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3 years ago

Two Jupiter-size planets are released from rest 1.40×10^11 m apart.. What are their speeds as they crash together?

1 answer:

7 0

M₁ = mass of planet #1
M₂ = mass of planet #2
M = total mass
R₁ = radius of planet #1
R₂ = radius of planet #2
d₁ = initial distance between planet centers
d₂ = final distance between planet centers
a = semimajor axis of plunge orbit
v₁ = relative speed of approach at distance d₁
v₂ = relative speed of approach at distance d₂

M₁ = M₂ = 1.8986e27 kilograms
M = M₁ + M₂ = 3.7972e27 kg
G = 6.6742e-11 m³ kg⁻¹ sec⁻²
GM = 2.5343e17 m³ sec⁻²
d₁ = 1.4e11 meters
a = d₁/2 = 7e10 meters
R₁ = R₂ = 7.1492e7 meters
d₂ = R₁ + R₂ = 1.42984e8 meters
v₁ = 0
v₂ = √[GM(2/d₂−1/a)]
<span>
v₂ = 59508.4 m/s </span>
<span>
The time to fall is 1337.7 days

.</span>I hope my answer has come to your help. Thank you for posting your question here in Brainly.

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The auditory canal behaves like a resonant tube to aid in hearing. One end terminated at the eardrum, while the other opens to t

kompoz [17]

Answer:

3.6 kHz

Explanation:

The pipes behave like a closed pipe . The end open is the end of the air canal outside the ear and the closed end is the eardrum.

The first harmonic will be as seen in the figure attached.

The length of the first harmonic will be λ/4.

λ/4=2.4 cm

λ=2.4 * 4=9.6 cm 0.096 m

Speed of Sound- 344 m/s(in air)

velocity(v) * Time Period(T) = Wavelength (λ)

Also, Time Period(T)= \frac{\textrm{1}}{\textrm{Frequency(f)}}

\frac{\textrm{Velocity}}{\textrm{Wavelength}}=\frac{\textrm{1}}{\textrm{Time Period}} =Frequency

Plugging in the values into the equation,

Frequency = $\frac{344}{0.096}$ Hz

= 3583.3 Hz≈3600 Hz= 3.6 kHz

Frequency= 3.6 kHz

7 0

2 years ago

Read 2 more answers

Suppose an object is moving 2.1 m/s north on a river, but the river is flowing to the east at a velocity of 1.2 m/s. What is the

rewona [7]

Answer:

2.4 m/s

Explanation:

Given:

Velocity of the object moving north = 2.1 m/s

Velocity of the river moving eastward = 1.2 m/s

The resultant velocity is the vector sum of the velocities of object and river.

Since the directions of velocity of object and river are perpendicular to each other, the magnitude of the resultant velocity is obtained using Pythagoras Theorem.

The velocities are the legs of the right angled triangle and the resultant velocity is the hypotenuse.

The magnitude of the resultant velocity (R) is given as:

$R^2=2.1^2+1.2^2\\\\R^2=4.41+1.44\\\\R=\sqrt{5.85}\\\\R=2.4\ m/s$

Therefore, the resultant velocity has a magnitude of 2.4 m/s.

8 0

2 years ago

The curvature of the helix r(t)equals(a cosine t )iplus(a sine t )jplusbt k (a,bgreater than or equals0) is kappaequalsStartF

4vir4ik [10]

Answer:

$\kappa = \frac{1}{2 b}$

Explanation:

The equation for kappa ( κ) is

$\kappa = \frac{a}{a^2 + b^2}$

we can find the maximum of kappa for a given value of b using derivation.

As b is fixed, we can use kappa as a function of a

$\kappa (a) = \frac{a}{a^2 + b^2}$

Now, the conditions to find a maximum at $a_0$ are:

$\frac{d \kappa(a)}{da} \left | _{a=a_0} = 0$

$\frac{d^2\kappa(a)}{da^2} \left | _{a=a_0} < 0$

Taking the first derivative:

$\frac{d}{da} \kappa = \frac{d}{da} (\frac{a}{a^2 + b^2})$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} \frac{d}{da}(a)+ a * \frac{d}{da} (\frac{1}{a^2 + b^2} )$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 + a * (-1) (\frac{1}{(a^2 + b^2)^2} ) \frac{d}{da} (a^2+b^2)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - a (\frac{1}{(a^2 + b^2)^2} ) (2* a)$

$\frac{d}{da} \kappa = \frac{1}{a^2 + b^2} * 1 - 2 a^2 (\frac{1}{(a^2 + b^2)^2} )$

$\frac{d}{da} \kappa = \frac{a^2+b^2}{(a^2 + b^2)^2} - 2 a^2 (\frac{1}{(a^2 + b^2)^2} )$

$\frac{d}{da} \kappa = \frac{1}{(a^2 + b^2)^2} (a^2+b^2 - 2 a^2)$

$\frac{d}{da} \kappa = \frac{b^2 - a^2}{(a^2 + b^2)^2}$

This clearly will be zero when

$a^2 = b^2$

as both are greater (or equal) than zero, this implies

$a=b$

The second derivative is

$\frac{d^2}{da^2} \kappa = \frac{d}{da} (\frac{b^2 - a^2}{(a^2 + b^2)^2} )$

$\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} \frac{d}{da} ( b^2 - a^2 ) + (b^2 - a^2) \frac{d}{da} ( \frac{1}{(a^2 + b^2)^2} )$

$\frac{d^2}{da^2} \kappa = \frac{1}{(a^2 + b^2)^2} ( -2 a ) + (b^2 - a^2) (-2) ( \frac{1}{(a^2 + b^2)^3} ) (2a)$

$\frac{d^2}{da^2} \kappa = \frac{-2 a}{(a^2 + b^2)^2} + (b^2 - a^2) (-2) ( \frac{1}{(a^2 + b^2)^3} ) (2a)$

We dcan skip solving the equation noting that, if a=b, then

$b^2 - a^2 = 0$

at this point, this give us only the first term

$\frac{d^2}{da^2} \kappa = \frac{- 2 a}{(a^2 + a^2)^2}$

if a is greater than zero, this means that the second derivative is negative, and the point is a minimum

the value of kappa is

$\kappa = \frac{b}{b^2 + b^2}$

$\kappa = \frac{b}{2* b^2}$

$\kappa = \frac{1}{2 b}$

3 0

2 years ago

An object accelerating may be changing in what two ways

Oksana_A [137]

Velocity and direction

8 0

2 years ago

If the earth can travel 1800 km in 1 minute what is it’s speed in km/s

Lady bird [3.3K]

Answer:

30km/s

Explanation:

1800/60=30

4 0

3 years ago