answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
eduard
2 years ago
15

A small airplane is sitting at rest on the ground. Its center of gravity is 2.58 m behind the nose of the airplane, the front wh

eel (nose wheel) is 0.800 m behind the nose, and the main wheels are 3.02 m behind the nose. What percentage of the airplane's weight is supported by the nose wheel?
Physics
1 answer:
otez555 [7]2 years ago
7 0

Answer:

The percentage of the weight supported by the front wheel is  A= 19.82 %

Explanation:

From the question we are told that

   The center of gravity of the plane to its nose  is  z = 2.58 m

    The distance of the front wheel of the plane to  its nose is l = 0.800\ m

     The distance of the main wheel of the plane to its nose is e =  3.02 \ m

At equilibrium  the Torque about the nose of the airplane is mathematically represented as

          mg (z- l) -  G_B *(e - l) = 0

Where m is the mass of the airplane

          G_B is the weight of the airplane supported by the main wheel  

       So  

             G_B =\frac{mg (z-l)}{(e - l)}

Substituting values

            G_B =\frac{mg (2.58 -0.8 )}{(3.02  - 0.80)}

           G_B = 0.8018 mg

Now the weight supported at the frontal wheel is mathematically evaluated as

           G_F = mg - G_B

Substituting values      

       G_F = mg - 0.8018mg    

      G_F = (1 - 0.8018) mg      

     G_F = 0.1982 mg    

Now the weight of the airplane is  =  mg

Thus percentage of this weight supported by the front wheel is  A = 0. 1982 *100 = 19.82 %

You might be interested in
A Turtle and a Snail are 360 meters apart, and they start to move towards each other at 3 p.m. If the Turtle is 11 times as fast
Neko [114]

Answer:

Snail's speed = \frac{30m}{2400s} = 0.0125m/s

Turtle's speed =  \frac{330m}{2400s} = 0.1375m/s

Explanation:

Let the snail's speed be x m/s

The turtle's speed then is 11x m/s

Speed = Distance ÷ Time

Since speed and distance are directly proportional;

The ratio of the distances snail and turtle cover before they meet is x:11x respectively.

Simplified, the ratio of snail distance : turtle distance = 1:11

So snail covers a distance of \frac{1}{12} × 360 = 30m

And turtle covers a distance of \frac{11}{12} × 360 = 330m

The time each took before they met is 40 × 60 = 2400 seconds

Snail's speed = \frac{30m}{2400s} = 0.0125m/s

Turtle's speed =  \frac{330m}{2400s} = 0.1375m/s

8 0
2 years ago
Read 2 more answers
The apartment’s explosion, reportedly caused by a gas leak, produced a violent release of gas and heat. the heat increased the _
uranmaximum [27]
<h2>Apartment Explosion Reported </h2>

The apartment’s explosion, reportedly caused by a gas leak, produced a violent release of gas and heat. The heat increased the temperature of the air in the room, which means an increase in the air's molecular kinetic energy.

When heat is provided then temperature increases and the molecules of substances move rapidly by increase of kinetic energy (K.E) temperature increases. It is understood that heat increases temperature.

6 0
2 years ago
Read 2 more answers
A 250 GeV beam of protons is fired over a distance of 1 km. If the initial size of the wave packet is 1 mm, find its final size
Margarita [4]

Answer:

The final size is approximately equal to the initial size due to a very small relative increase of 1.055\times 10^{- 7} in its size

Solution:

As per the question:

The energy of the proton beam, E = 250 GeV =250\times 10^{9}\times 1.6\times 10^{- 19} = 4\times 10^{- 8} J

Distance covered by photon, d = 1 km = 1000 m

Mass of proton, m_{p} = 1.67\times 10^{- 27} kg

The initial size of the wave packet, \Delta t_{o} = 1 mm = 1\times 10^{- 3} m

Now,

This is relativistic in nature

The rest mass energy associated with the proton is given by:

E = m_{p}c^{2}

E = 1.67\times 10^{- 27}\times (3\times 10^{8})^{2} = 1.503\times 10^{- 10} J

This energy of proton is \simeq 250 GeV

Thus the speed of the proton, v\simeq c

Now, the time taken to cover 1 km = 1000 m of the distance:

T = \frac{1000}{v}

T = \frac{1000}{c} = \frac{1000}{3\times 10^{8}} = 3.34\times 10^{- 6} s

Now, in accordance to the dispersion factor;

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{ht_{o}}{2\pi m_{p}\Delta t_{o}^{2}}

\frac{\delta t_{o}}{\Delta t_{o}} = \frac{6.626\times 10^{- 34}\times 3.34\times 10^{- 6}}{2\pi 1.67\times 10^{- 27}\times (10^{- 3})^{2} = 1.055\times 10^{- 7}

Thus the increase in wave packet's width is relatively quite small.

Hence, we can say that:

\Delta t_{o} = \Delta t

where

\Delta t = final width

3 0
2 years ago
There is an electric field in the region between the two plates. The magnitude of this electric field is ed. This imposes anothe
krok68 [10]

Answer:

it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

Explanation:

The configuration of parallel plates is called a capacitor and is widely used to create constant electric fields inside.

 To obtain this field it is essential that the charge on the plates are of the same magnitude, but in the opposite direction

This is so that the fields created by each plate can be added inside and subtracted from the outside of the plates

5 0
2 years ago
You have a light spring which obeys Hooke's law. This spring stretches 2.92 cm vertically when a 2.70 kg object is suspended fro
ehidna [41]

(a) 907.5 N/m

The force applied to the spring is equal to the weight of the object suspended on it, so:

F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N

The spring obeys Hook's law:

F=k\Delta x

where k is the spring constant and \Delta x is the stretching of the spring. Since we know \Delta x=2.92 cm=0.0292 m, we can re-arrange the equation to find the spring constant:

k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance \Delta x is equal to the elastic potential energy stored by the spring, given by:

W=U=\frac{1}{2}k\Delta x^2

Substituting k=907.5 N/m and \Delta x=8.80 cm=0.088 m, we find the amount of work that must be done:

W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J

5 0
2 years ago
Other questions:
  • Thermals created by warm air rising and cold air sinking are called___?
    11·1 answer
  • A geologist is studying the shore along a river. She finds a pile of rocks at the base of a riverbank. These broken rock pieces
    14·2 answers
  • A 55-kg box is being pushed a distance of 7.0 m across the floor by a force whose magnitude is 160 N. The force is parallel to t
    9·1 answer
  • Through how many volts of potential difference must an electron, initially at rest, be accelerated to achieve a wave length of 0
    13·1 answer
  • Give two ways of reversing the direction of the forces on the coil in the electric motor?​
    10·1 answer
  • A cart with mass m1 = 3.2 kg and initial velocity of v1,i = 2.1 m/s collides with another cart of mass M2 = 4.3 kg which is init
    7·1 answer
  • A stationary 1.67-kg object is struck by a stick. The object experiences a horizontal force given by F = at - bt2, where t is th
    13·1 answer
  • A government agency estimated that air bags have saved over 14,000 lives as of April 2004 in the United States. (They also state
    13·1 answer
  • Consider a point on a bicycle wheel as the wheel makes exactly four complete revolutions about a fixed axis. Compare the linear
    8·1 answer
  • A pesticide was applied to a population of roaches, and it was determined that the LD50 was 55mgkg. If the average mass of a roa
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!