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charle [14.2K]
2 years ago
6

A 1,492.3-kg airplane travels down the runway. Each of its four engines provides a force of

Physics
1 answer:
Fittoniya [83]2 years ago
3 0

The acceleration of the air plane is 3.879 \mathrm{m} / \mathrm{s}^{2}

<u>Explanation:</u>

Given:

The mass of the air plane = 1492.3 kg

Force of each four engine = 1447.5 N

So, the total force of four engines can be calculated as = 4(1447.5) = 5790 N

The force that acts on the object is equal to the product of mass (m) and its acceleration. It can express by the below formula,

                         \text {Force }(F)=m \times \text { acceleration }(a)

The above equation can be written as below to find acceleration,

                        a=\frac{F}{m}

Now. Substitute the given values, we get,

                        a=\frac{5790}{1492.3}=3.879 \mathrm{m} / \mathrm{s}^{2}

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3 0
3 years ago
To test the performance of its tires, a car
velikii [3]

<u>Answer</u>:

The coefficient of  static friction between the tires and the road is 1.987

<u>Explanation</u>:

<u>Given</u>:

Radius of the track, r =  516 m

Tangential Acceleration a_r=  3.89 m/s^2

Speed,v =  32.8 m/s

<u>To Find:</u>

The coefficient of  static friction between the tires and the road = ?

<u>Solution</u>:

The radial Acceleration is given by,

a_{R = \frac{v^2}{r}

a_{R = \frac{(32.8)^2}{516}

a_{R = \frac{(1075.84)}{516}

a_{R = 2.085 m/s^2

Now the total acceleration is

\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2

=>= \sqrt{ (a_r)^2+(a_R)^2}

=>\sqrt{ (3.89 )^2+( 2.085)^2}

=>\sqrt{ (15.1321)+(4.347)^2}

=>19.4791 m/s^2

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of  static friction is

\mu =\frac{f}{W}

From (1) and (2)

\mu =\frac{ma}{mg}

\mu =\frac{a}{g}

Substituting the values, we get

\mu =\frac{19.4791}{9.8}

\mu =1.987

8 0
3 years ago
A copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)
gavmur [86]
<h2>The increase in length = 1.87 x 10⁻²</h2>

Explanation:

When copper rod is heated , its length increases

The increase in length can be found by the relation

L = L₀ ( 1 + α ΔT )

here L is the increased length and L₀ is the original length

α  is the coefficient of linear expansion and ΔT is the increase in temperature .

The increase in length = L - L₀ = L₀ x α ΔT

Substituting all these value

Increase in length = 27.5 x 1.7 x 10⁻⁵ x 35.9

= 1.87 x 10⁻² m

5 0
3 years ago
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6 0
3 years ago
Does Earth´s magnetic field move?
AlekseyPX

Answer:

Yes it does.

Explanation:

"The North Magnetic Pole moves over time due to magnetic changes in Earth's core. " - Wikipedia.

It does move around as the magnetic north does.

6 0
3 years ago
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