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Stolb23 [73]
4 years ago
10

A book of mass 7 kg rests on a plank. You tilt one end of the plank and slowly increase the angle of the tilt. The coefficient o

f static friction between the book and the plank is 0.32. What is the maximum angle of tilt for which the book will remain stationary and not slide down the plank
Physics
2 answers:
VARVARA [1.3K]4 years ago
5 0

Answer:

17.74°

Explanation:

For an inclined surface,

F = μmgcosΦ................ Equation 1

F' = mgsinΦ................. Equation 2

Where F = friction force, F' = down slope force, m = mass of the book, g = acceleration due to gravity, μ = coefficient of static friction, Φ = maximum angle of tilt.

For the book to remain stationary and not slide.

F = F'

μmgcosΦ = mgsinΦ

μ = cosΦ/sinΦ

μ = tanФ

make Ф the subject of the equation

Ф = tan⁻¹(μ).................... Equation 3

Given: μ = 0.32

Substitute into equation 3

Ф = tan⁻¹(0.32)

Ф = 17.74°

MAXImum [283]4 years ago
5 0

To solve this problem we will apply the concepts related to the Friction force and the force induced by gravity. Since the displacement is in angular mode, the component of the horizontal force of friction will be equivalent to the component of the vertical force of gravity. For balance to exist, both must be equal

F_{fx} = F_{gy}

Where,

F_{fx} = \mu mg cos\theta

F_{gy} = mg sin \theta

m = Mass

g = Gravitational acceleration

\mu = Constant of friction

Then,

m g sin \theta = \mu m g cos \theta

sin \theta = mu cos \theta

tan \theta = mu

tan\theta = 0.32

\theta = 17.74\°

Therefore the maximum angle  of tilt for which the book will remain stationary and not slide down the plank is 17.74°

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pychu [463]

By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.

Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .

now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.

charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.

Now , use vector for solve it.

vector F_{net} = vector Fe + vector Fn,

⇒ |F_{net}| = \sqrt{}  F^{2} _{e } + F^{2}_n

⇒ Fe = Fs = KqQ/(1m)² = KqQ

⇒ F_{net} = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}

⇒ F_{net}= √2KqQ

Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.

To learn more about positive charges here

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8 0
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If an astronaut throws an object in space, the object’s speed will _____
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In fact, after the astronaut throws the object, no additional forces will act on it (since the object is in free space). According to Newton's second law:
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AlexFokin [52]

Answer:

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Explanation:

given,

Radius of earth = 6,370,000 m

we know,

1 km = 1000 m

1 m = 0.001 Km

6,370,000 m =  6,370,000 x 0.001

                       = 6,370 Km

The number 6,370 has 3 significant figure.

To transform this to an exponential number, it is necessary to move the decimal to the left so there is only one digit in front of the decimal point.

Representing the given number in scientific notation

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Answer:

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Hence, The truck's speed is 4.04 m/s.

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