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Akimi4 [234]
1 year ago
15

The escape velocity is defined to be the minimum speed with which an object of mass m must move to escape from the gravitational

attraction of a much larger body, such as a planet of total mass M. The escape velocity is a function of the distance of the object from the center of the planetR, but unless otherwise specified this distance is taken to be the radius of the planet because it addresses the question "How fast does my rocket have to go to escape from the surface of the planet?" What is the total mechanical energy Etotal of the object at a very large (i.e., infinite) distance from the planet? Find the escape velocity ve for an object of mass m that is initially at a distance R from the center of a planet of mass M. Assume that R≥Rplanet, the radius of the planet, and ignore air resistance. Express the escape velocity in terms of R, M, m, andG, the universal gravitational constant.
Physics
1 answer:
s344n2d4d5 [400]1 year ago
6 0

Answer:

v = √2G M_{earth} / R

Explanation:

For this problem we use energy conservation, the energy initiated is potential and kinetic and the final energy is only potential (infinite r)

        Eo = K + U = ½ m1 v² - G m1 m2 / r1

        Ef = - G m1 m2 / r2

When the body is at a distance R> Re, for the furthest point (r2) let's call it Rinf

       Eo = Ef

       ½ m1v² - G m1 M_{earth} / R = - G m1 M_{earth} / R

      v² = 2G M_{earth} (1 / R - 1 / Rinf)

If we do Rinf = infinity     1 / Rinf = 0

       v = √2G M_{earth} / R

      Ef = = - G m1 m2 / R

The mechanical energy is conserved  

 

      Em = -G m1  M_{earth} / R

      Em = - G m1  M_{earth} / R

     R = int        ⇒  Em = 0

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A popular ride at some playgrounds consists of two circular disks joined by rungs . The center of the unit is attached to a hori
weeeeeb [17]

Answer:

v = 36.667 m/s

Explanation:

Knowing the rotational inertia as

Lₙ = 550 kg * m²

r = 1.0 m

m = 30.0 kg

To determine the minimum speed v must have when she grabs the bottom

Lₙ =  I * ω

I = ¹/₂ * m * r²

I = ¹/₂ * 30.0 kg * 1.0² m

I = 15 kg * m²

Lₙ =  I * ω  ⇒ ω =  Lₙ / I

ω = [ 550 kg * m² /s  ] / ( 15 kg * m² )

ω = 36.667 rad /s

v = ω * r

v = 36.667 m/s

7 0
2 years ago
A metal cube with sides of length a=1cm is moving at velocity v0→=1m/sj^ across a uniform magnetic field B0→=5Tk^. The cube is o
Nitella [24]

Answer:

the magnitude of the electric field is 1.25 N/C

Explanation:

The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s

ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V

Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube

ε = ∫E.ds

ε = Eds since E is always parallel to the side of the cube

= E∫ds  ∫ds = 4L since we have 4 sides

= E(4L)

= 4EL

So,4EL = 0.05 V

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

= 0.05 V/0.04 m

= 1.25 V/m

= 1.25 N/C

So, the magnitude of the electric field is 1.25 N/C

7 0
1 year ago
An air compressor compresses 6 L of air at 120 kPa and 22°C to 1000 kPa and 400°C. Determine the flow work, in kJ/kg, required b
Mariana [72]

Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

6 0
2 years ago
Suppose the electric field in some region is found to be E = kr3 ˆr, in spherical coordinates (k is some constant). (a) Find the
Assoli18 [71]

Answer:

Part a)

\rho = 3\epsilon_0 k r^2

Part b)

Q = 4\pi \epsilon_0kR^5

Explanation:

Part a)

As we know that electric field intensity due to some given charge distribution is given as

E = kr^3 \hat r

now electric flux through a spherical surface of radius r is given as

\phi = E. A

\phi = kr^3(4\pi r^2)

now by Guass law we know that

E.A = \frac{q}{\epsilon_0}

q = 4\pi \epsilon_0kr^5

now volume charge density is given as

\rho = \frac{q}{\frac{4}{3}\pi r^3}

\rho = 3\epsilon_0 k r^2

Part b)

Total charge inside the radius R is given as

Q = 4\pi \epsilon_0kR^5

7 0
1 year ago
Identify two potential improvements to the opal extraction process and explain how these improvements could minimize harm to the
Orlov [11]

Answer

• Improving the environmental performances

• Developing Green Mining technology

Explanation

The effect to the environment caused by opal mining are; impact on soils and geology, clearing of native vegetation disrupting flora and fauna, change in land use and effects of air quality.

Opal mining is currently examining environmental impacts and adopting measures that mitigate the impacts making the process less destructive to the environment.

With the current commitment to sustainability, opal companies are investing funds for Green Mining as a positive way to impact the environment before and after mining.


7 0
2 years ago
Read 2 more answers
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