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erik [133]
2 years ago
13

A rocket is fired vertically upwards starting frkm rest. It accelerates at 30m/s for 4secs. At the end of 4secs it runs out of f

uel but continues to rise. How long does it rise with reference to firing position?
Physics
1 answer:
HACTEHA [7]2 years ago
8 0

Answer:

t = 16.5 s

Explanation:

First we apply first equation of motion to the accelerated motion of the rocket:

v_{f1} = v_{i1} + at_{1}

where,

vf₁ = final speed of rocket during accelerated motion = ?

vi₁ = initial speed of rocket during accelerated motion  = 0 m/s

a = acceleration of rocket during accelerated motion = 30 m/s²

t₁ = time taken during accelerated motion = 4 s

Therefore,

v_{f} = 0\ m/s + (30\ m/s^2)(4\ s)\\\\v_{f} = 120\ m/s

Now, we analyze the motion rocket when engine turns off. So, the rocket is now in free fall motion. Applying 1st equation of motion:

v_{f2} = v_{i2} + g t_{2}

where,

vf₂ = final speed of rocket after engine is off = 0 m/s

vi₂ = initial speed of rocket after engine is off  = Vf₁ = 120 m/s

g = acceleration of rocket after engine is off = - 9.8  m/s² (negative sign for upward motion)

t₂ = time taken after engine is off = ?

Therefore,

0\ m/s = 120\ m/s + (- 9.8\ m/s^2)(t_{2})\\\\t_{2} = \frac{120\ m/s}{9.8\ m/s^2}\\\\t_{2} = 12.25\ s

So, the time taken from the firing position till the stopping position is:

t = t_{1} + t_{2}\\\\t = 4 s + 12.5 s

<u>t = 16.5 s</u>

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Answer:

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2 years ago
The velocity profile in fully developed laminar flow in a circular pipe of inner radius R 5 2 cm, in m/s, is given by u(r) 5 4(1
xxMikexx [17]

The question is not clear and the complete clear question is;

The velocity profile in fully developed laminar flow In a circular pipe of inner radius R = 2 cm, in m/s, is given By u(r) = 4(1 - r²/R²). Determine the average and maximum Velocities in the pipe and the volume flow rate.

Answer:

A) V_max = 4 m/s

B) V_avg = 2 m/s

C) Flow rate = 0.00251 m³/s

Explanation:

A) We are given that;

u(r) = 4(1 - (r²/R²))

To obtain the maximum velocity, let's apply the maximum condition for a single-variable continual real valued problem to obtain;

(d/dr)(u(r)) = 0

Thus,

(d/dr)•4(1 - (r²/R²)) = 0

4(d/dr)(1 - (r²/R²)) = 0

If we differentiate, we have;

4(0 - (2r/R²)) = 0

-8r/R² = 0

Thus, r = 0 and with that, the maximum velocity is at the centre of the pipe.

Thus, for maximum velocity, let's put 0 for r in the U(r) function.

Thus,

V_max = 4(1 - 0²/R²) = 4 - 0 = 4 m/s

B) Average velocity is given by;

V_avg = V_max/2

V_avg = 4/2 = 2 m/s

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From question, r = 2cm = 0.02m

A = π x 0.02²

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ΔV = π x 0.02² x 2 = 0.00251 m³/s

8 0
2 years ago
A car starts from rest and accelerates uniformly at 2.0 m/s2 toward the north. A second car starts from rest 4.0 s later at the
yKpoI14uk [10]

Answer:

the correct solution is 13 s

Explanation:

This is a kinematic problem, let's use accelerated rectilinear motion relationships.

For the first car it has an accelerometer of 2.0 m/s²

       x = v₀₁ t + ½ a₁ t²

The second car leaves the same point, but 4.0 seconds later

       x = v₀₂ (t-4) + ½ a₂ (t-4)²

With this form we use the same time for both cars.

The initial speeds are zero for both vehicles leave the rest, at the point where they are located has the same position

        x = ½ a₁ t²

        x = ½ a₂ (t-4)²

Let's solve

       a₁  t² = a₂ (t-4)²

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      t² (1- 2.0 / 4.0) - 8 t +16

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Let's solve the second degree equation

     t = [16 ±√( 16² - 4 32)] / 2  

     t = ½ (16 ± 11,3)

Solutions

     t1 = 13.66 s

     t2 = 2.34 s

These are the mathematical solutions for the meeting point, but car 2 leaves after 4 seconds, so the only solution is 13.66 s

the correct solution is 13 s, if you have to select one the nearest 12s

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<h3>What connection exists between the empirical formula and the molecular formula?</h3>

You can determine the number of atoms of each element in a molecule using its molecular formula. These empirical formulations provide the most basic or reduced elemental ratio of a compound. The empirical formula and the molecular formula of a substance are same if the molecular formula can no longer be decreased.

To know more about atomic mass visit:-

brainly.com/question/17067547

#SPJ4

5 0
8 months ago
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