Can some one help me with this question its hard :( down below ill give brainliest

A brick is resting on a smooth wooden board that is at a 30° angle. What is one way to overcome the static friction that is hold

lubasha [3.4K]

Answer:

to overcome the out of friction we must increase the angle of the plane

Explanation:

To answer this exercise, let's propose the solution of the problem, write Newton's second law. We define a coordinate system where the x axis is parallel to the plane and the other axis is perpendicular to the plane.

X axis

fr - Wₓ = m a (1)

Y axis

N- $W_{y}$ = 0

N = W_{y}

let's use trigonometry to find the components of the weight

sin θ = Wₓ / W

cos θ = W_{y} / W

Wₓ = W sin θ

W_{y} = W cos θ

the friction force has the formula

fr = μ N

fr = μ Wy

fr = μ mg cos θ

from equation 1

at the point where the force equals the maximum friction force

in this case the block is still still so a = 0

F = fr

F = (μ mg) cos θ

We can see that the quantities in parentheses with constants, so as the angle increases, the applied force must be less.

This is the force that balances the friction force, any force slightly greater than F initiates the movement.

Consequently, to overcome the out of friction we must increase the angle of the plane

the correct answer is to increase the angle of the plane

4 0

3 years ago

for the primitive yo-yo in fig, to find the downward acceleration of the cylinder and the tension in the string. you can take th

mixas84 [53]

The downward acceleration of the solid cylinder at the given tension in the string is determined as 2Tr/MR.

<h3>Downward acceleration of the cylinder</h3>

The downward acceleration of the solid cylinder is determined from the principle of conservation of angular momentum as shown below;

Iα = Tr

where;

I is moment of inertia of the solid cylinder
α is angular acceleration of the cylinder
T is tension in the string
r is length of the string

α = Tr/I

$\frac{a}{R} = \frac{Tr}{I} \\\\\frac{a}{R} = \frac{Tr}{\frac{1}{2} MR^2}\\\\\frac{a}{R} =\frac{2Tr}{MR^2} \\\\a = \frac{2Tr}{MR}$

where;

a is the downward acceleration of the solid cylinder
R is radius of the cylinder

Thus, the downward acceleration of the solid cylinder at the given tension in the string is determined as 2Tr/MR.

Learn more about acceleration here: brainly.com/question/605631

#SPJ1

7 0

2 years ago

A car engine changes chemical potential energy into the blank energy of the moving car

kap26 [50]

<h3><u>Answer;</u></h3>

Kinetic energy

A car engine changes chemical potential energy into the <u>kinetic energy</u> of the moving car.

<h3><u>Explanation;</u></h3>

A car engine converts potential chemical energy stored in gasoline into thermal energy and then into kinetic mechanical energy.
When gasoline undergoes combustion it reacts with oxygen to produce carbon dioxide and water vapor.Gasoline is a mixture of octane and similar hydrocarbons and contains potential chemical energy.
The hot exhaust gases from the combustion of gasoline that are produced within the cylinder expand and exert pressure, moving the piston in the cylinder outward then inward as the gas is exhausted. Kinetic mechanical energy of the moving pistons is transferred to the drive shaft and eventually to the wheels, giving the car kinetic mechanical energy.