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Makovka662 [10]

1 year ago

15

The reactive force developed by a jet engine to push an airplane forward is called thrust, and the thrust developed by the engin

e of a boeing 777 is about 85400 lbf. express this thrust in n and kgf.

1 answer:

finlep [7]1 year ago

3 0

The value of the thrust developed by the engine of a boeing 777 in N and Kgf are ;

i) 376616N

ii) 38430Kgf

<h3>What is force?</h3>

Force is a push or a pull. The reactive force always serve to balance the applied force. We are here asked to convert the the thrust developed by the engine of a boeing 777 which is about 85400 lbf to the following units;

i) N

ii)kgf

Thus;

1 Ib = 0.45 Kg

1 lbf = 0.45 Kg * 9.8 m/s^2 = 4.41 N

We know that;

1 lbf = 4.41 N

85400 lbf = 85400 lbf. * 4.41 N/1 lbf

= 376616N

Again;

1 lbf = 0.45 Kgf

85400 lbf = 85400 lbf * 0.45 Kgf/1 lbf

= 38430Kgf

Learn more about force:brainly.com/question/13191643

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Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is $2.2 m/s^2$

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, $T^2 \propto r^3$ , where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

$M g sin \theta - T = Ma$ (1)

where

$Mg sin \theta$ is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

$g=9.8 m/s^2$ (acceleration of gravity)

$\theta=35^{\circ}$

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

$T-mg sin \theta = ma$ (2)

where

m = 3.5 kg (mass of the block)

$\theta=35^{\circ}$

From (2) we get

$T=mg sin \theta + ma$

Substituting into (1),

$M g sin \theta - mg sin \theta - ma = Ma$

Solving for a,

$a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2$

2b)

The tension in the string can be calculated using the equation

$T=mg sin \theta + ma$

where

m = 3.5 kg (mass of lighter block)

$g=9.8 m/s^2$

$\theta=35^{\circ}$

$a=2.2 m/s^2$ (acceleration found in part 2)

Substituting,

$T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N$

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The kinetic energy of an object is the energy due to its motion. It is calculated as

$K=\frac{1}{2}mv^2$

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m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

$U=mgh$

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

$K=\frac{1}{2}(0.5)(3)^2=2.25 J$

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