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Rudik [331]
3 years ago
5

In 1865, jules verne proposed sending men to the moon by firing a space capsule from a 220-m-long cannon with final speed of 10.

97 km/s. what would have been the unrealistically large acceleration experienced by the space travelers during their launch? (a human can stand an acceleration of 15g for a short time.)
Physics
1 answer:
Aneli [31]3 years ago
5 0
I'm going to assume that the 10.97 km/s is the velocity at the end of the cannon.

vi = 0
d = 220m
vf = 10.97 km/s = 10970 m/s
a = ???

vf^2 = vi^2 + 2*a*d
10970^2 = 0 + 2*a*220
10970^2 / 440 = a You are right this is a pretty big number.
2.73 * 10^5 = a


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Consider a roller coaster begins 15m above the ground. If the cart has a mass of 75kg, what is the velocity of the cart halfway
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v = 12.12 m/s

Explanation:

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The mass of the cart, m = 75 kg

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We need to find the velocity of the cart halfway to the ground. Let the velocity be v. Using the conservation of energy at this position, h = 15/2 = 7.5 m

mgh=\dfrac{1}{2}mv^2\\\\v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 7.5} \\\\v=12.12\ m/s

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a bal is launched upward with a velocity of v0 from the edge of a cliff of height D. it reaches a maximum height of H above its
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D/H =15

Explanation:

  • We can find first the peak height H, taking into consideration, that at the maximum height, the ball will reach momentarily to a stop.
  • At this point, we can find the value of H, applying the following kinematic equation:

       v_{f} ^{2} -v_{0} ^{2} = 2* g* H (1)

  • If vf=0, if we assume that the positive direction is upwards, we can find the value of H as follows:

       H = \frac{v_{0} ^{2} }{2*g} (2)

  • We can use the same equation, to find the value of D, as follows:

        v_{f} ^{2} -v_{1} ^{2} = 2* g* D (3)

  • In order to find v₁, we can use the same kinematic equation that we used to get H, but now, we know that v₀ = 0.
  • When we replace these values in (1), we find that  v₁ = -v₀.
  • Replacing in (3), we have:

        (4*v_{0})^{2} - (-v_{0}) ^{2}  = 2* g* D\\ \\ 15*v_{0}^{2}  = 2*g*D

  • Solving for  D:

       D = \frac{15*v_{0} ^{2} }{2*g}

  • From (2) we know that H can be expressed as follows:

       H = \frac{v_{0} ^{2} }{2*g}

  • ⇒ D = 15 * H

        \frac{D}{H} = 15

3 0
3 years ago
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