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3 years ago

Calculate the acceleration of an airplane that starts at rest and reaches a speed 45m/s in 9 seconds

Physics

2 answers:

PolarNik [594]3 years ago

6 0

Here's the explanation of HOW to find the answer. That's much more important than the answer.

The two BOLD lines are the parts you need to remember. They're the things you LEARN when you STUDY. If you remember them, then you can solve almost ANY acceleration problem.

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) minus (speed at the beginning)

Change in speed = (45 m/s) minus (zero)

Change in speed = 45 m/s .

Acceleration = (45 m/s) / (9 sec)

Acceleration = 5 m/s²

sasho [114]3 years ago

3 0

I believe the acceleration would be 5m/s

All you would need to do is divide the final speed by the time it took to get there. I am only about 80 sure this answer is correct, so take my advise only if you feel comfortable.

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Define friction State the law of frictionGive two factors that may affect friction

umka21 [38]

Friction:

Friction is the resistive force that acts on the body in a direction opposite to the direction of motion of the same object.

Laws of friction:

1. The frictional force acting on the object is proportional to the normal force acting on the object.

For example, if the object of mass m is moving along the horizontal surface, the normal force acting on the object is mg.

Thus, the frictional force acting on the object is,

$\begin{gathered} F_r\propto F_N \\ F_r=\mu F_N \end{gathered}$

where F_N is the normal force acting on the object,

$\mu\text{ is the coefficient of friction}$

The diagrammatic representation of the frictional force acting on the object is,

2. The frictional force acting on the object depends upon the nature of the surface in contact with the object.

3. The frictional force does not depend upon the area of contact of an object with the surface.

4. The kinetic friction acting on the object is independent of the velocity of the object.

5. The coefficient of static friction is more than the coefficient of kinetic friction.

Two factors on which the frictional force depends are:

1. The frictional force depends on the adhesion between the two bodies in contact.

2. The frictional force acting on the depends upon the roughness of the surface on which the object is moving.

3 0

1 year ago

Body of a 218 kg starta from rest and attains the velocity of 63 m/s in 9 second

Vanyuwa [196]

Answer:

567m

Explanation:

By using formula of velocity.

7 0

3 years ago

All about Henry Mosléy?<br>

Anni [7]

Answer:

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Explanation:

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2 years ago

What is the momentum of an 80 kg runner moving at the speed of 2.5 m/s? Use the

Vladimir79 [104]

Answer:

200 kgm/s

Explanation:

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8 0

2 years ago

(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.

faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let call "x₁" distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁ ( force between earth and the object )

F₁ = K * Mt * m₀/ ( x₁)² K is a gravitational constant

F₂ (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂ K*Mt*m₀ / x₁² = K*Ml*m₀ /x₂²

Or simplifying the expression

Mt/ x₁² = Ml/ x₂²

We know that x₁ + x₂ = 384000 Km then

x₁ = 384000 - x₂

Mt/( 384000 - x₂)² = Ml / x₂²

Mt * x₂² = Ml *( 384000 - x₂)²

We need to solve for x₂

Mt * x₂² = Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂² = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂² = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂² + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂ = -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11 / 1160

x₂ ₁ = -56.43*10∧5 + √3184*10∧10 + 254880*10∧10 / 1160

x₂ ₁ = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁ = -56.43*10∧5 + 508* 10∧5 / 1160

x₂ ₁ = 451.27*10∧5/1160

x₂ ₁ = 4512.7*10∧4 /1160

x₂ ₁ = 3.89*10∧4 km (distance between the moon and the object)

x₂ ₁ = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁ = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0

3 years ago