1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
inna [77]
2 years ago
5

What are two serious condition of the circulatory system?

Physics
2 answers:
maria [59]2 years ago
8 0

Answer:

heart failure, heart attack,

Explanation:

heart failure, heart attack,

GarryVolchara [31]2 years ago
4 0

Answer:

coronary artery disease, high blood pressure, heart attack, heart failure are some examples

You might be interested in
A filamentary conductor is formed into an equilateral triangle with sides of length carrying current i . find the magnetic field
arsen [322]

magnetic field due to a finite straight conductor is given by

B = \frac{\mu_0 i}{4\pi r}(sin\theta_1 + sin\theta_2)

here since it forms an equilateral triangle so we will have

\theta_1 = \theta_2 = 60 degre

also the perpendicular distance of the point from the wire is

r = \frac{a}{2\sqrt3}

now from the above equation magnetic field due to one wire is given by

B = \frac{\mu_0 i}{4\pi \frac{a}{2\sqrt3}(sin60 + sin60)

B = \frac{\mu_0 i*2\sqrt3}{4\pi a}(\sqrt3)

B = \frac{3\mu_0 i}{2\pi a}

now since in equilateral triangle there are three such wires so net magnetic field will be

B = \frac{9\mu_0 i}{2\pi a}

5 0
3 years ago
What happens to the density of a given substance if you increase the amount of the substance that you have?
Amanda [17]
Have you ever looked up the density of a substance ?  You ought to try it.  Go ahead. Pick a substance, then go online or open up an actual book and find its density.  You will never see any particular volume mentioned along with the density . . . because it doesn't matter.  The whole idea of density is that it describes the substance, no matter how much or how little you have of it.  The density of a tiny drop of water under a microscope is the same as the density of a supertanker-ful of water.
7 0
3 years ago
Read 2 more answers
Two isolated, concentric, conducting spherical shells have radii R1 = 0.500 m and R2 = 1.00 m, uniform charges q1=+2.00 µC and q
scZoUnD [109]

Complete Question

The diagram for this question is shown on the first uploaded image  

Answer:

a E =1.685*10^3 N/C

b E =36.69*10^3 N/C

c E = 0 N/C

d V = 6.7*10^3 V

e   V = 26.79*10^3V

f   V = 34.67 *10^3 V

g   V= 44.95*10^3 V

h    V= 44.95*10^3 V

i    V= 44.95*10^3 V

Explanation:

From the question we are given that

       The first charge q_1 = 2.00 \mu C = 2.00*10^{-6} C

       The second charge q_2 =1.00 \muC = 1.00*10^{-6}

      The first radius R_1 = 0.500m

      The second radius R_2 = 1.00m

 Generally \ Electric \ field = \frac{1}{4\pi\epsilon_0}\frac{q_1+\ q_2}{r^2}

And Potential \ Difference = \frac{1}{4\pi \epsilon_0}   [\frac{q_1 }{r}+\frac{q_2}{R_2} ]

The objective is to obtain the the magnitude of electric for different cases

And the potential difference for other cases

Considering a

                      r  = 4.00 m

           E = \frac{((2+1)*10^{-6})*8.99*10^9}{16}

                = 1.685*10^3 N/C

Considering b

           r = 0.700 m \ , R_2 > r > R_1

This implies that the electric field would be

            E = \frac{1}{4\pi \epsilon_0}\frac{q_1}{r^2}

             This because it the electric filed of the charge which is below it in distance that it would feel

            E = 8*99*10^9  \frac{2*10^{-6}}{0.4900}

               = 36.69*10^3 N/C

   Considering c

                      r  = 0.200 m

=>   r

 The electric field = 0

     This is because the both charge are above it in terms of distance so it wont feel the effect of their electric field

       Considering d

                  r  = 4.00 m

=> r > R_1 >r>R_2

Now the potential difference is

                  V =\frac{1}{4\pi \epsilon_0} \frac{q_1 + \ q_2}{r} = 8.99*10^9 * \frac{3*10^{-6}}{4} = 6.7*10^3 V

This so because the distance between the charge we are considering is further than the two charges given  

          Considering e

                       r = 1.00 m R_2 = r > R_1

                V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{1.00} \frac{1.00*10^{-6}}{1.00} ] = 26.79 *10^3 V

          Considering f

              r = 0.700 m \ , R_2 > r > R_1

                      V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.700} \frac{1.0*10^{-6}}{1.00} ] = 34.67 *10^3 V

          Considering g

             r =0.500\m , R_1 >r =R_1

   V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{r} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

          Considering h

                r =0.200\m , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

           Considering i    

   r =0\ m \ , R_1 >R_1>r

  V = \frac{1}{4\pi \epsilon_0} [\frac{q_1}{R_1} +\frac{q_2}{R_2}  ] = 8.99*10^9 * [\frac{2.00*10^{-6}}{0.500} \frac{1.0*10^{-6}}{1.00} ] = 44.95 *10^3 V

8 0
3 years ago
Apply the concept of density to explain why oil floats on water.
Sauron [17]
Anything less dense than water will float, like oil. Anything more dense than water will sink, like rock.
6 0
3 years ago
A test piolot flies with an acceleration of 5
Colt1911 [192]

On Earth, 1 g = 9.8 m/s² .

5 g = 5 · (9.8 m/s²)

5 g = 49 m/s²

5 0
3 years ago
Other questions:
  • A rubber balloon has become negatively charged from being rubbed with a wool cloth, and the charge is measured as 1.00 × 10−14 C
    13·1 answer
  • Friends who are your age and share the same interests are called
    9·2 answers
  • Find the location of the center of mass of the earth-moon system relative to the center of earth.
    9·1 answer
  • When an object is turning around, is it also at rest at the point?
    5·2 answers
  • In terms of the variables in the problem, determine the time, t, after the launch it takes the balloon to reach the target. Your
    6·1 answer
  • a ball is rolled uphill a distance of 3 meters before it slows,stops,and begins to roll back. the ball rolls downhill 6 meters b
    13·1 answer
  • Which state of matter would be described as a highly energized charge particles with moving extremely fast
    6·1 answer
  • What is the magnitude of force required to accelerate a car of mass 1.7 * 10 ^ 3 kg by 4.75 m/s
    13·1 answer
  • A A load of 1000 N can be lifted by applying
    6·2 answers
  • If we apply a potential difference of 4. 50 v between the ends of a wire that is 2. 50 m in length and 0. 654 mm in radius, the
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!