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blagie [28]
3 years ago
7

Based on this passage, what is campylobacter?

Physics
2 answers:
almond37 [142]3 years ago
8 0

Answer: a drug-resistant bacteria

Explanation: in the passage, we can read that:

"Campylobacter is estimated to cause approximately 1.3 million infections, 13,000 hospitalizations, and 120 deaths each year in the United States. "

from this, we can expect that the campylobacter is some form of bacteria or virus.

" The CDC is seeing resistance to ciprofloxacin in almost 25 percent of campylobacter tested and resistance to azithromycin in about 2 percent. "

And here we can see that the campylobacter shows resistance to some drugs.

Then the option that makes sense is:

2) a drug-resistant bacteria

alexdok [17]3 years ago
4 0
I believe the correct answer from the choices listed above is option 4. Base from the passage given above, it is very clear that the Campylobacter is a bacteria used to test in <span>identifying drug-resistant bacteria. Hope this answers the question. Have a nice day.</span>
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A gray kangaroo can bound across level ground with each jump carrying it 9.1 m from the takeoff point. Typically the kangaroo le
Alona [7]

Answer:

u = 10.63 m/s

h = 1.10 m

Explanation:

For Take-off speed ..

by using the standard range equation we have

R = u² sin2θ/g

R = 9.1 m

θ = 26º,

Initial velocity = u

solving for u

u² = \frac{Rg}{sin2\theta}

u^2 = \frac{9.1 x 9.80}{sin26}

u^2 = 113.17

u = 10.63 m/s

for Max height

using the standard h(max) equation ..

v^2 = (v_osin\theta)^2 -2gh

h =\frac{(v_o^2sin\theta)^2}{2g}

h  =  \frac{(113.17)(sin26)^2}{(2 x 9.80)}}

h = 1.10 m

7 0
3 years ago
n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at
Volgvan

Answer:

a)-1.014x 10^{-7J

b)3.296 x  10^{-7J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

U_{i = -GMaMb/ d

U_{i= - 6.67 x 10^{-11} x 47 x 110/ 3.4 => -1.014x 10^{-7J

b) at d= 0.8m (3.4-2.6) and U_{i=-1.014x 10^{-7J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

K_{i + U_{i= K_{f + U_{f

As sphere starts from rest and sphere A is fixed at its place, therefore K_{i is zero

U_{i= K_{f + U_{f

The final potential energy is

U_{f= - GMaMb/d

Solving for 'K_{f '

K_{f = U_{i + GMaMb/d => -1.014x 10^{-7 + 6.67 x 10^{-11} x 47 x 110/ 0.8

K_{f = 3.296 x  10^{-7J

6 0
2 years ago
A piece of aluminum foil has a known surface density of
bija089 [108]

The cube has 6 equal, square, foil faces. The mass of foil for each face is (380/6) milligrams.

The surface area of each piece is (380)/(6•11) cm^2.

The length of each side of the piece is √(380/6•11) cm

That's about 2.4 cm .

It's a cute little foil cube, just under 1-inch each way.


5 0
3 years ago
Which characteristics can be used to differentiate star systems? Select three options.
anyanavicka [17]

Answer:

1,3and4 just did and got correct

Explanation:

5 0
2 years ago
As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labe
spin [16.1K]

Answer:

1.19 m/s²

Explanation:

The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so

f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀    (1)

Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²

Equating (1) and (2) we ave

2(√mg/μ)/f = T²g/4π²

Making g subject of the formula

g = 2π√(2√(m/μ)/f)/T

The period T = 316 s/100 = 3.16 s

Substituting the other values into , we have

g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16

g = 2π√(2 × 35.877/200 Hz)/3.16

g = 2π√(71.753/200 Hz)/3.16

g = 2π√(0.358)/3.16

g = 2π × 0.599/3.16

g = 1.19 m/s²

6 0
3 years ago
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