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2 years ago

8

All aspects of the Kirby-Bauer test are standardized to assure reliability. What might be the consequence of pouring the plates

2mm deep instead of 4mm deep

1 answer:

EleoNora [17]2 years ago

7 0

Answer:

it would affect the distance the antiantibodies diffuse from the disk

Explanation:

You might be interested in

s) Use Cramer’s rule to solve the system below, and state the condition at which solution exists. ax+by = 1 cx+dy =−1

Zanzabum

(a) The solution to the system of equation is, x = (d + b)/(ad - cb) and y = (-a - c)/(ad - cb).

(b) The condition at which the solution exists is, ad - cb ≠ 0.

<h3>Solving the system of equation with Cramer's rule</h3>

ax + by = 1

cx + dy = -1

D = [a b]

[c d]

D = ad - cb

Dx = [1 b]

[-1 d]

Dx = d + b

Dy = [a 1]

[c - 1]

Dy = -a - c

x = Dx/D

x = (d + b)/(ad - cb)

y = Dy/D

y = (-a - c)/(ad - cb)

Cramer's rule applies to the case where the coefficient determinant is nonzero.

Thus, D ≠ 0 (ad - cb ≠ 0).

Learn more about Cramer's rule here: brainly.com/question/10445102

#SPJ1

4 0

2 years ago

A hollow, spherical shell with mass 2.00kg rolls without slipping down a slope angled at 38.0?.

mezya [45]

Answer:

$\mu = 0.31$

Explanation:

Given data:

mass = 2.00 kg

slope angle = 38.0

From figure

balancing force

$mgsin\theta - f = ma$ .....1

Balancing torque

$F_R = \frac{2}{3} mR^2 \alpha$ ......2

for pure rolling

$\alpha = \frac{a}{R}$

$F = \frac{2}{3} ma$

from 1 and 2nd equation

$mgsin\theta - \frac{2}{3}ma = ma$

$mgsin\theta = \frac{5}{3} ma$

$a = \frac{3}{5} g sin\theta$

$= \frac{3\theta 9.8 sin 38}{5} = 3.62 m/s^2$

$F =\mu N$

$= \frac{2}{3} ma$

$= \frac{2}{3} 2\times 3.62 = 4.83 N$

N =normal force $= mgsin\theta = 2 \times 9.8 sin 38 = 15.44 N$

$\mu \times 15.44 = 4.83$

solving for coefficent of friction we get

$\mu = 0.31$

4 0

3 years ago

"Mass burn technology uses unprocessed municipal solid waste to generate heat which is used to produce electricity."

Flura [38]

Answer:

True

Explanation:

Mass burn technology is a type of waste-to-energy technology commonly used in the mass-burn system, where unprocessed municipal solid waste is burned in a large incinerator with a boiler, to generate heat used in the production of electricity.

6 0

3 years ago

Read 2 more answers

An air conditioning system operating on reversed carnot cycle is required to remove heat from the house at a rate of 32kj/s to m

Brilliant_brown [7]

Answer:

(e) 1.64 kW

Explanation:

The Coefficient of Performance of the Reverse Carnot's Cycle is:

$COP = \frac{T_{L}}{T_{H}-T_{L}}$

$COP = \frac{293.15\,K}{308.15\,K-293.15\,K}$

$COP = 19.543$

Lastly, the power required to operate the air conditioning system is:

$\dot W = \frac{\dot Q_{L}}{COP}$

$\dot W = \frac{32\,kW}{19.543}$

$\dot W = 1.637\,kW$

Hence, the answer is E.

3 0

3 years ago

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str

Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$

$J=\frac{\pi}{32}\times (46)^4$

J = 207.6 $mm^4$

So the shear stress at point A is :

$\tau_A =\frac{Tc_A}{J}$

$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$

$\tau_A = 4913.29 \ MPa$

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

3 0

2 years ago