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lesantik [10]
3 years ago
13

One kilogram of the explosive TNT supplies explosive energy of about 4.2×10^6 J. Let's see what that energy can achieve in lifti

ng objects from Earth's surface to altitudes of hundreds of kilometers. We approach the issue through gravitational potential energy.
a. Assume the acceleration due to gravity is constant at g = 9.81 m/s2 from Earth's surface up to an altitude of 630 km above the surface. (A satellite at such an altitude would be in a "low Earth orbit.") What is the gravitational potential energy, in joules, of a 1.0-kg object at that altitude, taking the reference level for gravitational potential energy (U -0) at Earth's surface?
b. How many kilograms of TNT would need to be exploded, if the released energy were somehow used to raise a 10-kg object from Earth's surface to an altitude of 630 km at 100% efficiency, assuming constant g?
c. Again take the reference level at Earth's surface. Earth's radius and mass are 6.38x103 km and 5.97x1024 kg, respectively. Now calculate the gravitational potential energy, in joules, of a 1 0-kg object at an altitude of 630 km without assuming constant 25%
d. Without assuming constant g, how many kilograms of TNT would need to be exploded, if the released energy were somehow used to raise a 1-0-kg object from Earth's surface to an altitude of 630 km at 100% efficiency?
Physics
1 answer:
makkiz [27]3 years ago
3 0

Answer:

4,0000

Explanation:

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Oliga [24]

Answer:

The answer to the question is

The roller coaster will reach point B with a speed of 14.72 m/s

Explanation:

Considering both kinetic energy KE = 1/2×m×v² and potential energy PE = m×g×h

Where m = mass

g = acceleration due to gravity = 9.81 m/s²

h = starting height of the roller coaster

we have the given variables

h₁ = 36 m,

h₂ = 13 m,

h₃ = 30 m

v₁ = 1.00 m/s

Total energy at point 1 = 0.5·m·v₁² + m·g·h₁

= 0.5 m×1² + m×9.81×36

=353.66·m

Total energy at point 2 = 0.5·m·v₂² + m·g·h₂

= 0.5×m×v₂² + 9.81 × 13 × m = 0.5·m·v₂² + 127.53·m

The total energy at 1 and 2 are not equal due to the frictional force which must be considered

Total energy at point 2 = Total energy at point 1 + work done against friction

Friction work = F×d×cosθ = (\frac{1}{5} × mg)×60×cos 180 = -117.72m

0.5·m·v₂² + 127.53·m = 353.66·m -117.72m

0.5·m·v₂² = 108.41×m

v₂² = 216.82

v₂  =  14.72 m/s

The roller coaster will reach point B with a speed of 14.72 m/s

8 0
3 years ago
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