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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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3 years ago
1. Which of the following is not included
Harrizon [31]

Answer:

cartilage

Explanation:

8 0
3 years ago
A cylindrical capacitor has an inner conductor of radius 2.7 mmmm and an outer conductor of radius 3.1 mmmm. The two conductors
Mars2501 [29]

Answer:

(A) Capacitance per unit length = 4.02 \times 10^{-10}

(B) The magnitude of charge on both conductor is Q = 4.22 \times 10^{-19} C and the sign of charge on inner conductor is +Q and the sign on outer conductor is -Q

Explanation:

Given :

Radius of inner part of conductor  (R_{1}) = 2.7 \times 10^{-3} m

Radius of outer part of conductor  (R_{2}) = 3.1 \times 10^{-3} m

The length of the capacitor (l) = 3 \times 10^{-3} m

(A)

Capacitance is purely geometrical property. It depends only on length, radius of conductor.

From the formula of cylindrical capacitor,      

     C = \frac{2\pi\epsilon_{o} l }{ln\frac{R_{2} }{R_{1} } }

Where, \epsilon_{o} = 8.85 \times 10^{-12}

But we need capacitance per unit length so,

     \frac{C}{l}  = \frac{2\pi\epsilon_{o}  }{ln\frac{R_{2} }{R_{1} } }

capacitance per unit length = \frac{6.28 \times 8.85 \times 10^{-12} }{ln(1.148)} = 4.02 \times 10^{-10}

(B)

The charge on both conductors is given by,

     Q = C \Delta V

Where, C = capacitance of cylindrical capacitor and value of C = 12.06 \times 10^{-13} F, \Delta V = 350 \times 10^{-3} V

∴ Q = 4.22 \times 10^{-19} C

The magnitude of charge on both conductor is same as above but the sign of charge is different.

Charge on inner conductor is +Q and Charge on outer conductor is -Q.

8 0
3 years ago
Consider a highway composed of concrete. If the road is 4.5 km long, 30 m wide, and 0.70 m thick, what is its mass?
Aloiza [94]

Answer:

Explanation:

We need to assume that the density of the concrete is about 2350 Kg/m^3. And using the dimensions of the highway we can calculate the volume of the highway.

vh=4500 * 30 * 0,70\\vh=94500 m^3\\den=m/v\\m=den*v\\m=2350*94500\\\\m=222075 ton

5 0
3 years ago
(a) Write the energy equation for an elastic collision. (b) For an inelastic collision.
velikii [3]

Answer:

Explanation:

There are two types of collision.

(a) Elastic collision: When there is no loss of energy during the collision, then the collision is said to be elastic collision.

In case of elastic collision, the momentum is conserved, the kinetic energy is conserved and all the forces are conservative in nature.

The momentum of the system before collision = the momentum of system after collision

The kinetic energy of the system before collision = the kinetic energy after the collision

(b) Inelastic collision: When there is some loss of energy during the collision, then the collision is said to be inelastic collision.

In case of inelastic collision, the momentum is conserved, the kinetic energy is not conserved, the total mechanical energy is conserved and all the forces or some of the forces are non conservative in nature.

The momentum of the system before collision = the momentum of system after collision

The total mechanical energy of the system before collision = total mechanical of the system after the collision

5 0
3 years ago
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