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2 years ago

13

An airplane cruising at a constant velocity and altitude. Which of the following diagrams best represents the four forces of the

airplane?

1 answer:

Jobisdone [24]2 years ago

7 0

Answer:

You need to put the diagrams. repost and include the diagrams

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Assume you have a rocket in Earth orbit and want to go to Mars. The required change in velocity is ΔV≈9.6km/s . There are two op

Nostrana [21]

Answer: Part 1: Propellant Fraction (MR) = 8.76

Part 2: Propellant Fraction (MR) = 1.63

Explanation: The Ideal Rocket Equation is given by:

Δv = $v_{ex}.ln(\frac{m_{f}}{m_{e}} )$

Where:

$v_{ex}$ is relationship between exhaust velocity and specific impulse

$\frac{m_{f}}{m_{e}}$ is the porpellant fraction, also written as MR.

The relationship $v_{ex}$ is: $v_{ex} = g_{0}.Isp$

To determine the fraction:

Δv = $v_{ex}.ln(\frac{m_{f}}{m_{e}} )$

$ln(MR) = \frac{v}{v_{ex}}$

Knowing that change in velocity is Δv = 9.6km/s and $g_{0}$ = 9.81m/s²

<u>Note:</u> Velocity and gravity have different measures, so to cancel them out, transform km in m by multiplying velocity by 10³.

<u />

<u>Part 1</u>: Isp = 450s

$ln(MR) = \frac{v}{v_{ex}}$

ln(MR) = $\frac{9.6.10^{3}}{9.81.450}$

ln (MR) = 2.17

MR = $e^{2.17}$

MR = 8.76

<u>Part 2:</u> Isp = 2000s

$ln(MR) = \frac{v}{v_{ex}}$

ln (MR) = $\frac{9.6.10^{3}}{9.81.2.10^{3}}$

ln (MR) = 0.49

MR = $e^{0.49}$

MR = 1.63

8 0

2 years ago

If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

vovikov84 [41]

Complete Question

In an action movie, the villain is rescued from the ocean by grabbing onto the ladder hanging from a helicopter. He is so intent on gripping the ladder that he lets go of his briefcase of counterfeit money when he is 130 m above the water. If the briefcase hits the water 6.0 s later, what was the speed at which the helicopter was ascending?

Answer:

The speed of the helicopter is $u = 7.73 \ m/s$

Explanation:

From the question we are told that

The height at which he let go of the brief case is h = 130 m

The time taken before the the brief case hits the water is t = 6 s

Generally the initial speed of the briefcase (Which also the speed of the helicopter )before the man let go of it is mathematically evaluated using kinematic equation as

$s = h+ u t + 0.5 gt^2$

Here s is the distance covered by the bag at sea level which is zero

$0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $0 = 130+ u * (6) + 0.5 * (-9.8) * (6)^2$

=> $u = \frac{-130 + (0.5 * 9.8 * 6^2) }{6}$

=> $u = 7.73 \ m/s$

7 0

2 years ago

carbon-14 has a half-life of approximately 5,700 years. a fossil shell contain 25% of the original amount of its carbon-14. appr

denis23 [38]

The half-life equation $m=m_{0} (\frac{1}{2})^n$ in which <em>n </em>is equal to the number of half-lives that have passed can be altered to solve for <em>n.</em>

<em> $n = \frac{log(\frac{m}{m_{0}} )}{log(\frac{1}{2})}$ </em>

<em> $\frac{log(\frac{.25}{1} )}{log(\frac{1}{2})} = 2$ </em>

Then, the number of half-lives that passed can be multiplied by the length of a half-life to find the total time.

<em>2 * 5700 = </em>11400 yr

3 0

2 years ago

Imagine that the above hoop is a tire. the coefficient of static friction between rubber and concrete is typically at least 0.9.

Stels [109]

The hoop is attached.

Consider that the friction force is given by:
F = μ·N
   = μ·m·g·cosθ

We also know, considering the forces of the whole system, that:
F = -m·a + m·g·sinθ
and
a = (1/2)·<span>g·sinθ

Therefore:
</span>-(1/2)·m·g·sinθ + m·g·sinθ = <span>μ·m·g·cosθ
</span>(1/2)·m·g·sinθ = <span>μ·m·g·cosθ
</span>μ = (1/2)·m·g·sinθ / <span>m·g·cosθ
   = </span>(1/2)·tanθ

Now, solve for θ:
θ = tan⁻¹(2·μ)
   = tan⁻¹(2·0.9)
   = 61°

Therefore, the maximum angle <span>you could ride down without worrying about skidding is 61°.</span>

5 0

2 years ago

A car came to a stop from a speed of 35 m/s in a time of 8.1 seconds. What was the acceleration of the car?

uranmaximum [27]

Simply subtract the two velocities and divide by 8.1,

$\frac{0 - 35}{8.1} = - 4.32$

~~

I hope that helps you out!!

Any more questions, please feel free to ask me and I will gladly help you out!!

~Zoey

5 0

2 years ago

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