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4 years ago

Machine movement can be divided into what two main categories?

Physics

2 answers:

defon4 years ago

4 0

Answer:

Motion

Action

Explanation:

its right trust me !!!

Ivenika [448]4 years ago

3 0

Answer:

1.) Employee Driven Machines

2.) Self Driven Machines

Explanation:

A machine is a mechanical device that utilizes the application of forces to carry out an operation. It's operation can be "employee driven" (i.e it is operated by a person) or "self driven" (i.e it is operated by itself).

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A rock with mass m = 4.00 kg falls from rest in a viscous medium. The rock is acted on by a net constant downward force of F = 1

pogonyaev

Answer:

a) 4.35 m/s²

b) 2.73 m/s²

c) 7.25 m/s

d) 8.06 m/s

e) At t = 2 s

x = 16.5 m

v = 7.88 m/s

a = 0.099 m/s²

f) t = 0.743 s

Explanation:

Force balance on the rock

ma = 17.4 - F

4a = 17.4 - kv

4a = 17.4 - 2.16v

a) At the initial instant, F = kv = 0

4a = 17.4

a = 4.35 m/s²

b) When v = 3 m/s

4a = 17.4 - (2.16)(3) = 10.92

a = 2.73 m/s²

c) a₀ = 4.35 m/s²

0.1 a₀ = 0.435 m/s²

4a = 17.4 - 2.16v

4(0.435) = 17.4 - 2.16v

1.74 = 17.4 - 2.16v

2.16v = 15.66

v = 7.25 m/s

d) Terminal speed is when the body stops accelerating in the fluid

When a = 0

0 = 17.4 - 2.16v

2.16 v = 17.4

v = 8.06 m/s

e) 4a = 17.4 - 2.16v

a = 4.35 - 0.54 v

But a = dv/dt

(dv/dt) = 4.35 - 0.54v

∫ dv/(4.35 - 0.54v) = ∫ dt

Integrating the left hand side from 0 to v and the right hand side from 0 to t

- 1.852 In (4.35 - 0.54v) = t

In (4.35 - 0.54v) = - 0.54 t

4.35 - 0.54v = e⁻⁰•⁵⁴ᵗ

0.54v = 4.35 - e⁻⁰•⁵⁴ᵗ

v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ

Then, v = dx/dt

(dx/dt) = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ

dx = (8.06 - 0.54 e⁻⁰•⁵⁴ᵗ) dt

∫ dx = ∫ (8.06 - 0.54 e⁻⁰•⁵⁴ᵗ) dt

Integrating the left hand side from 0 to x and the right hand side from 0 to t

x = 8.06t + e⁻⁰•⁵⁴ᵗ

Acceleration too can be obtained as a function of time

since v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ and a = dv/dt

a = 0.54² e⁻⁰•⁵⁴ᵗ = 0.2916 e⁻⁰•⁵⁴ᵗ

At t = 2 s

Coordinate

x = 8.06t + e⁻⁰•⁵⁴ᵗ

x = (8.06)(2) + e^(-1.08) = 16.5 m down into the fluid.

Velocity

v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ

v = 8.06 - 0.54 e^(-1.08) = 7.88 m/s

Acceleration

a = 0.2916 e⁻⁰•⁵⁴ᵗ

a = 0.2916 e^(-1.08) = 0.099 m/s²

f) t = ? When v = 0.9 × 8.06 = 7.254 m/s

v = 8.06 - 0.54 e⁻⁰•⁵⁴ᵗ

7.254 = 8.06 - 0.54e⁻⁰•⁵⁴ᵗ

- 0.806 = - 0.54 e⁻⁰•⁵⁴ᵗ

e⁻⁰•⁵⁴ᵗ = 1.493

0.54t = In 1.493 = 0.401

t = 0.743 s.

6 0

3 years ago

Balance the following chemical equations

IrinaVladis [17]

Answer:

Enter an equation of a chemical reaction and click 'Balance'. The answer will appear below

Always use the upper case for the first character in the element name and the lower case for the second character.

To enter an electron into a chemical equation use {-} or e

To enter an ion specify charge after the compound in curly brackets: {+3} or {3+} or {3}. Example: Fe{3+} + I{-} =...

Substitute immutable groups in chemical compounds to avoid ambiguity. For instance equation C6H5C2H5 + O2 = C6H5OH + CO2...

Explanation:

4 0

4 years ago

The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0

4 years ago

What would the density of an object be with a mass of 0.8g

True [87]

The density of the object is 0.032 g/cm^3.

<h3>What is density?</h3>

The term density means the ratio of the mass to the volume of an object. Now we know that when an object is dense that it would sink to the bottom of a fluid.

Now given that;

Mass = 0.8 g

volume= 25 cm^3

Density = mass/volume

Density = 0.8 g / 25 cm^3

= 0.032 g/cm^3

Learn more about density:brainly.com/question/15164682

#SPJ1

MISSING PARTS

What is the density of an object having a mass of

0.8g and a volume of 25cm^3

4 0

2 years ago

A cyclist traveling at 30.0m/s along a straight road comes uniformly to stop in 5.00s. Determine the stopping acceleration, the

Amiraneli [1.4K]

Answer:

I. Stopping acceleration = 6 m/s²

II. Stopping distance, S = 75 meters

Explanation:

Given the following data;

Final velocity = 30 m/s

Time = 5 seconds

To find the stopping acceleration;

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

Substituting into the equation;

$a = \frac{30 - 0}{5}$

$a = \frac{30}{5}$

Acceleration = 6 m/s²

II. To find the stopping distance, we would use the third equation of motion;

$V^{2} = U^{2} + 2aS$

Where;

V represents the final velocity measured in meter per seconds.

U represents the initial velocity measured in meter per seconds.

a represents acceleration measured in meters per seconds square.

Substituting into the equation, we have;

30² = 0² + 2*6*S

900 = 12S

S = 900/12

S = 75 meters

7 0

3 years ago