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alexandr402 [8]
3 years ago
5

A space shuttle sits on the launch pad for 2.0 minutes, and then goes from rest to 4600 m/s in 8.0 minutes. Treat its motion as

straight-line motion. What is the average acceleration of the shuttle (a) during the first 2.0 minutes, (b) during the 8.0 minutes the shuttle moves, and (c) during the entire 10 minute period?
Physics
1 answer:
SpyIntel [72]3 years ago
8 0

Answer:

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Explanation:

a.)

    Acceleration (a) is defined as the time rate of change of velocity

                       a = \frac{v_{2} - v_{1} } {t}  

Given data

 Final velocity = v₂ = 0 m/s

 Initial velocity = v ₁ = 0 m/s

  As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,

                 a = 0 ms⁻²

b.)

     Given data

As the space shuttle start from rest, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 8 min = 480 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {480}

                     a = 9.58 ms⁻²

c.)

    Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 10 min = 600 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {600}

                     a = 7.67 ms⁻²

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Delicious77 [7]

Answer:

x = 0.775m

Explanation:

Conceptual analysis

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We apply Coulomb's law to calculate the electrical forces on q₃:

F_{23} = \frac{k*q_{2}*q_{3}}{x^2} (Electric force of q₂ over q₃)

F_{13} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (Electric force of q₁ over q₃)

Known data

q₁ = 15 μC = 15*10⁻⁶ C

q₂ = 6 μC = 6*10⁻⁶ C

Problem development

F₂₃ = F₁₃

\frac{k*q_{2}*q_{3}}{x^2} = \frac{k*q_{1}*q_{3}}{(2-x)^2} (We cancel k and q₃)

\frac{q_2}{x^2}=\frac{q_1}{(2-x)^2}

q₂(2-x)² = q₁x²

6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)

6(2-x)² = 15(x)²

6(4-4x+x²) = 15x²

24 - 24x + 6x² = 15x²

9x² + 24x - 24 = 0

The solution of the quadratic equation is:

x₁ = 0.775m

x₂ = -3.44m

x₁ meets the conditions for the forces to cancel in q₃

x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel

The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.

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3 years ago
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diamong [38]

Answer:

a= 23.65 ft/s²

Explanation:

given

r= 14.34m

ω=3.65rad/s

Ф=Ф₀ + ωt

t = Ф - Ф₀/ω

= (98-0)×\frac{\pi}{180}/3.65

98°= 1.71042 rad

1.7104/3.65

t= 0.47 s

r₁(not given)

assuming r₁ =20 in

r₁ = r₀ + ut(uniform motion)

u = r₁ - r₀/t

r₀ = 14.34 in= 1.195 ft

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a = 22.25ft/s²

acceleration of collar p related to the rod = 0

coriolis acceleration = 2ωu

= 2× 3.65×1 = 7.3 ft/s²

acceleration of collar p

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√(22.5² + 7.3²)

the magnitude of the acceleration of the collar P just as it reaches B in ft/s²

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An object moves in one dimensional motion with constant acceleration a = 4.2 m/s^2. At time t = 0 s, the object is at x0 = 3.9 m
solong [7]

Answer:

The object will move to Xfinal = 7.5m

Explanation:

By relating the final velocity of the object and its acceleration, I can obtain the time required to reach this velocity point:

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With the equation of the total space traveled and the previously determined time I can obtain the end point of the object on the x-axis:

Xfinal= X0 + /1/2) × a × (t^2) = 3.9m + (1/2) × 4.2( m/s^2) × ((1,7143 s) ^2) =

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mario62 [17]

Answer:25.61 m/s

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Given

truck is moving eastbound with a velocity of 16 m/s

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Velocity of truck relative to the SUV

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v_{ts}=16\hat{i}-(-20\hat{j})

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|v_{ts}|=25.61\ m/s                                                  

5 0
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Dimas [21]
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For problem C I don't know, haven't done that yet in my class. Sorry!
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