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Lisa [10]
3 years ago
14

A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 39.2 m/s^2. Th

e acceleration period lasts for time 10.0 ss until the fuel is exhausted. After that, the rocket is in free fall.
Required:
Find the maximum height ymax reached by the rocket.
Physics
1 answer:
Vanyuwa [196]3 years ago
4 0

Answer:

9800 m

Explanation:

During acceleration, given:

v₀ = 0 m/s

a = 39.2 m/s²

t = 10.0 s

Find: v and Δy

v = at + v₀

v = (39.2 m/s²) (10.0 s) + 0 m/s

v = 392 m/s

Δy = v₀ t + ½ at²

Δy = (0 m/s) (10.0 s) + ½ (39.2 m/s²) (10.0 s)²

Δy = 1960 m

During free fall, given:

v₀ = 392 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (392 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 7840 m

Therefore, h = 1960 m + 7840 m = 9800 m.

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Answer:

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k_{b} = 1.38\times 10^{- 23} J/k = Boltzmann's constant

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Solution:

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\lambda_{p} = \frac{h}{p_{p}}

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where

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Now,

Kinetic energy of a proton = thermal kinetic energy

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}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}

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