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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 solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its rad
maksim [4K]

To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as

\Upsilon = \frac{F/A}{\Delta L/L_0}

Where,

F = Force/Weight

A = Area

\Delta L= Compression

L_0= Original Length

According to the values given we have to

\Upsilon_{steel} = 200*10^9Pa

\Delta L = 5.6*10^{-7}m

L_0 = 3.2m

r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2

Replacing this values at our previous equation we have,

\Upsilon = \frac{F/A}{\Delta L/L_0}

200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}

F = 38272.5N

Therefore the Weight of the object is 3.82kN

4 0
1 year ago
Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. The heat capacity of Object A is
Oksi-84 [34.3K]

Answer:

Explanation:

Heat capacity A = 3 x heat capacity of B

initial temperature of A = 2 x initial temperature of B

TA = 2 TB

Let T be the final temperature of the system

Heat lost by A is equal to the heat gained by B

mass of A x specific heat of A x (TA - T) = mass of B x specific heat of B x ( T - TB)

heat capacity of A x ( TA - T) = heat capacity of B x ( T - TB)

3 x heat capacity of B x ( TA - T) = heat capacity of B x ( T - TB)

3 TA - 3 T = T - TB

6 TB + TB = 4 T

T = 1.75 TB

8 0
2 years ago
A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheri
MaRussiya [10]

Answer:

dz=19217687.07\ m

Explanation:

Given:

  • initial gauge pressure in the container, P_0=2.02\times 10^{5}\ Pa
  • atmospheric pressure at sea level, P_a=1.01\times 10^5\ Pa
  • initial volume, V_0=4.4\times 10^{-4}\ m^3
  • maximum pressure difference bearable by the container, dP_{max}=2.26\times 10^{5}\ Pa
  • density of the air, \rho_a=1.2\ kg.m^{-3}
  • density of sea water, \rho_s=1.2\ kg.m^{-3}

<u>The relation between the change in pressure with height is given as:</u>

\frac{dP_{max}}{dz} =\rho_a.g_n

where:

dz = height in the atmosphere

g_n= standard value of gravity

<em>Now putting the respective values:</em>

\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8

dz=19217.687\ km

dz=19217687.07\ m

Is the maximum height above the ground that the container can be lifted before bursting. (<em>Since the density of air and the density of sea water are assumed to be constant.</em>)

7 0
2 years ago
A 5.0-kg crate is resting on a horizontal plank. The coefficient of static friction is 0.50 and the coefficient of kinetic frict
Harlamova29_29 [7]

Answer:

The mass of the crate is 5kg.

We know that the force of friction can be obtained by:

F = N*k

where k is the coefficient of friction, where we use the static one if the object is at rest, and the kinetic one if the object os moving. N is the normal force

If we tilt the base making an angle of 30° with the horizontal, now the normal force against the plank will be equal to the fraction of the weight in the direction normal to the surface of the plank.

Knowing that the angle is 30°, then the fraction of the weight that pushes against the normal is Cos(30°)*W = cos(30°)*5kg*9.8m/s^2 = 42.4N

The fraction of the force in the parallel direction to the plank (the force that would accelerate the crate downwards) is:

F = sin(30°)*5k*9,8m/s = 24.5N

now, the statical friction force is:

Fs = 42.4N*0.5 = 21.2N

The statical force is less than the 24.5N, so the crate will move downwards, then the force that acts on the crate is the kinetic force of friction:

Fk = 42.4N*0.4 = 16.96N

Then, the total force that acts on the crate is:

total force = F - Fk = 24.5N - 16.69N = 7.54N and the direction of this force points downside along the parallel direction of the plank.

3 0
2 years ago
A container explodes and breaks into three fragments that fly off 120° apart from each other, with mass ratios 1: 4: 2. If the f
RSB [31]

Answer:

V₂ = 1.5 m/s

Explanation:

given,

speed of the first piece = 6 m/s

speed of the third piece = 3 m/s

speed of the second fragment = ?

mass ratios = 1 : 4 : 2

fragment break  fly off = 120°

α = β = γ  = 120°

sin α = sin β = sin γ = 0.866

using lammi's theorem

\dfrac{A}{sin\alpha}=\dfrac{B}{sin\beta}=\dfrac{C}{sin\gamma}

A,B and C is momentum of the fragments

\dfrac{m\times 6}{0.866}=\dfrac{4m\times v_2}{0.866}=\dfrac{2m\times 3}{0.866}

4 x V₂ = 2 x 3

V₂ = 1.5 m/s

3 0
2 years ago
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