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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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The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the
astra-53 [7]

Answer:

The other two small angles are 45° each

Explanation:

Given data in the problem:

The triangle is a right triangle

thus,

one of the angle is 90°

now,

let the other two angles be x and y

thus,

it is given that:

x = 2y - 45°

also in a triangle

sum of all the angles = 180°

thus,

x + y + 90° = 180°

or

x + y = 90°

now, substituting the  value of x from the above relation between x and y, we get

2y - 45° + y = 90°

or

3y = 135°

or

y = 45°

also,

x = 2y - 45°

or

x = 2 × (45°) - 45°

or

x = 45°

hence, <u>the other two small angles are 45° each</u>

4 0
2 years ago
De Vico Comet orbits the Sun every 74.0 years and has an orbital eccentricity of 0.96. Find the comet's average distance from th
Debora [2.8K]

Answer: The comet's average distance from the sun is 17.6AU

Explanation:

From Kepler's 3rd Law, P^2=a^3

Where P is period in years

and a is length of semi-major axis or the average distance of the comet to the sun.

Given the orbital period to be 74 years

74^2 =a^3

5476 = a^3

Cube root of 5476 =a

17.626 = a

Approximately a= 17.6 AU

5 0
2 years ago
Suppose that we are designing a cardiac pacemaker circuit. The circuit is required to deliver pulses of 1ms duration to the hear
olchik [2.2K]

Answer:

Minimum capacitance = 200 μF

Explanation:

From image B attached, we can calculate the current flowing through the capacitors.

Thus;

Since V=IR; I = V/R = 5/500 = 0.01 A

Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V

So minimum capacitance will be determined from;

i(t) = C(dv/dt)

So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]

C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF

8 0
2 years ago
16) A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s.
KiRa [710]

Answer:

57.6Joules

Explanation:

Rotational kinetic energy of a body can be determined using the expression

Rotational kinetic energy = 1/2Iω²where;

I is the moment of inertia around axis of rotation. = 5kgm/s²

ω is the angular velocity = ?

Note that torque (T) = I¶ where;

¶ is the angular acceleration.

I is the moment of inertia

¶ = T/I

¶ = 3.0/5.0

¶ = 0.6rad/s²

Angular acceleration (¶) = ∆ω/∆t

∆ω = ¶∆t

ω = 0.6×8

ω = 4.8rad/s

Therefore, rotational kinetic energy = 1/2×5×4.8²

= 5×4.8×2.4

= 57.6Joules

6 0
2 years ago
Read 2 more answers
A 120-g block of copper is taken from a kiln and quickly placed into a beaker of negligible heat capacity containing 300 g of wa
gavmur [86]

Answer : The correct option is, (d) 535^oC

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of copper = 0.10cal/g^oC

c_2 = specific heat of water = 1.00cal/g^oC

m_1 = mass of copper = 120 g

m_2 = mass of water = 300 g

T_f = final temperature of mixture = 35^oC

T_1 = initial temperature of copper = ?

T_2 = initial temperature of water = 15^oC  

Now put all the given values in the above formula, we get:

120g\times 0.10cal/g^oC\times (35-T_1)^oC=-300g\times 1.00cal/g^oC\times (35-15)^oC

T_1=535^oC

Therefore, the temperature of the kiln was, 535^oC

7 0
2 years ago
Read 2 more answers
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