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suter [353]
3 years ago
13

A downed pilot fires a flare from a flare gun. The flare an initial speed of 250 m/s and is fired at an angle of 35° to the grou

nd. Find the maximum height the flare will reach. How long does it take for the flare to reach its maximum altitude ? What will be the final vertical velocity ?
Physics
2 answers:
den301095 [7]3 years ago
7 0

Answer:

to reach the maximum altitude? is 15 sec

Explanation:

bagirrra123 [75]3 years ago
7 0

Answer:

Maximum Altitude: vi = 250 m/s At the maximum height q = 35° vy,f = vy,i − g∆t = 0 g = 9.81 m/s2 vy,i = vi (sin q) = g∆t ∆t =  vi (si g n q)  =  (250 9 m .8 / 1 s) m (s / i s n 2  35°) ∆t = 15 s

Vertical Velocity: Vyf = 439.09

Explanation:

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Answer:

State your hypothesis as concisely, and to the point, as possible. A hypothesis is usually written in a form where it proposes that, if something is done, then something else will occur. Usually, you don't want to state a hypothesis as a question. You believe in something, and you're seeking to prove it.

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3 years ago
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A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru
r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

          = 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

s = u t +\dfrac{1}{2}at^2

-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0
3 years ago
I NEED HELP ASAP WHAT IS THE DISTANCE MOVED IN 15 SECONDS
Oduvanchick [21]

Answer:

5 I think

Explanation:

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2 years ago
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For each of the problems below, you will need to draw a graph to find the solution.
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Answer:

Final velocity (v) of an object equals initial velocity (u) of that object plus acceleration (a) of the object times the elapsed time (t) from u to v. Use standard gravity, a = 9.80665 m/s2, for equations involving the Earth's gravitational force as the acceleration rate of an object.

Explanation:

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7. What is the velocity of an object that has a mass of 4.5 kilograms and
labwork [276]

Answer:

266.66 m/s

Explanation:

p=mv

1200=4.5v

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