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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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4 0
3 years ago
Assume that the equator of the Earth is 24,000 miles long (that’s its circumference). Now, pretend that you are standing somewhe
slega [8]

Answer:

1000 mph

Explanation:

P = Perimeter = 24000 mi

r = Radius of the equator

t = Time taken to complete one rotation = 24 h

Perimeter of a circle is given by

P=2\pi r\\\Rightarrow r=\dfrac{P}{2\pi}\\\Rightarrow r=\dfrac{24000}{2\pi}\ mi

Angular speed is given by

\omega=\dfrac{2\pi}{t}\\\Rightarrow \omega=\dfrac{2\pi}{24}

Velocity if given by

v=r\omega\\\Rightarrow v=\dfrac{24000}{2\pi}\times \dfrac{2\pi}{24}\\\Rightarrow v=1000\ mph

The person would be going at a speed of 1000 mph

3 0
3 years ago
PLEASE PLEASE HELP!!!!!! thank you soo much!!! *will give brainliest*
stiks02 [169]

Answer:

F=1.13\,*\,10^{15} (answer a)

Explanation:

Recall the formula for the Coulomb force:

F=k\frac{q_1*q_2}{d^2}

which in our case gives:

F=k\frac{q_1*q_2}{d^2} \\F=9*10^9\,\frac{63*45}{0.15^2} \\F\approx 1134000\,*\,10^9\\F\approx 1.13\,*\,10^{15}

which agrees with answer a)

6 0
2 years ago
You hold glider A of mass 0.125 kg and glider B of mass 0.375 kg at rest on an air track with a compressed spring of negligible
umka2103 [35]

Answer:

0.2 m/s

Explanation:

m_A = Mass of glider A= 0.125 kg

m_B = Mass of glider B = 0.375 kg

u_A = Initial Velocity of glider A = 0 m/s

u_B = Initial Velocity of glider B = 0 m/s

v_A = Final Velocity of glider A = 0.6 m/s

v_B = Final Velocity of glider B

As linear momentum is conserved

m_{A}u_{A}+m_{B}u_{B}=m_{A}v_{A}+m_{B}v_{B}\\\Rightarrow v_B=\frac{m_{A}u_{A}+m_{B}u_{B}-m_{A}v_{A}}{m_{B}}\\\Rightarrow v_b=\frac{0+0-0.125\times 0.6}{0.375}\\\Rightarrow v_b=-0.2\ m/s

Magnitude of the glider B at this time is 0.2 m/s

8 0
3 years ago
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