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Dmitriy789 [7]
3 years ago
6

What's newtons second law​

Physics
1 answer:
sweet-ann [11.9K]3 years ago
4 0

Answer:

According to Newton s Second Law of Motion, also known as the Law of Force and Acceleration, a force upon an object causes it to accelerate according to the formula net force = mass x acceleration.

Explanation:

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

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One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren
katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

\sigma = \frac{F}{A}

Where,

\sigma =Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

\sigma_y = \frac{F_y}{A}

where,

\sigma_y =Shear strength

F_y = Shear Force

A_0 =Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}

Resolving for F,

F= 1399.5N

PART B) We need here to apply the shear strength equation, then

\sigma_y = \frac{F_y}{A}

210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}

F_y = 945N

In such a way that the material is more resistant to tensile strength than shear force.

6 0
4 years ago
A disk initially rotated counterclockwise at 1.0 rad/s, but has a counterclockwise angular acceleration of 0.50 rad/s^2 for 2 se
Mumz [18]

Answer:

Initial angular position of the disc is 3 radian

Explanation:

As we know that the angular acceleration of the disc is given as

\alpha = 0.50 rad/s^2

initial angular speed is given as

\omega = 1 rad/s

now we know that angular displacement of the disc is given as

\theta = \omega t + \frac{1}{2}\alpha t^2

\theta = 1(2) + \frac{1}{2}(0.5)(2^2)

\theta = 3 rad

now we have

\theta_f - \theta_i = 3 rad

6 rad - \theta_i = 3

\theta_i = 3 ard

8 0
3 years ago
Which of the following describes the oscillation of the medium in transverse waves?
svp [43]

Wave An oscillation that transfers energy and momentum.

Mechanical wave A disturbance of matter that travels along a medium. Examples include waves on a string, sound, and water waves.

Wave speed Speed at which the wave disturbance moves. Depends only on the properties of the medium. Also called the propagation speed.

Transverse wave Oscillations where particles are displaced perpendicular to the wave direction.

Longitudinal wave Oscillations where particles are displaced parallel to the wave direction.

In a transverse wave, perpendicular to the direction the wave travels, the particles are displaced. Examples of transverse waves include on a string vibrations and on the water surface ripples. By moving the slinky up and down vertically, we can create a horizontal transverse wave.

In a longitudinal wave, parallel to the direction the wave travels, the particles are displaced. Compressions that move along a slinky are an example of longitudinal waves. By pushing and pulling the slinky horizontally, we can make a horizontal longitudinal wave.

Common mistakes and misconceptions

Sometimes people forget that wave velocity is not the same as the velocity of the medium particles. How fast the disturbance travels through a medium is the wave speed. The velocity of the particle is how fast a particle moves about its position of equilibrium.

6 0
3 years ago
PLEASEEEE HELP MEEEEE!!!!
Reika [66]

Answer:

D. Yes

m

m mb

m

m

m

n

nn

n

n

n

j

j

jj

3 0
3 years ago
A 55.0-kg box rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300. What h
MariettaO [177]

Answer:

161.86 N

Explanation:

mass of box m= 55.0 kg

weight of the box, mg= 55×9.81

g here is acceleration due to gravity =9.81 m/sec^2

coefficient of friction between the box and the surface μ= 0.3

the friction force F_s= μmg= 0.3×55×9.81

=161.86 N

to move the ball horizontal force required is 161.86 N

8 0
3 years ago
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