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malfutka [58]
2 years ago
8

A space probe is built with a mass of 1700 pound-mass [lbm] before launch on Earth. The probe is powered by four ion thrusters,

each capable of generating 225 millinewtons [mN] of thrust. Using Newton’s second law, the acceleration (a) of the craft is equal to the force (F) divided by the mass (m): a = F/m The velocity (v) of an object increases as an object accelerates. If the object starts at rest, the final velocity is given by the acceleration multiplied by the length of time the acceleration is applied to the object (t): v = at = (Ft)/m Using this equation, how many weeks will the thrusters have to operate for the probe, initially at rest, to reach a velocity of 420 miles per minute [mi/min]? You may assume the initial velocity of the probe is zero miles per minute [0 mi/min]
Physics
2 answers:
Wittaler [7]2 years ago
7 0

Answer:

The 64 weeks

Explanation:

Thinking process:

First we gather the data:

1700lbm   = 771.107 kg

225 milliNewtons = 0.225 N

Final velocity = 420 mil/min = 11 265.4 m/s

Let the initial velocity, u₀ = 0 m/s

Final velocity = v m/s

Acceleration is given by the formula, a = \frac{F}{m} \\   =\frac{0.225}{771.107} \\   = 0.0002918 m/s^{2}

However, we know that the final velocity, v is given by v = u + at

where u = 0

           a = 0.0002918

           t = ?

           v = 11 265.4 m/s

substituting:

11 265.4 = 0 + (0.0002918) (t)

dividing both sides gives, t = 38 606 579 s

                                           1 day = 86 400 s

                                           t = 446.83 days

                                             = 63.8 weeks

                                             = 64 weeks

Ahat [919]2 years ago
3 0

mass of the space probe is given as

m = 1700 lbm = 771.12 kg

each thrust will apply a net force

F = 225 milli N = 0.225 N

now we will have

v = at

v = \frac{F}{m}*t

given that we have final speed v = 420 miles/minute

we need to convert it into m/s

v = 420\frac{miles}{min}*\frac{1609 m}{1 miles}*\frac{1 min}{60 s}

v = 11263 m/s

now from above equation

11263 = \frac{0.225}{771.12}* t

t = 3.86 * 10^7 s

t = 10722.3 hours = 446.8 days = 63.8 weeks

<em>so it will require 63.8 weeks to reach the given speed</em>

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Answer:

a) The gravitational potential energy before the cart rolls down the incline is 24.6 J.

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d) It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

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g) The work done by the gravitational force is 24.6 J.

Explanation:

Hi there!

a) The gravitational potential energy is calculated using the following equation:

EP = m · g · h

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g = acceleration due to gravity.

h = height at which the object is located.

The height of the inclined plane can be calculated using trigonomoetry:

sin 14.0° = height / lenght

sin 14.0° = height / 4.50 m

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b) Please, see the attached figure for a graphical description of the problem and the forces acting on the cart. The force that causes the cart to accelerate down the incline is the horizontal component of the weight (Fwx in the figure). The magnitude of this force can be obtained using trigonometry:

sin 14° = Fwx / Fw

The weight of the cart (Fw) is calculated as follows:

Fw = m · g

Fw = 2.30 kg · 9.81 m/s²

Fw = 22.6 N

Then, the x-component of the weight will be:

FW · sin 14° = Fwx

22.6 N · sin 14° = Fwx

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The magnitude of the force that causes the cart to roll down is 5.47 N.

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x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the cart at time t.

x0 = initial position.

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a = acceleration.

t = time.

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x = 1/2 · a · t²

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The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

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KE = 1/2 · m · v²

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The gain of kinetic energy is equal to the loss of gravitational potential energy.

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W = Fw · d

Where "d" is the distance traveled in the direction of the force, that is the height of the inclined plane:

W = 22.6 N · 1.09 m = 24.6 J.

The work done by the gravitational force is 24.6 J.

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Complete Question

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