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kherson [118]
3 years ago
12

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)?
Physics
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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Natali5045456 [20]
Equation of motion:

y_{f}=y_{o}+v_{o}t+ \frac{1}{2} at^{2}

Since initial velocity is zero, the second term goes away:

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y_{f}-y_{o}= \frac{1}{2} at^{2}

y_{f}-y_{o}=5.4m

5.4m= \frac{1}{2} at^{2}

\frac{2(5.4m) }{a} = t^{2}

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\frac{2(5.4m) }{9.81 \frac{m}{ s^{2} } } = t^{2}

1.1 s^{2} = t^{2}\sqrt{1.1 s^{2}} =  \sqrt{t^{2}}

<u><em>t = 1.05s</em></u>
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3 years ago
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two barges full of salted toad guts have a collision. the red barge has a mass of 150000kg and is traveling northwest at 0.25m/s
solniwko [45]

The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.

Final velocity(v3)  of the red barge is calculated by following formula

m1×v1+ m2×v2= (m1+m2)v3

Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s

150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3

37500+ 320000= 1150000×v3

357500= 1150000×v3

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<h3>What is elastic collision velocity? </h3>
  • The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.

For more information on elastic collision velocity kindly visit to

brainly.com/question/29051562

#SPJ9

6 0
1 year ago
Please help!!!
Lubov Fominskaja [6]

question: Please help!!!

If a bottle is being squeezed with a force of 10 Newtons and the area of the bottle is 15

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Answer:

1025.64 N/m²

Explanation:

Pressure: This can be defined as the ratio of force to area. The si unit of pressure is N/m².

From the question,

P = F/A........................ Equation 1

Where F = Force, A = Area.

Given: F = 10 Newtons, A = 15 Squared Inches = (15×0.00065) = 0.00975 m²

Substitute these values into equation 1

P = 10/0.00975

P = 1025.64 N/m²

Hence the pressure of the bottle is 1025.64 N/m²

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3 years ago
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