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Ymorist [56]
3 years ago
6

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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A car mass of 1.2 x 10 kilograms starts from rest and attains a speed of 20 meters/seconds in 5 seconds. What net force acted on
lorasvet [3.4K]

Net force on the car=F=4.8 x 10³ N

Explanation:

mass of car= 1.2 x 10³ Kg

initial velocity= Vi=0

Final velocity= Vf= 20 m/s

time = t= 5 s

Using kinematic equation,

Vf= Vi + at

20= 0 + a (5)

5 a=20

a= 20/5

a= 4 m/s²

Now force is given by F = ma

F= 1.2 x 10³ (4)

F=4.8 x 10³ N

7 0
2 years ago
The force that causes an impulse and a force that causes a change in momentum are?
Aleks [24]
I don't understand the question but impulse and momentum is the same. So maybe is the force same too
3 0
2 years ago
Read 2 more answers
Review
sleet_krkn [62]

Answer:

yes

Explanation:

8 0
3 years ago
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You find an unmarked blue laser on your way to physics class. When you get to class you realize you can determine the wavelength
Ray Of Light [21]

Answer:

wavelength \lambda = 437.27 nm

Explanation:

given data

first bright fringe = 2.96 mm

slit separation = 0.325 mm

distance D = 2.20 m

solution

we know that this is double slit experiment

so we apply here Fringe width formula that is

β = \frac{D\lambda}{d}    ....................1

\lambda is Wavelength of light and  D is Distance between screen and slit and d is slit width

so put here value and we get

\lambda = \frac{2.96*0.325*10^{-6}}{2.20}    

\lambda = 437.27 × 10^{-9} m

wavelength \lambda = 437.27 nm

4 0
2 years ago
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
3 years ago
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