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2 years ago

List some reasons why growth characteristics are more useful on agar plates than on agar slants

2 answers:

5 0

Usually, in culturing of the bacteria we have a slant and then portion f it is transferred to the agar plate. The growth characteristics are more useful in the agar plates because it is where we really do the observation because bacteria in slants are still to be transferred in the agar plates.

joja [24]2 years ago

3 0

Answer:

There are the few reasons for growth characterization are more useful on agar plates than on agar slant are discussed below.

Explanation:

Generally, agar plates are very larger and streaking out the bacteria can be possible in it. By doing this, you can find the analyzation of individual colonies of their morphology with the other uses.

Agar slants are usually used for the short time period of stock cultures of bacteria.

Agar slants can be usually used to find the motility of bacteria by doing a single steak inoculation up the center of slant.

Agar plates are used used in stock cultures growing that can be refrigerated after the incubation and for several weeks it can be maintained.

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Scientists studying an anomalous magnetic field find that it is inducing a circular electric field in a plane perpendicular to t

yarga [219]

Answer

The rate at which the magnetic field is changing is $[\frac{dB}{dt} ] = 0.000467 T/s$

Explanation

From the question we are told that

The electric field strength is $E = 3.5mV/m = 3.5 *10^{-3} \ V/m$

The radius is $r = 1.5 \ m$

The rate of change of the magnetic field is mathematically represented as

$\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}$

Where $dl$ is change of a unit length

$\frac{d \phi}{dt} = A * \frac{dB}{dt}$

Where A is the area which is mathematically represented as

$A = \pi r^2$

$E \int\limits^{} { dl} = ( \pi r^2) (\frac{dB}{dt} )$

$E L = ( \pi r^2) (\frac{dB}{dt} )$

where L is the circumference of the circle which is mathematically represented as

$L = 2 \pi r$

$E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]$

$E = \frac{r}{2} [\frac{dB}{dt} ]$

$[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }$

substituting values

$[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }$

$[\frac{dB}{dt} ] = 0.000467 T/s$

8 0

2 years ago

Water is a colorless and odorless liquid. It can exist in solid, liquid, and gas states. It boils at 100 degrees C and melts at

BARSIC [14]

Answer: Option (c) is the correct answer.

Explanation:

Physical properties are the properties in which there is no change in chemical composition of a substance. On the other hand, chemical properties are the properties which change the chemical composition of a substance.

For example, when water boils at $100 ^{o}C$ then it changes into vapor state whereas when water freezes at $0^{0}C$ then it changes state from liquid to solid.

This means only physical state of water is changing and there is no change in chemical composition of water.

Hence, we can conclude that best option describing given information is that these are the physical changes water undergoes.

4 0

2 years ago

A shuttle on Earth has a mass of 4.5 E 5 kg. Compare its weight on Earth to its weight while in orbit at a height of 6.3 E 5 met

faltersainse [42]

Answer:

83%

Explanation:

On the surface, the weight is:

W = GMm / R²

where G is the gravitational constant, M is the mass of the Earth, m is the mass of the shuttle, and R is the radius of the Earth.

In orbit, the weight is:

w = GMm / (R+h)²

where h is the height of the shuttle above the surface of the Earth.

The ratio is:

w/W = R² / (R+h)²

w/W = (R / (R+h))²

Given that R = 6.4×10⁶ m and h = 6.3×10⁵ m:

w/W = (6.4×10⁶ / 7.03×10⁶)²

w/W = 0.83

The shuttle in orbit retains 83% of its weight on Earth.

4 0

2 years ago

Suppose you want to make a scale model of a hydrogen atom. You choose, for the nucleus, a small ball bearing with a radius of 1.

RoseWind [281]

Answer:

A) x _electron = 0.66 10² m , B) x _Eart = 1.13 10² m , C) d_sphere = 1.37 10⁻² mm

Explanation:

A) Let's use a ball for the nucleus, the electron is at a farther distance the sphere for the electron must be at a distance of

Let's use proportions rule

x_ electron = 0.529 10⁻¹⁰ /1.2 10⁻¹⁵ 1.5

x _electron = 0.66 10⁵ mm = 0.66 10² m

B) the radii of the Earth and the sun are

$R_{E}$ = 6.37 10⁶ m

tex]R_{Sum}[/tex] = 6.96 10⁸ m

Distance = 1.5 10¹¹ m

x_Earth = 1.5 10¹¹ / 6.96 10⁸ 1.5

x _Eart = 1.13 10² m

C) The radius of a sphere that represents the earth, if the sphere that represents the sun is 1.5 mm, let's use another rule of proportions

d_sphere = 1.5 / 6.96 10⁸ 6.37 10⁶

d_sphere = 1.37 10⁻² mm

5 0

2 years ago

The 20-lb cabinet is subjected to the force F = (3 + 2t) lb, where t is in seconds. If the cabinet is initially moving down the

Andre45 [30]

Answer:

t₁ = 0.95 s

Explanation:

In this chaos we must use the definition of Newton's second law

F = m a = m dv / dt

dv = F dt / m

Let's replace and integrate, let's take the upward direction of the plane as positive, the force is positive

dv = ∫ (3 + 2t) dt / m

v = (3 t + 2 t²/ 2) /m

Let's evaluate between the lower limit t = 0 v = -6 ft / s (going down) to the upper limit t = t and v = 0

0 - (-6) = (3 (t- 0) + (t² -0)) / m

t² + 3t -6m = 0

Let's look for the mass

W = mg

m = W / g

m = 20/32

m = 0.625 slug

Let's solve the second degree equation

t² + 3t -3.75 = 0

t = (-3 ± √ (32 + 4 1 3.75)) / 2

t = (-3 ± 4,899) / 2

t₁ = 0.95 s

t₂ = -3.95 s

We take the positive time

6 0

2 years ago