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ipn [44]
3 years ago
7

The concept of critical density is just like the idea of escape velocity for a projectile launched from the earth. An object lau

nched with a velocity less than the escape velocity will fall back to the earth, and an object launched with one greater than the escape velocity can escape the earth's gravitational field. Find an expression for vesc, the escape velocity of a projectile launched from the earth's surface. Express your answer in terms of the universal gravitational constant G, the mass of the earth Me, and the radius of the earth Re.
Physics
1 answer:
ikadub [295]3 years ago
5 0

Answer:

v = √ 2 G M/ R_{e}

Explanation:

To find the escape velocity we can use the concept of mechanical energy, where the initial point is the surface of the earth and the end point is at the maximum distance from the projectile to the Earth.

Initial

        Em₀ = K + U₀

Final

        Em_{f} =  U_{f}

The kinetic energy is k = ½ m v²

The gravitational potential energy is U = - G m M / r

r is the distance measured from the center of the Earth

How energy is conserved

       Em₀ =  Em_{f}

       ½ mv² - GmM / R_{e} = -GmM / r

       v² = 2 G M (1 /  R_{e} – 1 / r)

       v = √ 2GM (1 / R_{e} – 1 / r)

The escape velocity is that necessary to take the rocket to an infinite distance (r = ∞), whereby 1 /∞ = 0

        v = √ 2GM /  R_{e}

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A particle with charge 6 mC moving in a region where only electric forces act on it has a kinetic energy of 1.9000000000000001 J
Vesna [10]

Answer:

16.9000000000000001 J

Explanation:

From the given information:

Let the initial kinetic energy from point A be K_A = 1.9000000000000001 J

and the final kinetic energy from point B be K_B = ???

The charge particle Q = 6 mC = 6 × 10⁻³ C

The change in the electric potential from point B to A;

i.e. V_B - V_A = -2.5 × 10³ V

According to the work-energy theorem:

-Q × ΔV = ΔK

-Q \times ( V_B - V_A) = (K_B - K_A)

-(6\times 10^{-3}\ C) \times ( -2.5 \times 10^3) = (K_B - 1.9000000000000001 \ J)

15 = (K_B - 1.9000000000000001 \ J)

K_B = 15+ 1.9000000000000001 \ J

\mathbf{K_B =1 6.9000000000000001 \ J}

3 0
3 years ago
What are two parts of an atom
seraphim [82]

Answer:

Nucleus And electron cloud

Explanation:

Hope this helps

7 0
3 years ago
Tina wants to install one unit that both heats and cools. Which should Tina install?
madreJ [45]
A heat pump? it might be furnace but i think its heat pump
8 0
3 years ago
Read 2 more answers
One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau
suter [353]

Answer:

actual efficiency is  47.78 %

Carnot efficiency  is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency  and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy  = 7.0 × 10^9 cal  (4.184 J/1 cal)  = 29.29 × 10^9 J

actual efficiency  =  1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency  =  0.4778

actual efficiency is  47.78 %

and

Carnot efficiency  is = 1 - ( 703 / 2173 )

so Carnot efficiency  is  = 0.67648

Carnot efficiency  is 67.65 %

and

power output  = work / time

power output  =  1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

5 0
3 years ago
Two identical 9.10-g metal spheres (small enough to be treated as particles) are hung from separate 300-mm strings attached to t
Musya8 [376]

Answer:

n = 1.266\times 10^{12}

Explanation:

Given data:

mass of sphere is 10 g

Angle between string and vertical axis is \theta = 13 degree

thickness of string  300 mm = 0.3 m

sin\theta =\frac{2}{0.3 m}

r =0.3 sin 13 = 0.067 m

Fe = \frac{ kq_1 q-2}{d^2}

Fe = \frac{kq^2}{(2r)^2} = mg tan\theta

q^2 =  mg tan\theta \frac{(2r)^2}{k}

    = 0.0091 \times 9.8 tan13 \times \frac{(2\times 0.067)^2}{9\times 10^9}

q^2 = 4.10\times 10^{-14}

q = 2.026 \times 10^{-7} C

q = ne

n = \frac{1.6\times 10^{-19}}{2.02\times 10^{-7}}

n = 1.266\times 10^{12}

3 0
3 years ago
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