Answer:
Explanation:
Given that,
Weight of jet
W = 2.25 × 10^6 N
It is at rest on the run way.
Two rear wheels are 16m behind the front wheel
Center of gravity of plane 10.6m behind the front wheel
A. Normal force entered on the ground by front wheel.
Taking moment about the the about the real wheel.
Check attachment for better understanding
So,
Clock wise moment = anti-clockwise moment
W × 5.4 = N × 16
2.25 × 10^6 × 5.4 = 16•N
N = 2.25 × 10^6 × 5.4 / 16
N = 7.594 × 10^5 N
B. Normal force on each of the rear two wheels.
Using the second principle of equilibrium body.
Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces
ΣFy = 0
Nr + Nr + N — W = 0
2•Nr = W—N
2•Nr = 2.25 × 10^6 — 7.594 × 10^5
2•Nr = 1.491 × 10^6
Nr = 1.491 × 10^6 / 2
Nr = 7.453 × 10^5 N
Answer: 20.4752789138x x 10^23 atoms
To count how many atoms in moles you need to know Avogadro's number. Avogadro's number dictate that for every mole there is 6.022140857 × 10^23 molecule/atoms.
Then 3.4 moles of helium will be 3.4x 6.022140857 x 10^23 atoms= 20.4752789138x x 10^23 atoms
The power in horsepower is 40.1 hp
Explanation:
We start by calculating the work done by the airplane during the climb, which is equal to its change in gravitational potential energy:

where
mg = 11,000 N is the weight of the airplane
is the change in height
Substituting,

Now we can calculate the power delivered, which is given by

where
is the work done
is the time taken
Substituting,

Finally, we can convert the power into horsepower (hp), keeping in mind that

Therefore,

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A) The resultant force is 30.4 N at 
B) The resultant force is 18.7 N at 
Explanation:
A)
In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.
The two forces are:
at
above x-axis
at
above y-axis
Resolving each force:


So, the components of the resultant are:

And the magnitude of the resultant is:

And the direction is:

B)
In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

So we have:

So, the components of the resultant this time are:

And the magnitude is:

And the direction is:

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Answer:
Explanation:
Let the radius of track required be r.
Centripetal force will be provided by frictional force which will be equal to
m v²/ r
Frictional force = mg x μ
So
m v² /r = mg μ
r = v² / μ g =
v = 29 km /h = 8.05 m /s
r =( 8.05 x 8.05 ) /( .32 x 9.8 ) = 20.66 m