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SCORPION-xisa [38]
3 years ago
8

An airplane that travels 550 mph in still air encounters a 50​-mph headwind. how long will it take the plane to travel 1250 mi i

nto the​ wind?
Physics
1 answer:
Natali [406]3 years ago
5 0

As the headwind comes at the head of the plane, therefore 50 mph headwind slows the plane down.

Thus actual speed of plane with headwind, v=550- 50=500 mph

Now using the formula , velocity= \frac{distance}{time}

 500 mph = \frac{1250 mile}{time}

time = \frac{1250 mil}{500 mph}

time = 2.5 hour

Hence plane will take time to complete the distance 1250 mile in 2.5 hours

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A vector in the xy plane has components -14.0 units in the x-direction and 30.0 units in the y-direction. What is the magnitude
OverLord2011 [107]

\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}

We would calculate the magnitude by applying pythagorean theorem:

\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}

\longrightarrow \sf{Magnitude = 33.12}

\longrightarrow \sf{The \:  vector \: is \: (- 14, 30)}

The angle between two vectors is given by the formula:

\sf{\longrightarrow \small  \cos \emptyset =  \dfrac{(a1b1 + a2b2)}{ \sqrt{(a1)^2 + (a2)^2√(b1)^2 + (b2)^2} } }

In two dimensional, the x axis of vector form is:

\small\sf{\longrightarrow (b1, b2) = (1, 0) }

\sf{\longrightarrow \small \cos \:  \emptyset =  \dfrac{(14 * 1 + 30 x 0)}{( \sqrt{(-14)^2 + (30)^2)(√(1)^2 + (0)^2)} } }

\small\longrightarrow \sf{ \dfrac{14}{33.12} }

\small\longrightarrow \sf{\emptyset  \:  = arcCos (\dfrac{ - 14}{33.12} )}

\small\longrightarrow \sf{\emptyset= 115^\circ}

\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}

\small\bm{The \: angle  \: between  \: the  \: vector \: }

\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}

6 0
2 years ago
Steel is made of atoms of iron and carbon. Would iron
Nadusha1986 [10]

Still need the answer??

7 0
3 years ago
Read 2 more answers
What type of tool seems to be the mist accurate for measuring liquids?
Anit [1.1K]

Volumetric flasks are most accurate

6 0
3 years ago
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A 0.60-kilogram softball initially at rest is hit with a bat. The ball is in contact with the bat for 0.20 second and leaves the
Furkat [3]

Answer:

75 Newtons.

Explanation:

From Newton's second law of motion,

F = m(v-u)/t................... Equation 1

Where F = force exerted by the ball on the bat, m = mass of the ball, v = final velocity of the ball, u = initial velocity of the ball, t = time

Given: m = 0.6 kilogram, u = 0 meter per seconds (at rest), v = 25 meters per seconds, t = 0.2 seconds.

Substitute into equation 1

F = 0.6(25-0)/0.2

F = 3(25)

F = 75 Newton.

Hence the magnitude of the average force exerted by the ball on the bat = 75 Newtons.

8 0
4 years ago
At time t1 = 14 s, a car is located at 99, 80, 27 m and has velocity 4, 0, −3 m/s. At time t2 = 18 s, what is the position of th
Korvikt [17]

Answer:

115, 80, 15m

Explanation

t1 = 14s

t2 = 18s

change in time = 4s (18-14)

r(final) = r(initial) + (average velocity) x (change in time)

multiply the average velocity with the change in time

= (4, 0, -3) x 4 = 16, 0, -12

now we'll add this value to the initial position of the car

(99, 80, 27)m + (16, 0, -12)m = (115, 80, 15)m

8 0
4 years ago
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