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Pavel [41]
3 years ago
12

X rays are used in hospital to locate in the patients bones.If the x rays of wavelength of 2×10 to the power negative nine m tra

vel at a speed of 3×10 to the power eight m/s in air, what will be it's frequency?​
Physics
1 answer:
jasenka [17]3 years ago
8 0

Answer:1.5×10 to the power of 17(unit-Hertz/H)

Explanation:V=F×Wavelength

F=V/Wavelength=3×10 to power/2×10 to power of -9=1.5×10 to power of 17

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A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3
4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

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

where

\mu_0 is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT

8 0
3 years ago
The sun's core _____. A) is made of solid helium B) makes up most of the sun's size C) creates energy by fusing hydrogen into he
VashaNatasha [74]
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6 0
3 years ago
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If half of the weight of a flatbed truck is supported by its two drive wheels, what is the maximum acceleration it can achieve o
Scilla [17]

Answer:

Maximum acceleration will be equal to 3.43m/sec^2

Explanation:

We have given coefficient of kinetic friction \mu _k=0.7

And coefficient of static friction \mu _s=1

Acceleration due to gravity g=9.8m/sec^2

When truck moves maximum force will be equal to F=\mu _kmg

It is given that half of the weight is supported by its drive wheels

So force required =\frac{\mu _kmg}{2}

From newtons law maximum acceleration will be equal to a=\frac{\frac{\mu _kmg}{2}}{m}=\frac{\mu _kg}{2}=\frac{0.7\times 9.8}{2}=3.43m/sec^2

8 0
3 years ago
How wind Form?<br> Explain please
snow_lady [41]
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<u><em>Good luck! </em></u></span><u><em></em></u>
6 0
3 years ago
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