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3 years ago

A jet airliner moving initially at 858 mph

Physics

1 answer:

Vika [28.1K]3 years ago

8 0

Answer:

1546.1 mph

Explanation:

Let's take the east direction as positive x-direction and north as positive y-direction.

The components of the initial velocity of the jet are:

$v_x = 858 mph\\v_y = 0$

while the components of the wind's velocity are

$a_x = (777)(cos 38^{\circ})=612.2 mph\\a_y = (777)(sin 38^{\circ})=478.4 mph$

So, the components of the new velocity of the jet are:

$v'_x = v_x + a_x = 858+612.2 =1470.2 mph\\v'_y = v_y+a_y = 0+478.4=478.4 mph$

And therefore, the new speed is the magnitude of the new velocity, so:

$v'=\sqrt{v'_x^2+v'_y^2 }=\sqrt{(1470.2)^2+(478.4)^2}=1546.1 mph$

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

5 0

3 years ago

A horizontal force of 400 N is exerted on a 2.0-kg ball as it rotates (at

frutty [35]

Answer:

the speed of the ball is 10 m/s

Explanation:

Given;

magnitude of exerted force, F = 400 N

mass of the ball, m = 2 kg

radius of the circle, r = 0.5

The speed of the ball is calculated by applying centripetal force formula;

$F = \frac{mv^2}{r} \\\\v^2 = \frac{Fr}{m}\\\\v = \sqrt{\frac{Fr}{m}}\\\\ v = \sqrt{\frac{400*0.5}{2}}\\\\v = 10 \ m/s$

Therefore, the speed of the ball is 10 m/s

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4 years ago

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Can you please answer my question correctly ASAP

lana66690 [7]

Answer:

1) C, 2) C, 3) B, 4) D, 5) A/B, 6) D, 7) D, 8) A, 9) B, 10) A

Explanation:

1) Seconds, minutes and days are time units. (Correct answer: C)

2) The displacement is shortest distance between the point of origin and the point of destination. (Correct answer: C)

3) The velocity is the rate of change of position in time. (Correct answer: B)

4) Graphs are a very useful ressource to infer characteristics inherent to a studied physical variable. (Correct answer: D)

5) There are two possible answers, since both milimeters and meters are units for distance. (Correct answers: A/B)

6) The distance is the arc length of the entire path that object travelled. (Answer: D)

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8) In this case, we use a distance-time graph, where travelled distance is the dependent variable and time is the independent variable. (Answer: A)

9) The point of origin is the initial position of the object. (Answer: B)

10) The point of destination is the final position of the object. (Answer: A)

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3 years ago

you need to measure the length of your driveway. which measuring tool would you produce accurate results that you would expectto

lesantik [10]

I’d assume a tape measure?

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4 years ago

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Consider the reaction.

Free_Kalibri [48]

2.1 x 102

Is the correct solution for this problem

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3 years ago

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