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3 years ago

What are used for manufacturing paper along with wood and chemicals

Physics

1 answer:

Tju [1.3M]3 years ago

8 0

Answer:

Paper and pulp are made from cellulosic fibers and other plant materials. Some synthetic materials may be used to impart special qualities to the finished product. Paper is made from wood fibers, but rags, flax, cotton linters, and bagasse are also used in some papers.

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) Suppose a particle travels along a straight line with velocity v(t) = t 2 e −3t meters per second after t seconds. How far doe

pshichka [43]

Answer:

x(t=3s) = 0.07 m to the nearest hundredth

Explanation:

v(t) = t² e⁻³ᵗ

Find displacement after t = 3 s.

Recall, velocity, v = (dx/dt)

v = (dx/dt) = t² e⁻³ᵗ

dx = t² e⁻³ᵗ dt

∫ dx = ∫ t² e⁻³ᵗ dt

This integration will be done using the integration by parts method.

Integration by parts is done this way...

∫ u dv = uv - ∫ v du

Comparing ∫ t² e⁻³ᵗ dt to ∫ u dv

u = t²

∫ dv = ∫ e⁻³ᵗ dt

u = t²

(du/dt) = 2t

du = 2t dt

∫ dv = ∫ e⁻³ᵗ dt

v = (-e⁻³ᵗ/3)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ (-e⁻³ᵗ/3) 2t dt

= (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt

But the integral (∫ 2t (-e⁻³ᵗ/3) dt) is another integration by parts problem.

∫ u dv = uv - ∫ v du

u = 2t

∫ dv = ∫ (-e⁻³ᵗ/3) dt

u = 2t

(du/dt) = 2

du = 2 dt

∫ dv = ∫ (-e⁻³ᵗ/3) dt

v = (e⁻³ᵗ/9)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ 2t (-e⁻³ᵗ/3) dt = 2t (e⁻³ᵗ/9) - ∫ 2 (e⁻³ᵗ/9) dt = 2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)

Putting this back into the main integration by parts equation

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt = (-t²e⁻³ᵗ/3) - [2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)]

x(t) = ∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k (k = constant of integration)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k

At t = 0 s, v(0) = 0, hence, x(0) = 0

0 = 0 - 0 - (2/27) + k

k = (2/27)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + (2/27)

At t = 3 s

x(3) = (-9e⁻⁹/3) - (6e⁻⁹/9) - (2e⁻⁹/27) + (2/27)

x(3) = -0.0003702294 - 0.0000822732 - 0.0000091415 + 0.0740740741 = 0.07361243 m = 0.07 m to the nearest hundredth.

7 0

3 years ago

Read 2 more answers

Un proyectil es lanzado horizontalmente desde una altura de 12 metro con una velocidad de 80 m/sg. a.Calcular el tiempo de vuelo

Naily [24]

Answer:

t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s

Explanation:

Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso

y = y₀ + $v_{oy}$ t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

0 = y₀ – ½ g t²

t= √ (2 y₀/g)

calculemos

t= √ ( 2 12 / 9,8)

t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida

x = v₀ₓ t

x = 80 1,56

x= 124,8 m

La velocidad de impacto cuando toca el suelo

vx = v₀ₓ = 80 ms

$v_{y}$ = v_{oy} – gt

v_{y} = - 9,8 1,56

v_{y} = - 15,288 m/s

la velocidad es

v = (80 i^ - 15,288 j ) m/s

Traducttion

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( $v_{oy}$ = 0)

0 =y₀I - ½ g t²

t = √ (2y₀ / g)

let's calculate

t = √ (2 12 / 9.8)

t = 1.56 s

Projectile range is the horizontal distance traveled

x = v₀ₓ t

x = 80 1.56

x = 124.8 m

Impact speed when it hits the ground

vₓ = v₀ₓ = 80 ms

v_{y} = v_{oy} - gt

v_{y} = - 9.8 1.56

v_{y} = - 15,288 m / s

the speed is

v = (80 i ^ - 15,288 j) m / s

7 0

3 years ago

A child swings on a playground swing. How many times does the child swing through the swing's equilibrium position during the co

Karolina [17]

<h2>The child swing through the swing's equilibrium position 6 times during the course of 3 periods.</h2>

Explanation:

One period means time taken to complete one revolution.

In case of swings in one period time it travels the same position through two times.

Here we need to find how many times does the child swing through the swing's equilibrium position during the course of 3 period(s) of motion.

For 1 period = 2 times

For 3 periods = 3 x For 1 period

For 3 periods = 3 x 2 times

For 3 periods = 6 times

The child swing through the swing's equilibrium position 6 times during the course of 3 periods.

3 0

3 years ago

a bus starting from rest moves with a uniform acceleration of 0.1 metre per second square for 2 minutes find the speed acquired

blondinia [14]

S ?
U 0m/s
V ?
A 0.1m/s^2
T 2min (120 sec)

S=ut+0.5at^2
S=0(120 sec)+0.5(0.1m/s^2)(120 sec)^2
S=720m

Distance double 720m*2=1440m

V^2=u^2+2as
V^2=(0)^2+2(0.1 m/s^2)(1440m)
V^2=288
V= square root of 288=12 root 2=16.97 to 2 decimal places

6 0

3 years ago

irga5000 [103]

Answer:

Explanation:

solution is found below

6 0

3 years ago