During a medical evaluation, the doctor can _________

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

EleoNora [17]

Answer:

it would affect the distance the antiantibodies diffuse from the disk

Explanation:

7 0

2 years ago

if you had 100 B size sheets and you cut them into A size sheets, how many sheets of A size paper would you have

castortr0y [4]

Answer:

200

Explanation:

A size sheets (also known as letter size) are 8.5 inches by 11 inches.

B size sheets (also known as ledger size) are 11 inches by 17 inches.

One B size sheet is twice as large as a A size sheet. So if you have 100 B size sheets and cut each one in half, you'll get 200 A size sheets.

8 0

3 years ago

Your local hospital is considering the following solution options to address the issues of congestion and equipment failures at

kiruha [24]

Jsjhjrhwjdbwjwjrueiworuuwud

4 0

2 years ago

The fracture strength of glass may be increased by etching away a thin surface layer. It is believed that the etching may alter

Korvikt [17]

Answer:

the ratio of the etched to the original crack tip radius is 30.24

Explanation:

Given the data in the question;

we determine the initial fracture stress using the following expression;

(σf)₁ = 2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ ----- let this be equation 1

where; (σ₀)₁ is the initial fracture strength

( $p_t$ )₁ is the original crack tip radius

α₁ is the original crack length.

first, we determine the final crack length;

α₂ = α₁ - 16% of α₁

α₂ = α₁ - ( 0.16 × α₁)

α₂ = α₁ - 0.16α₁

α₂ = 0.84α₁

next, we calculate the final fracture stress;

the fracture strength is increased by a factor of 6;

(σ₀)₂ = 6( σ₀ )₁

Now, expression for the final fracture stress

(σf)₂ = 2(σ₀)₂ $[$ α₂/( $p_t$ )₂ $]^{1/2$ ------- let this be equation 2

where ( $p_t$ )₂ is the etched crack tip radius

value of fracture stress of glass is constant

Now, we substitute 2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ from equation for (σf)₂ in equation 2.

0.84α₁ for α₂.

6( σ₀ )₁ for (σ₀)₂.

∴

2(σ₀)₁ $[$ α₁/( $p_t$ )₁ $]^{1/2$ = 2(6( σ₀ )₁) $[$ 0.84α₁/( $p_t$ )₂ $]^{1/2$

divide both sides by 2(σ₀)₁

$[$ α₁/( $p_t$ )₁ $]^{1/2$ = 6 $[$ 0.84α₁/( $p_t$ )₂ $]^{1/2$

$[$ 1/( $p_t$ )₁ $]^{1/2$ = 6 $[$ 0.84/( $p_t$ )₂ $]^{1/2$

$[$ 1/( $p_t$ )₁ $]$ = 36 $[$ 0.84/( $p_t$ )₂ $]$

1 / ( $p_t$ )₁ = 30.24 / ( $p_t$ )₂

( $p_t$ )₂ = 30.24( $p_t$ )₁

( $p_t$ )₂/( $p_t$ )₁ = 30.24

Therefore, the ratio of the etched to the original crack tip radius is 30.24

6 0

3 years ago

A police officer in a patrol car parked in a 70 km/h speed zone observes a passing automobile traveling at a slow, constant spee

Ludmilka [50]

Answer:

S = 0.5 km

velocity of motorist = 42.857 km/h

Explanation:

given data

speed = 70 km/h

accelerates uniformly = 90 km/h

time = 8 s

overtakes motorist = 42 s

solution

we know initial velocity u1 of police = 0

final velocity u2 = 90 km/h = 25 mps

we apply here equation of motion

u2 = u1 + at

so acceleration a will be

a = $\frac{25-0}{8}$

a = 3.125 m/s²

so

distance will be

S1 = 0.5 × a × t²

S1 = 100 m = 0.1 km

and

S2 = u2 × t

S2 = 25 × 16

S2 = 400 m = 0.4 km

so total distance travel by police

S = S1 + S2

S = 0.1 + 0.4

S = 0.5 km

and

when motorist travel with uniform velocity

than total time = 42 s

so velocity of motorist will be

velocity of motorist = $\frac{S}{t}$

velocity of motorist = $\frac{500}{42}$

velocity of motorist = 42.857 km/h

3 0

3 years ago

During a medical evaluation, the doctor can __________.