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Pani-rosa [81]
2 years ago
11

A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s

Physics
1 answer:
Natasha2012 [34]2 years ago
6 0

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

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harina [27]

Answer:

Force = 186 N

Explanation:

Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :

Torque = \vec{r} \times \vec{F}

Where \vec{r} is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector), \vec{F} is the applied force.

The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench.  By doing this we have that r (the magnitud of the position vector) is 35cm.

Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :

35cm * \dfrac{1m}{100cm} = 0.35m

Now using the torque formula:

Torque = \vec{r} \times \vec{F}

Torque = rFsin(\theta)}

Where \theta is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector  \theta = 90°, thus :

Torque = rFsin(\theta)}

65 N m = (0.35m)Fsin(90°)}

F = \dfrac{65N m} {(0.35m)sin(90°)}

F = \dfrac{65N m} {(0.35m)sin(90°)}

F = \dfrac{1300} {7}N

so the force is approximately 186 N.

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Albert presses a book against a wall with his hand. As Albert gets tired, he exerts less force, but the book remains in the same
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Answer:

maximum static frictional force of the wall on the book (Increasing)

normal force of the wall on the book (Decreasing)

weight of the book (Not changing)

Explanation:

Now according to Newton's third law of motion

"Every action has equal but opposite reaction"

By the data given in question, Albert was pressing the book against the wall.Now, Albert started to reduce his force up against the wall.

First we have to consider all the forces applied on book in this scenario.

1. Weight of book acting downwards (y-axis)

2. Friction between book and wall acting upward (y-axis)

3. Albert's force on book against wall (x-axis)

4. Normal reaction of wall against Albert's force (x-axis)

Now, when Albert reduced his force, new scenario will be

1. Weight will be remain constant as it is W = mg

Neither mass nor acceleration due to gravity changed, so weight acting upon the book will remain same.

2. When Albert reduced force, normal reaction of wall reduced against it according to Newton's third law of motion

3. Now notice that friction is a force which acts in accordance with the applied force. For example if a box is placed at floor, no friction is applied, but when you drag the box, friction starts to act and increases until its limit comes. So, when Albert reduced his force, weight will try to pull the book and maximum static friction will increase to hinder the movement of book downwards.

Notice that maximum static friction will hinder the book from movement, since Albert reduced his force, but wight didn't pull the book, which means that maximum static friction increased to hinder downward motion.

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Answer:

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Explanation:

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