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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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A mass of 148 g stretches a spring 13 cm. The mass is set in motion from its equlibrium position with a downward velocity of 10
Charra [1.4K]

Answer:

u(t)=1.15 \sin (8.68t)cm

0.3619sec

Explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s

Where g=980 cm/s^2

u(t)=Acos8.68 t+Bsin 8.68t

u(0)=0

Substitute the value

A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t

Substitute u'(0)=10

8.68B=10

B=\frac{10}{8.68}=1.15

Substitute the values

u(t)=1.15 \sin (8.68t)cm

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec

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3 years ago
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70x3=210 mph... its that easy
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HELP ASAP <br> Describe when contact metamorphism occurs?
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