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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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Answer:

The work done by the gravel to stop the truck is 520.44 kJ

Explanation:

<u>Step 1</u>: Data given

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Velocity of  the truck = 20.68 m/s

distance = 26.6 meters

<u>Step 2:</u> Calculate initial kinetic energy

sin 9.5° = 0.165

h = ℓ*sin 9.5° = 26.6*0.165= 4.39 m

Ek = 1/2m*Vo² = 1/2*3047.8*20.68² = 651714.7 Joule = 651.7 kJ  = initial kinetic energy

<u>Step 3: </u>Calculate potential energy

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5 0
3 years ago
If 50 km thick crust having an average density of 3.0 g/cm3 has a surface elevation of 2.5 km above sea level, what would you pr
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Answer:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

Explanation:

To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.

The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.

Thus let the density of the material be Pm

50*3= 47.5*Pm

Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube

Thus with an average density of 2.8gram per centimeter cube

50*2.8= (50-x)*3.16

(50-x)= (50*2.8)/3.16

50-x=44.3

x=50-44.3= 5.7

5 0
3 years ago
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