All categories

3 years ago

The property of matter that resists changes in motion is

Physics

1 answer:

vredina [299]3 years ago

6 0

Answer:

Inertia

Explanation:

Inertia is the property that any physical object has of remaining in its state of relative motion. Therefore, it is the resistance that opposes matter to modify its state of motion, which includes changes in speed or changes in the direction of movement.

You might be interested in

Indica qué es una propiedad específica de la materia. Además explica por qué son útiles las propiedades específicas de la materi

Katyanochek1 [597]

Answer:

Check Explanation

Comprobar explicación

Explanation:

English Translation

Indicate what a specific property of matter is. Also explain why the specific properties of matter are useful compared to the general ones.

Solution

The specific properties of matter are properties that describes the intensive properties of the system. They are properties that do not depend on or change with the extent or size of the system. They are usually obtained by dividing the generalised properties or extensive properties by the extent or size of matter to make them independent of size/extent/Mass.

Examples of specific properties include specific heat capacity, specific volume etc. They usually have units of general units/Mass units.

The specific properties of matter are more important than the general ones because

- They help in general comparisons of the properties of different materials. They are used to rank, classify and compare properties of different materials.

- They are used in reference table/data to easily record easily accessible properties of matter. It helps to record standards that are general and independent of sizes/extents/Mass, thereby keeping the reference table/data/chart precise and concise.

- They provide us with values that are easy to memorize and remember, unlike trying to cram the different properties of different masses/sizes of matter.

In Spanish/En español

Las propiedades específicas de la materia son propiedades que describen las propiedades intensivas del sistema. Son propiedades que no dependen ni cambian con la extensión o el tamaño del sistema. Por lo general, se obtienen dividiendo las propiedades generalizadas o las propiedades extensivas por la extensión o el tamaño de la materia para hacerlas independientes del tamaño / extensión / masa.

Los ejemplos de propiedades específicas incluyen capacidad calorífica específica, volumen específico, etc. Usualmente tienen unidades de unidades generales / unidades de masa.

Las propiedades específicas de la materia son más importantes que las generales porque

- Ayudan en las comparaciones generales de las propiedades de diferentes materiales. Se utilizan para clasificar, clasificar y comparar propiedades de diferentes materiales.

- Se utilizan en la tabla / datos de referencia para registrar fácilmente propiedades de materia fácilmente accesibles. Ayuda a registrar estándares que son generales e independientes de tamaños / extensiones / masa, manteniendo así la tabla / datos / tabla de referencia precisa y concisa.

- Nos proporcionan valores que son fáciles de memorizar y recordar, a diferencia de tratar de agrupar las diferentes propiedades de diferentes masas / tamaños de materia.

Hope this Helps!!!

¡¡¡Espero que esto ayude!!!

7 0

3 years ago

A sled plus passenger with total mass m = 53.1 kg is pulled a distance d = 25.3 m across a horizontal, snow-packed surface for w

Alex Ar [27]

Answer:

Explanation:

Force of friction

F = μ mg

μ is coefficient of friction , m is mass and g is acceleration due to gravity .

If f be the force applied to pull the sled , the horizontal component of force should be equal to frictional force

The vertical component of applied force will reduce the normal force or reaction force from the ground

Reaction force R = mg - f sin28.3

frictional force = μ R where μ is coefficient of friction

frictional force = μ x (mg - f sin28.3 )

This force should be equal to horizontal component of f

μ x (mg - f sin28.3 ) = f cos 28.3

μ x mg = f μsin28.3 + f cos 28.3

f = μ x mg / (μsin28.3 + cos 28.3 )

a )

work done by pulling force = force x displacement

f cos28.3 x d

μ x mg d cos28.3 / (μsin28.3 + cos 28.3 )

b ) Putting the given values

= .155 x 53.1 x 9.8 x 25.3 cos28.3 / ( .155 x sin28.3 + cos 28.3 )

= 1796.76 / (.073 + .88 )

= 1885.37 J

c )

Work done by frictional force

= frictional force x displacement

= - μ x (mg - f sin28.3 ) x d

= - μ x mgd + f μsin28.3 x d

= - μ x mgd + μsin28.3 x d x μ x mg / (μsin28.3 + cos 28.3 )

d )

Putting the values in the equation above

- .155 x 53.1 x 9.8 x 25.3 +

.155 x .474 x 25.3 x .155 x 53.1 x 9.8 /( .155 x .474 + .88)

= -2040.67 + 149.92 / .95347

= -2040.67 + 157.23

= -1883.44 J .

6 0

3 years ago

If the final position vector of a moving object has a smaller magnitude than the initial position vector, then the change in the

BARSIC [14]

Explanation:

The given statement is absolutely true. this is because magnitude of a vector is always non negative, it can not be zero unless its a zero vector. So, in the given question, final position vector of a moving object has a smaller magnitude than the initial position vector, so, magnitude is neither zero nor negative. Hence, it has a positive magnitude.

5 0

3 years ago

Each of 134 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first bl

poizon [28]

Answer:

A. T=126N

B. T=63N

Explanation:

To determine the tension in each given blocks, we first determine the acceleration of each block. It obvious that each mass will move with the same acceleration since the string connecting them is massless.

Hence using the equation of force we have

F=ma

Where m=total mass of blocks,

a=acceleration

F= force applied in this case the tension in the string.

For a 134 identical masses with an applied force of 134N, the acceleration of each mass can be computed as

134=134m*a

a=134/134m

a=(1/m )m/s²

a. To calculate the tension in the string between the 126 and 127 block, we use the equation below

T=ma

Since the number of blocks before the string is 126, we multiply the mass of each block by 126.

Hence the tension can be computed as

T=126m*a

Since a=1/m then

T=126m*1/m

T=126N

B.To calculate the tension in the string between the 63 and 64 block, we use the equation below

T=ma

Since the number of blocks before the string is 63, we multiply the mass of each block by 63.

Hence the tension can be computed as

T=63m*a

Since a=1/m then

T=63m*1/m

T=63N

8 0

4 years ago

As a projectile thrown upward moves in its parabolic path, at what point along its path are the velocity and acceleration vector

Ivahew [28]

Answer: Maximum height

Explanation:

Suppose a projectile is thrown at an angle of $\theta$ with velocity u

As the projectile moves upward, its vertical component of velocity decreases due to gravity. While the horizontal component remains the same as there is no acceleration in the horizontal direction.

At maximum height vertical component of velocity is zero. At this instant horizontal component of velocity is perpendicular to the acceleration due to gravity. This can be explained with the diagram.

5 0

3 years ago