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ra1l [238]
3 years ago
15

You are on an airplane traveling with a constant velocity at an altitude of 20,000m. What is the acceleration of gravity at that

altitude? The earth's radius is 6.37 x 10^6m.. . A) 9.81 m/s^2. B) 9.78 m/s^2. C) 9.75 m/s^2. D) 9.72 m/s^2. E) 9.69 m/s^2.
Physics
2 answers:
-BARSIC- [3]3 years ago
3 0
The correct answer would be A.The acceleration of gravity is <span>9.81 m/s^2. This is calculated by using the equation:

a = GM/r</span>²

where G is the gravitational constant 6.6726 x 10-11N-m2/kg2, M is the mass of the planet (<span>5.9736E+24 kg) and r is the radius of the planet.
</span>
AleksAgata [21]3 years ago
3 0

Answer:

C) 9.75m/s^2

Explanation:

Acceleration of gravity is calculated with the following formula:

g=\frac{GM}{r^2}

Where G It is the universal gravitational constant

G=6.674x10^{-11}Nm^2/kg^2

and M is the mass of the earth:

M=5.972x10^{24} kg

r is the distance from the center of earth wich will be in this case the earth's radius plus the altitude of the airplane:

r=6.37x10^{6} m + 20000m=6.39x10^{6}m

thus, the acceleration g is:

g=\frac{(6.674x10^{-11}Nm^2/kg^2)(5.972x10^{24} kg)}{(6.39x10^{6}m)^2} =9.75m/s^2

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Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ =  5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁  + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6    + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

I_{disk} = \frac{1}{2} mr^2\\\\I_{disk}  = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk}  = 0.305 \ kgm^2

The moment of inertia of the rod about its center is calculated as follows;

I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2

The initial rotational kinetic energy of the disk and rod;

K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \  \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i=  \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \  \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J

The final rotational kinetic energy of the disk-rod system is calculated as follows;

K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \  \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J

The loss in rotational kinetic energy due to the collision is calculated as follows;

\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J  \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J

Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.

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