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Allisa [31]
3 years ago
11

A 4.0×1010kg asteroid is heading directly toward the center of the earth at a steady 16 km/s. To save the planet, astronauts str

ap a giant rocket to the asteroid perpendicular to its direction of travel. The rocket generates 5.0×109N of thrust. The rocket is fired when the asteroid is 4.0×106km away from earth. You can ignore the earth’s gravitational force on the asteroid and their rotation about the sun. Part A: If the mission fails, how many hours is it until the asteroid impacts the earth? Part B: The radius of the earth is 6400 km. By what minimum angle must the asteroid be deflected to just miss the earth? Part C: What is the actual angle of deflection if the rocket fires at full thrust for 300 s before running out of fuel?
Physics
1 answer:
blondinia [14]3 years ago
8 0

Answer:

a. t = 69.4 hr = 2.89 days

b. theta = 91.67*10^-3 degrees

c.  deflection_angle = 0.134 degrees

Explanation:

a).

The asteroid impacts the earth in t

t = x/v = (4.0*10^6 km)/(16 km/sec)

t = 2.5 * 10^5 sec

t = 69.4 hr = 2.89 days

b).

tan(theta) = 6400 km/(4.0*10^6 km)

tan(theta) = 1.6*10^-3

theta = arctan(1.6*10^-3)

theta = 1.6*10^-3 radians  (for small angles, tan(theta) ~= theta)

theta = 91.67*10^-3 degrees

c).

v_minimum = 6400 km/(2.5 * 10^5 sec)

v_minimum = 25.6 m/s

Using F = m*a, we can calculate the acceleration of the asteroid due to the rocket's thrust:

5.0*10^9 N = 4.0*10^10 kg * a

a = (5.0*10^9 N)/(4.0*10^10 kg)  

a = 0.125 m/s^2

The transverse velocity after 300 seconds of this acceleration is:

v_transverse = a*t = 0.125 m/s^2 * 300 s

v_transverse = 37.5 m/s = 37.5*10^-3 km/s

tan(deflection_angle) = v_transverse/(20 km/s)

tan(deflection_angle) = (37.5*10^-3 km/s)/(16 km/s) = 2.34^-3

deflection_angle = arctan(2.34*10^-3)  

deflection_angle = 2.34*10^-3 radians = 0.134 degrees

v_transverse/(16 km/s) > (6400km)/(5.0*10^6 km)  

(note that the right hand side if this inequality is tan(theta) calculated above)

v_transverse > 23.704 m/

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3 years ago
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Answer:

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2 years ago
A particular balloon can be stretched to a maximum surface area of 1257 cm2. The balloon is filled with 3.1 L of helium gas at a
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Answer:

The ballon will brust at

<em>Pmax = 518 Torr ≈ 0.687 Atm </em>

<em />

<em />

Explanation:

Hello!

To solve this problem we are going to use the ideal gass law

PV = nRT

Where n (number of moles) and R are constants (in the present case)

Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2} --- (*)

Our initial state is:

P1 = 754 torr

V1 = 3.1 L

T1 = 294 K

If we consider the final state at which the ballon will explode, then:

P2 = Pmax

V2 = Vmax

T2 = 273 K

We also know that the maximum surface area is: 1257 cm^2

If we consider a spherical ballon, we can obtain the maximum radius:

R_{max} = \sqrt{\frac{A_{max}}{4 \pi}}

Rmax = 10.001 cm

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V_{max} = \frac{4}{3} \pi R_{max}^3

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P_{max} = P_1 \frac{V_1T_2}{V_2T_1}

Therefore:

Pmax= P1 * (0.687)

That is:

Pmax = 518 Torr

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Answer:

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