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3 years ago

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Suppose you designed a spacecraft to work by photon pressure. The sail was a completely absorbing fabric of area 1.0 km2 and you

directed a laser beam of wavelength 650 nm onto it at a rate of 1 mol of photons per second from a base on the moon. The spacecraft has a mass of 1.0 kg. Given that, after a period of acceleration from standstill, speed = (force/mass) x time, how many minutes would it take for the craft to accelerate to a speed of 1.0 m/s (about 2.2 mph)?

1 answer:

Alekssandra [29.7K]3 years ago

4 0

Answer:

(a) F = 6.14 *10⁻⁴ N

(b) P = 6.14* 10⁻¹⁰ Pa

(c) t = 27.2 min

Explanation:

Area of sail A = 1.0 km² = 1.0 * 10⁶m²

Wavelength of light λ = 650 nm = 650 * 10⁻⁹ m

Rate of impact of photons R = 1 mol/s = 6.022 * 10²³ photons/s

(a)

Momentum of each photon is Ρ = h/λ = 6.625 * 10⁻³⁴ / 650 * 10⁻⁹

= 1.0192 * 10⁻²⁷ kg.m/s

Since the photons are absorbed completely, in every collision the above momentum is transferred to the sail.

Momentum transferred to the sail per second is product of rate of impact of photons and momentum transferred by each photon.

dp/dt = R * h/ λ

This is the force acting on the sail.

F = R * h/λ = 6.022 * 10²³ * 1.0192 * 10⁻²⁷ = 6.14 *10⁻⁴ N

F = 6.14 *10⁻⁴ N

b)

Pressure exerted by the radiation on the sail = Force acting on the sail / Area of the sail

P = F/A = 6.14 * 10⁻⁴ / 1.0 * 10⁶ = 6.14* 10⁻¹⁰ Pa

P = 6.14* 10⁻¹⁰ Pa

c)

Acceleration of spacecraft a = F/m = 6.14 * 10⁻⁴ /1.0 = 6.14 * 10⁻⁴m/s²

As the spacecraft starts from rest, initial speed u=0,m/s ,

final speed is u = 1.0 m/s after time t

v = u+at

t = 1.0 - 0/ 6.14 * 10⁻⁴ = 1629s = 27.2 min

t = 27.2 min

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3 years ago

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AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

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We can cancel the common factor of $m$ which leaves us with

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Lets solve for $v_f$

$gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}$

Subtract $gh_f$ from both sides of the equation.

$gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}$

Multiply both sides of the equation by 2.

$2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}$

Simplify the left side.

Apply the distributive property.

$2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}$

Cancel the common factor of 2.

$2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}$

Take the square root of both sides of the equation to eliminate the exponent on the right side.

${v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}$

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${v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)$

Anything multiplied by 0 is 0.

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1 year ago

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Nataly_w [17]

Answer:

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Explanation:

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Now let's use the continuity equation

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3 years ago

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*Used copy paste from my own answer as it is a repeated question, no copied work*

3. A
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4. A
This is another direct linear relation as P=IV.
5. D
The relation between P, R, and V is P=, so P is inversely proportional to R.
6.B
The relation between P,I, and R is , so P is directly proportional to the square of I.

Please note that y:x relations are always straight lines while relations are parabolic lines.

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3 years ago

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Pani-rosa [81]

Answer:

Part a)

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Part b)

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Explanation:

Part a)

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Part b)

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3 years ago