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4 years ago

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1.A test rocket is launched vertically from ground level (y = 0 m), at time t = 0.0 s. The rocket engine provides constant upwar

d acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 49 m and acquired a velocity of 30 m/s How long did the burn phase last 2.What is the max height the rocket will reach

1 answer:

Charra [1.4K]4 years ago

6 0

Answer:

1. t = 3.27 seconds

2. y = 147.3 m

Explanation:

Newton's Laws of Motions.

y = v₁t + 1/2 at²

a = (v₂-v₁)/t

where

y = the vertical distance travelled

v₁ = the initial velocity

v₂ = the final velocity

t = the time

a = the acceleration

final velocity is equal to 0.

So, v₂ = 0.

a = (v₂-v₁)/t

a = (0-30)/t

a = -30/t

plugin values into the first equation:

y = v₁t + 1/2 at²

49 = 30t + 1/2 (-30/t)t²

49 = 30t -15t

49 = 15 t

t = 49/15

t = 3.27 seconds

2.

y = v₁t + 1/2 at²

a = -30/3.27

a = 9.2

y = 30(3.27) + 1/2(9.2) 3.27²

y = 147.3 m

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3 years ago

A bicyclist starts from rest and accelerates at a rate of 2.3 m/s^2 until it reaches a speed of 23 m/s. It then slows down at a

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Answer:

33 seconds.

Explanation:

The equation for speed with constant acceleration at time t its:

$V(t) \ = \ V_0 \ + \ a \ t$

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<h3>First half of the problem</h3>

Starting at rest, the initial speed will be zero, so

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$V(t_{f1}) = 23 \frac{m}{s}$

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$23 \frac{m}{s} = 2.3 \ \frac{m}{s^2} t_{f1}$

$\frac{23 \frac{m}{s}}{2.3 \ \frac{m}{s^2}} = t_{f1}$

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So, for the first half of the problem we got a time of 10 seconds.

<h3>Second half of the problem</h3>

Now, the initial speed will be

$V_0 = 23 \frac{m}{s}$ ,

the acceleration

$a=-1.0 \frac{m}{s^2}$ ,

with a minus sign cause its slowing down, the final speed will be

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$\frac{23 \frac{m}{s}}{1.0 \frac{m}{s^2}} = t_{f2}$

$23 s = t_{f2}$

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<h3>Total time</h3>

$t_total = t_{f1} + t_{f2} = 33 \ s$

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3 years ago

A 6.2 kg ladder, 1.97 m long, rests on two sawhorses. Sawhorse A is 0.64 m from one end of the ladder, and sawhorse B is 0.17 m

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Answer:

42.69 N and 18.07 N

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Distance of sawhorse B from other end=0.17 m

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Where $g=9.8 m/s^2$

Torque applied on Sawhorse A= $0.345F_a$

Torque applied on Sawhorse B= $0.815F_b$

In equilibrium

$0.345F_a=0.815F_b$

$F_b=\frac{0.345}{0.815}F_a$

Total force= $F_a+F_b$

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$\frac{0.815F_a+0.345F_a}{0.815}=60.76$

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$F_a=\frac{60.76\times 0.815}{1.16}=42.69 N$

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