP I DONT GET IT

Physics

1 answer:

GaryK [48]3 years ago

6 0

Answer:

0.1 L

Explanation:

From the question given above, we obtained the following data:

Initial volume (V₁) = 0.05 L

Initial Pressure (P₁) = 207 KPa

Final pressure (P₂) = 101 KPa

Final volume (V₂) =?

We can obtain the new volume (i.e the final volume) of the gas by using the Boyle's law equation as illustrated below:

P₁V₁ = P₂V₂

207 × 0.05 = 101 × V₂

10.35 = 101 × V₂

Divide both side by 101

V₂ = 10.35 / 101

V₂ = 0.1 L

Thus, the new volume of the gas is 0.1 L

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PLEASE HELP WITH THIS QUESTION.

Sophie [7]

Answer:

#2 and #3 respectively

Explanation:

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3 years ago

Estimate the kinetic energy of the Mars with respect to the Sun as the sum of the terms, that due to its daily rotation about it

kotykmax [81]

Answer:

K = 2.07 10³⁹ J

Explanation:

This problem must be solved using rotational kinematics.

Kinetic energy has the form

K = ½ m v²

Linear velocity is related to angular velocity.

v = w r

replace

K = ½ m R² w²

With this equation we can find the total kinetic energy of Mars, formed by the rotation energy plus the translational energy

K = Kr + Kt

Where Kr is the energy by the rotation on its axis and Kt is the energy by the rotation around the sun.

Let's reduce to SI units

Rotation T₁ = 24.7 h (3600 s / 1 h) = 88920 s

R₁ = 3.4 10⁶ m

translation T₂ = 686 day (24 h / 1 day) (3600 s / 1 h) = 5.927 10⁷ s

R₂ = 2.3 10⁸ km (1000 m / 1 km) = 2.3 10¹¹ m

The angular velocity is the angle rotated (radians) between the time taken, in this case as the order gives us the angle is 2pi rad (360º). Remember that the equations work only in radians

Rotation

wr = 2π / T₁

wr = 2 π / 88920

wr = 7.066 10⁻⁵ rad / s

Translation

wt = 2 π / T₂

wt = 2 π / 5,927 10⁷

wt = 1.06 10⁻⁷-7 rad/s

Let's explicitly write the equation of kinetic energy and calculate

K = ½ m R₁² wr² + ½ m R₂² wt²2

K = ½ m (R₁² wr² + R₂² wt²)

K = ½ 6.4 10²³ [(3.4 10⁶)² 7.066² + (2.3 10¹¹)² (1.06 10⁻⁷)²]

K = 3.2 10²³ [61.49 10¹² + 6.414 10¹⁵]

K = 3.2 10²³ [61.49 10¹² + 6414 10¹²]

K = 3.2 10²³ (6475 10¹²)

K = 2.07 10³⁹ J

5 0

3 years ago

Which statement accurately describes malleability? HELP ASAP!!!

vovangra [49]

Answer:

[]

Explanation:

Malleability is when a certain substance can be easily pressed, bent, or pulled out of shape permanently. An example of a malleable substance is gold, as it can be permanently pressed into a sheet and unable to return to its original state.

8 0

3 years ago

Read 2 more answers

Mars’s moon Phobos orbits the planet at a distance of 9380 km from its center, and it takes 7 hours and 39 minutes to complete o

sukhopar [10]

Answer: 0.107

Explanation:

We can solve this problem with Kepler's Third Law of Planetary motion:

$T^{2}=4 \pi^{2} \frac{r^{3}}{G M_{MARS}}$ (1)

Where:

$T=7 h 39 min$ is the orbital period of Phobos around Mars

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M_{MARS}$ is the mass of Mars

$r=9380 km \frac{1000 m}{1 km}=9,380,000 m$ is the semimajor axis of the orbit Phobos describes around Mars (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

Well, firstly we have to convert the orbital period to seconds:

$T=7 h 39 min=(7 h \frac{3600 s}{1 h}) + (39 min \frac{60 s}{1 min})=25200 s + 2340 s=27540 s$

Now, we have to find $M_{MARS}$ from (1):

$M_{MARS}=\frac{4 \pi^{2} r^{3}}{G T^{2}}$ (2)

$M_{MARS}=\frac{4 \pi^{2} (9,380,000 m)^{3}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}) (27540 s)^{2}}$ (3)

$M_{MARS}=6.436(10)^{23} kg$ (4) This is the mass of Mars

On the other hand, it is known the mass of the Earth is:

$M_{EARTH}=5.972(10)^{24} kg$ (5)

Then, if we want to know the ratio of Mars’s mass to the mass of the earth, we have to divide $M_{MARS}$ by $M_{EARTH}$ :

$\frac{M_{MARS}}{M_{EARTH}}=\frac{6.436(10)^{23} kg}{5.972(10)^{24} kg}$

Finally:

$\frac{M_{MARS}}{M_{EARTH}}=0.107$

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4 years ago

Two planetoids with radii R1 and R2 and local accelerations g1 and g2 are separated by a centre-to-centre distance D, and are me

balu736 [363]

Answer:

Therefore, the gravitational zero points between two planetoids lie at a distance of 3000 m from the center of planetoid 1.

Explanation:

From Newton’s gravitation formula, the expression of the mass (M) of the planet of radius R is given as,

$F_{G} = mg_{1}\\ \left ( \frac{GMm}{{R_{1}}^{2}} \right )=mg_{1}\\\\M= \frac{gR^{2}}{G}\rightarrow \left ( 1 \right )$

Let's take x to be the distance of the zero gravitational points from the center of the planetoid 1.

Thus, the distance of the zero gravitational points from the center of the planetoid 2 is (D-x).

At zero gravitational point, the gravitational force between the planets and the rocket must be equal.

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3 years ago