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Yuki888 [10]
2 years ago
9

A 1000 kg weather rocket is launched straight up. the rocket motor provides a constant acceleration for 16 s, then the motor sto

ps. the rocket altitude 20 s after launch is 5100m. you can ignore any effects of air resistance. what was the rocket's acceleration during the first 16 s
Physics
1 answer:
8_murik_8 [283]2 years ago
5 0

The rocket's acceleration for the first 16 s is <u>27 m/s².</u>

The rocket starts from rest and experiences uniform upward acceleration a for a time t₁ = 16 s. During this period it travels upwards a distance s₁.

Use the equation of motion

s=ut+\frac{1}{2} t^2

Substitute s₁ for s, 0 m/s for u, 16 s for t =t₁,  and write an equation for s₁ in terms of a.

s=ut+\frac{1}{2} t^2\\ s_1=(0 m/s)(16s)+\frac{1}{2} a(16s)^2\\ s_1=128a........(1)

The engines are switched off after 16 s and for the next 4 s, the rocket travels upwards under the acceleration due to gravity g, which is directed down wards.

Write an expression of the velocity v of the rocket at the end of 16 s.

Use the equation of motion,

v=u+at

Substitute 0 m/s for u and 16 s for t =t₁,

v=u+at\\ v=(0m/s)+a(16 s)\\ v=16a........(2)

The rocket has a speed of v =16a at the beginning of its motion after its engines are switched off.

Determine the distance s₂ the rocket travels under the action of acceleration due to gravity g.

Use the equation of motion

s=ut+\frac{1}{2} t^2

Substitute s₂ for s, 16 a for u, 4 s for t and -9.8 m/s² for g.

s=ut+\frac{1}{2} t^2\\ s_2=(16a)(4 s)-(9.8 m/s^2)(4 s)^2\\ s_2=(64a)-(78.4 m).......(3)

The total distance s traveled by the rocket is given by,

s=s_1+s_2

Add equations (1) and (3) and substitute 5100 m for s.

s=s_1+s_2\\ 5100m=128a+64a-(78.4 m)\\ 192 a = 5178.4\\ a= 26.97 m/s^2

The acceleration of the rocket in the first 16 s is <u>27 m/s^2(2sf)</u>


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Answer:

<h3>0.445</h3>

Explanation:

In friction, the coefficient of friction formula is expressed as;

\mu = \frac{F_f}{R}

Ff is the frictional force = Wsinθ

R is the reaction = Wcosθ

Substitute inti the equation;

\mu = \frac{Wsin \theta}{W cos\theta} \\\mu = \frac{sin \theta}{cos\theta} \\\mu = tan \theta

Given

θ = 24°

\mu = tan 24^0\\\mu = 0.445\\

Hence the coefficient of kinetic friction between the box and the ramp is 0.445

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You are riding a bicycle. If you apply a forward force of 150 N, and you and the bicycle have a combined mass of 90 kg, what wil
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<h3>Hello There!!</h3>

<h3><u>Given</u>,</h3>

Force(F) = 150N

Mass(m) = 90kg

<h3><u>To </u><u>Find,</u></h3>

Acceleration(a) = ?

<h3><u>We know,</u></h3>

F= m×a

150 = 90 \times  \text{a} \\  \\  \text{a} =  \frac{150}{90}  \\  \fbox{cancelling by 3} \\  \\   \text{ a}  = \cancel \frac{150}{90} \\  \\ \text{ a}  =  \frac{5}{3}  = 1.67 \text{m/s} {}^{2}

\therefore  \text{Option A= 1.67 m/s² is the correct answer}

<h3>Hope this helps</h3>
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2 years ago
To determine the muzzle velocity of a bullet fired from a rifle, you shoot the 2.47-g bullet into a 2.43-kg wooden block. The bl
Elza [17]

Answer:

The velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 2.47 g = 0.00247 kg

mass of the wooden block, m₂ = 2.43 kg

initial velocity of the wooden block, u₂ = 0

height reached by the bullet-block system after collision = 0.295 cm = 0.00295 m

let the initial velocity of the bullet on leaving the gun's barrel = v₁

let final velocity of the bullet-wooden block system after collision = v₂

Apply the principle of conservation of linear momentum;

Total initial momentum = Total final momentum

m₁v₁ + m₂u₂ = v₂(m₁ + m₂)

0.00247v₁  + 2.43 x 0  =  v₂(2.43 + 0.00247)

0.00247v₁ = 2.4325v₂ -------(1)

The kinetic energy of the bullet-block system after collision;

K.E = ¹/₂(m₁ + m₂)v₂²

K.E = ¹/₂ (2.4325)v₂²

The potential energy of the bullet-block system after collision;

P.E = mgh

P.E = (2.4325)(9.8)(0.00295)

P.E = 0.07032

Apply the principle of conservation of mechanical energy;

K.E = P.E

¹/₂ (2.4325)v₂² = 0.07032

1.21625 v₂²  = 0.07032

v₂²  = 0.07032  / 1.21625

v₂² = 0.0578

v₂ = √0.0578

v₂ = 0.24 m/s

Substitute v₂ in equation (1), to obtain the initial velocity of the bullet;

0.00247v₁ = 2.4325v₂

0.00247v₁ = 2.4325 (0.24)

0.00247v₁ = 0.5838

v₁ = 0.5838 / 0.00247

v₁ = 236.36 m/s

Therefore, the velocity of the bullet on leaving the gun's barrel is 236.36 m/s.

5 0
2 years ago
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