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Allisa [31]
2 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]2 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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Explanation:

It is given that,

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An object in space is initially stationary relative to the Earth. Then, a force begins acting on the object, starting with a for
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Answer:

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3 years ago
A 1.8 kg uniform rod with a length of 90 cm is attached at one end to a frictionless pivot. It is free to rotate about the pivot
Leokris [45]

Answer:

a. 32.67 rad/s²  b. 29.4 m/s²

Explanation:

a. The initial angular acceleration of the rod

Since torque τ = Iα = WL (since the weight of the rod W is the only force acting on the rod , so it gives it a torque, τ at distance L from the pivot )where I = rotational inertia of uniform rod about pivot = mL²/3 (moment of inertia about an axis through one end of the rod), α = initial angular acceleration, W = weight of rod = mg where m = mass of rod = 1.8 kg and g = acceleration due to gravity = 9.8 m/s² and L = length of rod = 90 cm = 0.9 m.

So, Iα = WL

mL²α/3 = mgL

dividing through by mL, we have

Lα/3 = g

multiplying both sides by 3, we have

Lα = 3g

dividing both sides by L, we have

α = 3g/L

Substituting the values of the variables, we have

α = 3g/L

= 3 × 9.8 m/s²/0.9 m

= 29.4/0.9 rad/s²

= 32.67 rad/s²

b. The initial linear acceleration of the right end of the rod?

The linear acceleration at the initial point is tangential, so a = Lα = 0.9 m × 32.67 rad/s² = 29.4 m/s²

5 0
3 years ago
A football wide receiver rushes 16 m straight down the playing field in 2.9 s (in the positive direction). He is then hit and pu
Bumek [7]

Answer:

a) v1 = 5.52m/s

b) v2 = -1.52m/s

c) v3 = 4.62m/s

d) vt = 3.85m/s

Explanation:

The velocity of the football wide receiver is his displacement per unit time.

Velocity v = (displacement d)/time t

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v2 = -2.5/1.65

v2 = -1.52m/s

c) v3 = d3/t3

d3 = 24m

t3 = 5.2s

v3 = 24/5.2

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d) vt = dt/tt

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tt = 2.9 + 1.65 + 5.2 = 9.75s

vt = 37.5/9.75

vt = 3.85m/s

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