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kobusy [5.1K]
3 years ago
9

3 ways bacteria can be good

Physics
1 answer:
maksim [4K]3 years ago
3 0
Well there is good bacteria and bad bacteria, good bacteria keeps your body healthy
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An air conditioner running with R-134a on a cycle executed under the saturationdome between the pressure limits of 0.8 MPa and 0
ohaa [14]

Answer:

The COP of the system is = 4.6

Explanation:

Given data

Higher pressure  = 1.8 M pa

Lower pressure = 0.12 M pa

Now we have to find out high & ow temperatures at these pressure limits.

Higher temperature corresponding to pressure 1.8 M pa

T_{H} = 62.9 °c = 335.9 K

Lower temperature corresponding to pressure 0.2 M pa

T_{L} = - 10.1 °c = 262.9 K

COP of the system is given by

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

COP = \frac{335.9}{335.9 -262.9}

COP = 4.6

Therefore the COP of the system is = 4.6

8 0
3 years ago
While practicing S-turns, a consistently smaller half-circle is made on one side of the road than on the other, and this turn is
erica [24]

Answer:

The answer is "4-5-6 because the bank is growing too quickly in the beginning of its turn".

Explanation:

The S-turns is also known as the reference technique, in which the ground track of the aircraft on both sides of a defined ground-based straight-line distance represents 2 different but equivalent circles.  

Throughout the S-turns, on either side of the road, a progressively smaller half-circle is formed, and this turn does not stop until the road or reference line is crossed during the early part of the turn, the twists increase too quickly.

5 0
3 years ago
Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
SSSSS [86.1K]

Answer:

{F_{tot} = -1.092*10^{-2}N

Explanation:

The question: What is the net force exerted by these two charges on a third charge q_3 = 47.0 nC placed between q_1 and q_2 at x_3 = -1.240 mm ?

<u>Your answer may be positive or negative, depending on the direction of the force.</u>

Solution:

The coulomb force is given by the equation

F =k\dfrac{q_1 q_2}{d^2}.

where d is the separation between the charges q_1 and q_2.

Now, in our case

q_1=-13.5nC=-13.5*10^{-9}C

q_2=-1.735nC=-1.735*10^{-9}C

q_3= 47*10^{-9}C

The separation between charges q_3 and q_1 is

d_{31}= (-1.240mm)-(-1.735mm)=0.495mm=4.95*10^{-4} m

Therefore, the force between them is

F_{31} = (9*10^9NmC^{-2})\dfrac{47*10^{-9}*(-13.5*10^{-9})}{4.95*10^{-4}} =-0.01152N

and it is directed in the negative x-direction.

The separation between charges q_2  and q_3 is

d_{23} =-1.240mm-0=-1.240*10^{-3}m

therefore, the force between them is

F_{23} =(9*10^9Nm^2C^{-2})\dfrac{47*10^{-9}C*(-1.735*10^{-9}C)}{-1.240*10^{-3}m}=5.94*10^{-4}N

Therefore the total force on charge q_3 is

F_{tot} =5.94*10^{-4}-0.01152N = -0.010926N\\\\\boxed{F_{tot} = -1.092*10^{-2}N}

8 0
3 years ago
What is the second step in the scientific method
beks73 [17]
The steps<span> of the </span>scientific method<span> are to: Ask a Question. Do Background Research. Construct a Hypothesis. Test Your Hypothesis by Doing an Experiment. Analyze Your Data and Draw a Conclusion.</span>
7 0
4 years ago
Read 2 more answers
How high does a rocket have to go above Earth's surface before its weight is half of what it is on Earth?
Contact [7]

Answer:

h=1.6\times 10^6\ m

Explanation:

As we know that the acceleration due to gravity decreases with height.

At certain height it will get to the half of its value on the surface of the earth.

As we know that the weight on the surface of the earth is given as:

w=m.g

where:

m = mass of the object

g = acceleration due to gravity of the substance

Since mass of the substance is constant so the variation is weight is possible only due to change in the acceleration due to gravity.

<u>We know that the variation of the acceleration due gravity with height is given as:</u>

g_{_h}=g\times (1-\frac{2h}{R} )

where:

g_{_h}= value to acceleration due to gravity at height h

g = acceleration due to gravity at the earth's surface

h = height of the object

R = radius of the earth = 6400\ km

according to question the weight becomes half, so,:

4.9=9.8\times (1-\frac{2h}{6400\times10^3} )

h=1.6\times 10^6\ m is the height a rocket has to go above Earth's surface before its weight is half of what it is on Earth.

4 0
3 years ago
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