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poizon [28]
2 years ago
7

Astronomers have discovered a new planet called "Xandar" beyond the orbit of Pluto (No, not really but I need a fake planet for

this problem!). It has a satellite called "Jim" that orbits with a period of 12.9 days at a distance of 72,600 km. What is the mass of "Xandar"?
Physics
1 answer:
Burka [1]2 years ago
3 0

Answer:

m = 1.82E+23 kg

Explanation:

G = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

mG = universal gravitational constant = 6.67E-11 N·m²/kg²

r = radius of orbit = 72,600 km = 7.26E+07 m

C = circumference of orbit = 2πr = 4.56E+08 m

P = period of orbit = 12.9 d = 1,114,560 s

v = orbital velocity of satellite Jim = C/P = 409 m/s

m = mass of Xandar = to be determined

v = √(Gm/r)

v² = [√(Gm/r)]²

v² = Gm/r

rv² = Gm

rv²/G = m

m = rv²/G

m = 1.82E+23 kg

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Fittoniya [83]

Answer:

Q_{lost} per mole of pentane = 3157.53 kJ/mol

Explanation:

Given:

Mass of pentane, m = 0.468 gram

Molar mass of pentane, M = 72.15

Now, mol of pentane, n = mass/M = 0.468/72.15 = 0.00648 mol of C5H12

Now,

ΔT = 23.65 - 20.45 = 3.2°C

Heat capacity of the calorimeter, C = 2.21 kJ/°C

Specific heat capacity of the water, Cp = 4.184  J/g.°C

Now,

the heat gained = the heat lost

Q_{gained} = -Q_{lost}

also,

Q_{gained} = Q_{water} + Q_{calorimeter}

Q_{water} = m\times C\times(T_f-T_i)

or

Q_{water} = 1000\times4.184\times(23.65-20.45) = 13388.8\ J

and

Q_{calorimeter} = C\times\Delta T = 2.21\times1000\times3.2 = 7072\ J

Now,

Q_{total} = 13388.8 +7072 = 20460.8\ J

we have,

Q_{lost} = -Q_{gain} = - 20460.8\ J (Here negative sign depicts the release of the heat)

Q_{lost} per mole of pentane =-20460.8/(0.00648 ) = 3157.53 kJ/mol

3 0
2 years ago
Some car manufacturers claim that their vehicles could climb a slope of 42 ∘. For this to be possible, what must be the minimum
Paladinen [302]

Answer:

D. 0.9

Explanation:

Calculating minimum coefficient of static friction, we first resolve the forces (normal and frictional) acting on the vehicle at an angle to the horizontal into their x and y components. After this, we can now substitute the values of x and y components into equation of static friction. Diagrammatic illustration is attached.

Resolving into x component:

                        ∑F_{x} = F_{s} - mgsin\alpha =0

                          F_{s} = mgsin\alpha     ------(1)

Resolving into y component:

                        ∑F_{y} = F_{n} - mgcos\alpha =0

                          F_{n} = mgcos\alpha      ------(2)

Static frictional force, F_{s} \leq μ F_{n}       ------(3)

substituting F_{s} from equation (1) and F_{n} from equation (2) into equation (3)

                         mgsin\alpha \leq μ mgcos\alpha

                         sin\alpha \leq μ cos\alpha

                         μ \geq \frac {sin\alpha}{cos\alpha}

                         μ \geq tan\alpha

The angle the vehicles make with the horizontal α = 42°

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6 0
2 years ago
The end of a stopped pipe is to be cut off so that the pipe will be open. If the stopped pipe has a total length L, what fractio
Alexxandr [17]

Answer:

4/10 of L.

Explanation:

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The fundamental mode of a stopped pipe is also called its fundamental frequency, and is f₁=v/4L.

Where f₁=fundamental frequency, v= velocity of sound, L= Length of pipe.

The fifth harmonic of the stopped pipe f₅ =5v/4L .................(1)

For an open pipe,

Fundamental  mode is also called fundamental frequency f₁₀=v/2l₀ .......... (2)

Where f₁₀ = fundamental frequency of a closed pipe, v= velocity of sound and l₀=length of the resulting open pipe.

from the question, the fundamental mode of the resulting open pipe = The fifth harmonic of the original stopped pipe.

∴ f₅=f₁₀

⇒5v/4L = v/2l₀

Equating v from both side of the equation,

⇒ 5/4L = 1/2l₀

Cross multiplying the equation,

5×2l₀ = 4L× 1

10l₀ = 4L

Dividing both side of the equation by the coefficient of l₀ i.e 10

10l₀/10 = 4L/10

∴ l₀ = 4/10(L)

∴ 4/10 of L must be cut off

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zhuklara [117]

Answer:

<em>You would use the kinematic formula:</em>

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Explanation:

The upwards vertical motion is ruled by the equation:

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Where:

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       y_0\text{ is the initial position: }y_0=0

       t=2s

       g\text{ is the gravitational acceleration: }\approx 9.8m/s^2

       V_{0y}\text{ is the initial vertical velocity}

Naming Δy = y - y₀, the equation becomes:

      \Delta y=V_{0y}\times t-g\times t^2/2

Then, you just need to substitute with Δy = 0.1m, t = 2s, and g = 9.8m/s², ans solve for the intital vertical velocity.

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