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3 years ago

6

700 miles away, to try out his luck. The ride there with the wind takes four hours. A few months later almost broke he heads bac

k, on the same plane and the same wind. The flight then, against the wind, takes five hours. What is the speed of the plane and what is the speed of the wind?

1 answer:

Black_prince [1.1K]3 years ago

8 0

Answer:

speed of plane = 157.5 mph

speed o wind = 17.5 mph

Explanation:

Distance traveled to move from one position to other is given as

$d = 700 miles$

time taken by it

$t = 4 hr$

so here the speed is given as

$v_p + v_w = \frac{700}{4}$

$v_p + v_w = 175 mph$

during his return journey we have

$v_p - v_w = \frac{700}{5}$

$v_p - v_w = 140 mph$

now from above two equations we have

$v_p = 157.5 mph$

$v_w = 17.5 mph$

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Richard is driving home to visit his parents. 135 mi of the trip are on the interstate highway where the speed limit is 65 mph .

Elis [28]

<h2>Answer:</h2>

He saves 13.2 minutes

<h2>Explanation:</h2>

Hey! The question is incomplete, but it can be found on the internet. The question is:

How many minutes did he save?

Let's call:

$t_{1}:Time \ at \ speed \ 65mph \\ \\ t_{2}:Time \ at \ speed \ 73mph \\ \\ v_{1}=65mph \\ \\ v_{2}=73mph$

We know that the 135 miles are on the interstate highway where the speed limit is 65 mph. From this, we can calculate the time it takes to drive on this highway. Assuming Richard maintains constant the speed:

$v=\frac{d}{t} \\ \\ d:distance \\ \\ t:time \\ \\ v:velocity \\ \\ t_{1}=\frac{d}{v_{1}} \\ \\ t=\frac{135}{65} \\ \\ t_{1}=2.07 \ hours$

Today he is running late and decides to take his chances by driving at 73 mph, so the new time it takes to take the trip is:

$t_{2}=\frac{135}{73} \\ \\ t_{2}=1.85 \ hours$

So he saves the time $t_{s}$ :

$t_{s}=t_{1}-t_{2}=2.07-1.85=0.22 \ hours$

In minutes:

$t_{s}=0.22h\left(\frac{60min}{1h}\right) \\ \\ \boxed{t_{s}=13.2min}$

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3 years ago

the fireman wishes to direct the flow of water from his hose to the fire at b. determine two possible angles u1 and u2 at which

Lelechka [254]

The two possible angles obtained by using the qudratic equation are;

θ $_{1}$ = 15.10° and θ2 = 73.51°

Given, speed of water = $v_{A}$ = 50ft/s

For the motion along x direction, time period can be calculated as follows:

$s_{x} = (v_{A}) _x_{} } t$

35 = (50 × cosθ) t

t = 0.64 / cosθ

For the motion in y direction, an equation can be obtained as follows:

$s_{y} = (v_{A})_{y} t +\frac{1}{2} (a_{y} )t^{2}$

$s_{y} = (-v_{A}$ $sin$ θ) $} t +\frac{1}{2} (a_{y} )t^{2}$

Plugging in the values we get:

$-20 = (-50_$ $sin$ θ) $} t +\frac{1}{2} (-32.2} )t^{2}$

-20 = -32tanθ - 10.304 $sec^{2}$ θ

Upon solving the above quadratic equation, we get,

tanθ = 0.27 , -3.38

Therefore,

tanθ $_{1}$ = 0.27

θ $_{1}$ = 15.10°

and, tanθ $_{2}$ = -3.38

θ $_{2}$ = 73.51

Learn more about quadratic equation here:

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1 year ago

Calculate the total energy of 2.0 kg object moving horizontally at 10 m/s 50 meters above the surface

mina [271]

We have:

Total Energy: KE + GPE
KE (Kinetic Energy) = $\frac{1}{2} m*v^2$
GPE (Gravitational Potential Energy) = $m*g*h$

Data:
m (mass) = 2.0 Kg
v (speed) = 10 m/s
h (height) = 50 m
Use: g (gravity) = 10 m/s²

Formula:

Total Energy: KE + GPE
$TE = \frac{1}{2} m*v^2 + m*g*h$

Solving:
$TE = \frac{1}{2} m*v^2 + m*g*h$
$TE = \frac{1}{2} *2.0*10^2 + 2.0*10*50$
$TE = \frac{2.0*100}{2} + 1000$
$TE = \frac{200}{2} + 1000$
$TE = 100 + 1000$
$\boxed{\boxed{TE = 1100\:Joule}}\end{array}}\qquad\quad\checkmark$

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3 years ago

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A converging lens brings rays of light together at a focal point. the bending of light rays is the result of

disa [49]

The bending of light rays is the result of refraction of light passing through the lens. A converging lens is curved on both sides such that the rays coming out of it come together at a point (converge). The point at which the right rays meet after refraction is called the focal point which is a real in the convex lens.

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3 years ago

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the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible. what are the bes

slamgirl [31]

Answer:

1.) Micrometres screw gauge

2.) Tape rule.

Explanation:

Given that the diameter and the length of a thin wire, approximately 1m in length, are measured as accurately as possible.

what are the best instruments to use ?

To measure the diameter of a thin wire, the best instrument to use is known as micrometres screw gauge.

And to measure the length of a thin wire up to 1 m, the measuring device can be tape rule or long metre rule.

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3 years ago