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valkas [14]
2 years ago
12

What are used for manufacturing paper along with chemical

Physics
1 answer:
Airida [17]2 years ago
8 0

Answer:

Dry-strength additives, or dry-strengthening agents, are chemicals that improve paper strength normal conditions. These improve the paper's compression strength, bursting strength, tensile breaking strength, and delamination resistance. Typical chemicals used include cationic starch and polyacrylamide derivatives.

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The origin of an x axis is placed at the center of a nonconducting solid sphere of radius R that carries a charge +qsphere distr
MA_775_DIABLO [31]

Answer:

q=49Q/64

and

x =16R/15

Explanation:

See  attached figure.

E_{Q}= E due to sphere

E_{q}= E due to particule

E_{total}=E_{Q}-E_{q}=0  (1)

according to the law of gauss and superposition Law:

E_{Q}=E_{1}+E_{2}=E_{2} ; electric field due to the small sphere with r1=R/4

E_{Q}=kq_{2}/(r_{1}^{2})=

q_{2}=density*4/3*pi*r_{1}^{3}=Q/(4/3*pi*R^{3})*4/3*pi*r_{1}^{3}=Q*r_{1}^{3}/R^{3}

then: E_{Q}=kq_{2}/(r_{1}^{2})=k*Q*r_{1}^{3}/(R^{3}*r_{1}^{2}) = kQ/(4*R^{2})  (2)

on the other hand, for the particule:

E_{q}=kq/(r_{p}^{2})

r_{p}=2R-R/4=7R/4   ⇒    E_{q}=16kq/(49R^{2})   (3)

We replace (2) y (3) in (1):

E_{total}=E_{Q}-E_{q}=0=kQ/(4*R^{2}) - 49kq/(16R^{2})

q=49Q/64

--------------------

if R<x<2R   AND E_{total}=E_{Q}-E_{q}=0

E_{total}=E_{Q}-E_{q}=0=kQ/(x^{2}) - kq/(2R-x^{2})

remember that  q=49Q/64

then:

Q(2R-x^{2})=49/64*x^{2}

solving:

x_{1} =16R/15

x_{2} =16R

but: R<x<2R  

so : x =16R/15

7 0
3 years ago
When two waves in the same medium hit each other, the resulting displacement of the medium is
Irina-Kira [14]

Answer:

1. Either larger or smaller than the displacement of either wave acting alone, depending on the signs of the displacements of the two waves.

7 0
3 years ago
The gas in a balloon has T=280 K and V=0.0279 m^3. if the temperature increases to 320 K at constant pressure, what is the new v
victus00 [196]

Answer:

0.0319 m³

Explanation:

Use ideal gas law:

PV = nRT

where P is pressure, V is volume, n is amount of gas, R is the gas constant, and T is temperature.

Since P, n, and R are held constant:

n₁ R / P₁ = n₂ R₂ / P₂

Which means:

V₁ / T₁ = V₂ / T₂

Plugging in:

0.0279 m³ / 280 K = V / 320 K

V = 0.0319 m³

8 0
3 years ago
An experimental tungsten light bulb filament has a length of 5 cm and a diameter of 0.074 cm. The filament is basically just a w
adell [148]

Answer:

power emitted is 1.75 W

Explanation:

given data

length l = 5 cm = 5 ×10^{-2} m

diameter d = 0.074 cm = 74 ×10^{-5} m

total filament emissivity = 0.300

temperature = 3068 K

to find out

power emitted

solution

we find first area that is π×d×L

area = π×d×L

area = π×74 ×10^{-5}×5 ×10^{-2}

area = 1162.3892  ×10^{-5} m²

so here power emitted  is express as

power emitted  = E × σ × area × (temperature)^4

put here all value

power emitted  = 0.300× 5.67 × 1162.3892  ×10^{-5}  × (3068)^4

power emitted = 1.75 W

5 0
3 years ago
After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. when the box reaches position
KiRa [710]

As we know by work energy theorem

total work done = change in kinetic energy

so here we can say that wok done on the box will be equal to the change in kinetic energy of the system

W_p = KE_f - KE_i

initial the box is at rest at position x = x1

so initial kinetic energy will be ZERO

at final position x = x2 final kinetic energy is given as

KE_f = \frac{1}{2}mv_1^2

now work done is given as

W_p = \frac{1}{2}mv_1^2 - 0

so we can say

W_p = \frac{1}{2}mv_1^2

so above is the work done on the box to slide it from x1 to x2

3 0
3 years ago
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