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3 years ago

In which city did the wright brothers take off on their very first flight

Physics

1 answer:

Nezavi [6.7K]3 years ago

7 0

<h2>Answer: Kitty Hawk, North Carolina </h2>

The Wright brothers, Wilbur and Orville, were pioneers of aviation, since they flew in a device heavier than air, which was inconceivable at that time.

Their first successful flight was on December 17th, 1903 in Kitty Hawk, North Carolina, which lasted only 12 seconds in which their plane (the Flyer I, with 341 kg, 6.4 m long and a wingspan of 12.3 m) traveled 37 m without touching the ground. This was achieved through the help of an external catapult that "threw" them into the air.

It should be noted that the Wright brothers only studied until high school, however, their passion for solving the problem of the human inability to fly, their perseverance and experience acquired over the years in their bicycle company, led them to reach that goal. An achievement that marked the beginning of the aviation era.

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2) Two railway tracks are parallel to west east direction. Along one track, train A moves with a speed of 45 m/s from

OLEGan [10]

Answer:

$Relative\ Velocity = 105m/s$

Explanation:

Given

$V_A = 45m/s$

$V_B = 60m/s$

Required

Determine the speed of B w.r.t A

The question implies that, we determine the relative velocity of B w.r.t A

Because both trains are moving towards one another, the required velocity is a $sum\ of\ velocities\ of$ both trains:

This is shown below:

$Relative\ Velocity = V_A + V_B$

$Relative\ Velocity = 45m/s + 60m/s$

$Relative\ Velocity = 105m/s$

4 0

3 years ago

A person on a merry-go-round makes a complete revolution in about 3.17 s.

GREYUIT [131]

Answer:

second one

Explanation:

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8 0

3 years ago

A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0

MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

$\omega=\frac{2\pi}{T}$

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

$\omega=\frac{2\pi}{1 s}=6.28 rad/s$

Now we can find the linear speed of the ladybug, which is given by

$v=\omega r$

where:

$\omega=6.28 rad/s$ is the angular speed

r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation

Substituting, we find

$v=(6.28)(0.65)=4.1 m/s$

Learn more about angular speed:

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7 0

4 years ago

If an oscillating mass has a frequency of 1.25 Hz, it makes 100 oscillations in

KatRina [158]

Answer:

Time, t = 80 seconds

Explanation:

Given that,

The frequency of the oscillating mass, f = 1.25 Hz

Number of oscillations, n = 100

We need to find the time in which it makes 100 oscillations. We know that the frequency of an object is number of oscillations per unit time. It is given by :

$f=\dfrac{n}{t}$