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bazaltina [42]
3 years ago
6

The body contains many small currents caused by the motion of ions in the organs and cells. Measurements of the magnetic field a

round the chest due to currents in the heart give values of about 10.0 μG . Although the actual currents are rather complicated, we can gain a rough understanding of their magnitude if we model them as a long, straight wire.If the surface of the chest is 5.40 cm from this current, how large is the current in the heart?
Physics
1 answer:
RoseWind [281]3 years ago
8 0

Answer:

270 μA

Explanation:

Use the magnetic field due to long, straight wire and solve for current I.

B= \frac{\mu_0I}{2\pi r}

I= \frac{2\pi r B}{\mu_0}

plug in the values

I= \frac{2\pi(5.40\times10^{-2})\times10\times10^{-6}\times10^{-4}}{4\pi\times10^{-7}}

= 2.7×10^{-4)×10^6

=270 μA

The current that flows in the  heart is 270 μA

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16. For this table of data, how should the y-axis be labeled (with units)?
vampirchik [111]

Answer:

The y-axis should be labelled as W in Newtons (kg·m/s²)

Explanation:

The given data is presented here as follows;

Mass (kg)            {}        Newtons (kg·m/s²)

3.2                      {}           31.381

4.6             {}                    45.1111

6.1              {}                    59.821

7.4              {}                    72.569

9                {}                     89.241

10.4              {}                   101.989

10.9              {}                  106.892

From the table, it can be seen that there is a nearly linear relationship between the  amount of Newtons and the  mass, as the slope of the data has a relatively constant slope

Therefore, the data can be said to be a function of Weight in Newtons to the mass in kilograms such that the weight depends on the mass as follows;

W(m) in Newtons = Mass, m in kg × g

Where;

g is the constant of proportionality

Therefore, the y-axis component which is the dependent variable is the function, W(m) = Weight of the body while the x-axis component which is the independent variable is the mass. m

The graph of the data is created with Microsoft Excel give the slope which is the constant of proportionality, g = 9.8379, which is the acceleration due to gravity g ≈ 9.8 m/s²

We therefore label the y-axis as W in Newtons (kg·m/s²)

6 0
3 years ago
Water enters a cylindrical tank through two pipes at rates of 250 and 100 gal/min. If the level of the water in the tank remains
tester [92]

Answer:

The total amount of water that enters the tank is:

250 gal/min + 100 gal/min = 350gal/min.

Then, if the level of the water remains constant, this means that the water leaves the tank at a rate of 350gal/min.

We know that the diameter of the pipe is 8 inches, then the area of the pipe is:

A = pi*(d/2)^2 = 3.14*(4in)^2 = 50.24in^2

now, the flow can be calculated as:

Q = v*A = (velocity*area)

if we want to write our velocity in inches per minute, then we need to write the entering flow in cubic inches:

1 gallon = 231 in^3

then:

350gal/min = (350*231) in^3/min = 80,850 in^3/min.

Then the water that leaves the tank must be the same amoun, we have:

Q = 80,850 in^3/min. = v*A = v*50.24in^2

v =  (80,850in^3/min)/50.24in^2 = 1609.3 in/min.

The velocity of the flow leaving the tank is 1609.3 in/min.

3 0
3 years ago
A vacuum tube can be used to__. A. change alternating current into direct current B. increase the strength of a signal.. C. turn
vesna_86 [32]

The correct answer of this question is :  A) Change alternating current into direct current.

EXPLANATION :

As per the question, we are given vacuum tube. Vacuum tube can be of various types. Normally it contains two electrodes called cathode and anode which are enclosed in an evacuated glass chamber . There are also other types of vacuum tubes which contain extra electrodes like control grid .

The vacuum tube can be used as a rectifier. It means that it can be used as an electronic device which will convert alternating current into direct current. It may be a half wave rectifier or a full wave rectifier. Actually the direct current obtained during the rectification of alternating current is pulsating in nature.

Hence, the correct answer is that a vacuum tube can be used to change alternating current into direct current.


4 0
3 years ago
Read 2 more answers
A particular HeNe laser beam at 633 nm has a spot size of 0.8 mm. Assuming a
Kobotan [32]

Answer:

Explanation:

A

8 0
3 years ago
A coin is dropped in a 15.0 m deep well.
labwork [276]

Answer:

t = 1.75

t = 0.04

Explanation:

a)

For part 1 we want to use a kenamatic equation with constant acceleration:

X = 1/2*a*t^2

isolate time

t = sqrt(2X / a)

Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2

t = sqrt(2*15m / 9.8m/s^2)

t = 1.75 s

b)

The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:

v = d / t

isolate time

t = d / v

plug in known variables

t = 15m / 340m/s

t = 0.04 s

7 0
3 years ago
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