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2 years ago

A chemical process with an e-factor of 1 creates less waste than an e-factor of 25?

Physics

1 answer:

morpeh [17]2 years ago

8 0

A chemical process with an e-factor of 1 creates less waste than an e-factor of 25 is True.

<h3>What do you mean by E-factor?</h3>

The actual amount of waste produced during the process—which is referred to as anything other than the targeted product—is known as the E factor. It considers the chemical yield as well as reagents, solvent losses, all process aids, and, in theory, fuel (although this is often difficult to quantify). There is one exception: I often did not include water in the E factor calculation. For instance, only the inorganic salts and organic chemicals present in an aqueous waste stream are taken into account; the water itself is left out. In many instances, the inclusion of water used in the process might result in very high E factors, which can make it challenging to compare processes in a meaningful way.

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A ball swings in a horizontal circle. If the radius and the frequency are both doubled, by what factor does the tension in the s

kherson [118]

Answer:

The tension increases by a factor of 8

Explanation:

We know that the tension, T in the string equals the centripetal force on the ball. So

T = mrω² = mr(2πf)² = 4mrπ²f² where m = mass of ball, r = radius of circle and f = frequency of rotation

If the radius and frequency are doubled, then r = 2r and f = 2f. So, the new tension is T' = 4mr'π²f'² = 4m(2r)π²(2f)² = 4 × 2 × 4mrπ²f² = 8T

Since T' = 8T,

So T'/T = 8.

So the tension increases by a factor of 8

4 0

3 years ago

Pangaea is the name given for "all land" of Earth. This name is used by Geologists to describe their theory of Earth's original

Monica [59]

Pangaea one major super-continent.

3 0

3 years ago

Two spherical balloons are filled with water. The first balloon has a radius of 3 cm, and the second has a radius of 6 cm. How m

Marianna [84]

The volume of water in the larger balloon is 8 times greater than in the smaller balloon

We'll begin by calculating the volume of each balloon.

<h3>For smaller balloon:</h3>

Radius (r) = 3 cm
Pi (π) = 3.14
Volume (V) =?

V = 4/3 πr³

V = 4/3 × 3.14 × 3³

<h3>For larger balloon:</h3>

Radius (r) = 6 cm
Pi (π) = 3.14
Volume (V) =?

V = 4/3 πr³

V = 4/3 × 3.14 × 6³

Finally, we shall determine how much greater the larger balloon is to the smaller balloon

Volume of smaller balloon = 113.04 cm³
Volume of larger balloon = 904.32 cm³
Greatness =?

Greatness => large / small

Large / small = 904.32 / 113.04

Large / small = 8

Cross multiply

<h3>Large = 8 × small </h3>

Therefore, the larger balloon is 8 times greater than the smaller balloon.

Learn more on volume of sphere: brainly.com/question/9178703

4 0

3 years ago

Read 2 more answers

Suppose you have a 0.750-kg object on a horizontal surface connected to a spring that has a force constant of 150 N/m. There is

marshall27 [118]

Answer:

$x=0.0049\ m= 4.9\ mm$

$d=0.01153\ m=11.53\ mm$

Explanation:

Given:

mass of the object, $m=0.75\ kg$

elastic constant of the connected spring, $k=150\ N.m^{-1}$

coefficient of static friction between the object and the surface, $\mu_s=0.1$

(a)

Let x be the maximum distance of stretch without moving the mass.

<em>The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.</em>

$f=k.x$

$\mu_s.N=k.x$

where:

N = m.g = the normal reaction force acting on the body under steady state.

$0.1\times (9.8\times 0.75)=150\times x$

$x=0.0049\ m= 4.9\ mm$

(b)

Now, according to the question:

Amplitude of oscillation, $A= 0.0098\ m$

coefficient of kinetic friction between the object and the surface, $\mu_k=0.085$

Let d be the total distance the object travels before stopping.

<em>Now, the energy stored in the spring due to vibration of amplitude:</em>

$U=\frac{1}{2} k.A^2$

<u><em>This energy will be equal to the work done by the kinetic friction to stop it.</em></u>

$U=F_k.d$

$\frac{1}{2} k.A^2=\mu_k.N.d$

$0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d$

$d=0.01153\ m=11.53\ mm$

<em>is the total distance does it travel before stopping.</em>

7 0

3 years ago

Using the equation T=2π√L/g solve for L

weqwewe [10]

You said T = 2 · π · √(L / g)

(I think that's the formula for the full-swing period of a simple pendulum.)

Divide each side by 2π : T/2π = √(L / g)

Square each side: (T/2π)² = L / g

Multiply each side by ' g ' : L = g · (T/2π)²

4 0

3 years ago