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Vesna [10]
3 years ago
7

A projectile of mass m is fired straight upward from the surface of an airless planet of radius R and mass M with an initial spe

ed equal to the escape speed vesc (meaning the projectile will just barely escape the planet's gravity -- it will asymptotically approach infinite distance and zero speed.) What is the correct expression for the projectile's kinetic energy when it is a distance 9R from the planet's center (8R from the surface). Ignore the gravity of the Sun and other astronomical bodies. KE (at r = 9R) is:a. GMm/9Rb. GMm/8Rc. 1/2mvesc^2d. -GMm/8Re. None of these
Physics
1 answer:
motikmotik3 years ago
6 0

Answer:

K = G Mm / 9R

Explanation:

Expression for escape velocity V_e = \sqrt{\frac{2GM}{R} }

Kinetic energy at the surface = 1/2 m V_e ²

= 1/2 x m x 2GM/R

GMm/R

Potential energy at the surface

= - GMm/R

Total energy = 0

At height 9R ( 8R from the surface )

potential energy

= - G Mm / 9R

Kinetic energy = K

Total energy will be zero according to law of conservation of mechanical energy

so

K  - G Mm / 9R = 0

K = G Mm / 9R

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A mixture of helium and oxygen is used in scuba diving tanks to help prevent ""the bends"". 46 L helium and 12 L oxygen are comb
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Answer

given,

For helium

Volume,V = 46 L

Pressure,P = 1 atm

Temperature,T = 25°C  =  273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₁ = ?

number of moles

we know

P V = n R T

n_1 =\dfrac{46 \times 1}{0.0821\times 298}

  n₁ = 1.89 moles

For oxygen

Volume,V = 12 L

Pressure,P = 1 atm

Temperature,T = 25°C  =  273 +25 = 298 K

R=0.0821 L . atm /mole.K

n₂ = ?

number of moles

we know

P V = n R T

n_2 =\dfrac{12 \times 1}{0.0821\times 298}

  n₂ = 0.49 moles

Total volume of tank = 5 L

temperature of tank = 298 K

Partial pressure of helium

P_1=\dfrac{n_1 R T}{V}

P_1=\dfrac{1.89\times 0.0821\times 298}{5}

     P₁ = 9.25 atm

Partial pressure of oxygen

P_2=\dfrac{n_2 R T}{V}

P_2=\dfrac{0.49\times 0.0821\times 298}{5}

    P₂ = 2.39 atm

total pressure

    P = P₁ + P₂

    P = 9.25 + 2.39

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2 years ago
A projectile is shot at an angle 45 degrees to the horizontalnear the surface of the earth but in the absence of air resistance.
juin [17]

Answer:

Explanation:

Given

For first case

launch angle \theta =45^{\circ}C

at highest point h=150 m/s

150=u\cos 45

u=\frac{150}{\cos 45}=212.132 m/s

For second case

\theta _2=37^{\circ}C

at highest Point velocity is u\cos \theta _2

=212.132\times \cos 37

=169.41 m/s

as there is no acceleration in x direction therefore horizontal velocity is same          

7 0
3 years ago
Read 2 more answers
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