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3 years ago

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A particular type of fundamental particle decays by transforming into an electron e- and a positron e (same amount of charge as

e- but with positive sign). Suppose the decaying particle is at rest in a uniform magnetic field B of magnitude 3.53 mT and the e- and e move away from the decay point in paths lying in a plane perpendicular to B. How long after the decay do the e- and e collide

1 answer:

Solnce55 [7]3 years ago

7 0

Answer:

Time after which it collide again is

$T = 1.01 \times 10^{-8} s$

Explanation:

As we know that the neutral particle decays into two particles i.e. electron and positron

As we know that both particles are of same mass but opposite charge

So both particles will revolve in the circle in opposite sense

So both particles will collide again after one complete revolution

So the time period of revolution is given as

$T = \frac{2\pi m}{qB}$

so we whave

$T = \frac{2\pi(9.1 \times 10^{-31})}{(1.6 \times 10^{-19})(3.53 \times 10^{-3})}$

$T = 1.01 \times 10^{-8} s$

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A common method to measure thermal conductivity of a biomaterial is to insert a long metallic probe axially into the center of a

tia_tia [17]

Answer:

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

Explanation:

Let suppose that thermal conduction is uniform and one-dimensional, the conduction heat transfer ( $\dot Q$ ), measured in watts, in the hollow cylinder is:

$\dot Q = \frac{2\cdot k\cdot L}{\ln \left(\frac{D_{o}}{D_{i}} \right)}\cdot (T_{i}-T_{o})$

Where:

$k$ - Thermal conductivity, measured in watts per meter-Celsius.

$L$ - Length of the cylinder, measured in meters.

$D_{i}$ - Inner diameter, measured in meters.

$D_{o}$ - Outer diameter, measured in meters.

$T_{i}$ - Temperature at inner surface, measured in Celsius.

$T_{o}$ - Temperature at outer surface, measured in Celsius.

Now we clear the thermal conductivity in the equation:

$k = \frac{\dot Q}{2\cdot L\cdot (T_{i}-T_{o})}\cdot \ln\left(\frac{D_{o}}{D_{i}} \right)$

If we know that $\dot Q = 40.8\,W$ , $L = 0.6\,m$ , $T_{i} = 50\,^{\circ}C$ , $T_{o} = 20\,^{\circ}C$ , $D_{i} = 0.01\,m$ and $D_{o} = 0.04\,m$ , the thermal conductivity of the biomaterial is:

$k = \left[\frac{40.8\,W}{2\cdot (0.6\,m)\cdot (50\,^{\circ}C-20\,^{\circ}C)}\right]\cdot \ln \left(\frac{0.04\,m}{0.01\,m} \right)$

$k \approx 1.571\,\frac{W}{m\cdot ^{\circ}C}$

The thermal conductivity of the biomaterial is approximately 1.571 watts per meter-Celsius.

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3 years ago

How can you make the potential energy as high as possible in a magnetic field between one electromagnet and one piece of iron?

harkovskaia [24]

In step 1, to increase the potential energy, the iron will move towards the electromagnet.

In step 2, to increase the potential energy, the iron will move towards the electromagnet.

<h3>Potential energy of a system of magnetic dipole</h3>

The potential energy of a system of dipole depends on the orientation of the dipole in the magnetic field.

$U = \mu B$

where;

$\mu$ is the dipole moment
B is the magnetic field

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

$U = \mu\times (\frac{\mu_0 I}{2\pi r} )$

Increase in the distance (r) reduces the potential energy. Thus, we can conclude the following;

In step 1, to increase the potential energy, the iron will move towards the electromagnet.
In step 2, when the iron is rotated 180, it will still maintain the original position, to increase the potential energy, the iron will move towards the electromagnet.

Learn more about potential energy in magnetic field here: brainly.com/question/14383738

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2 years ago

Which vector has an x-component with a length of 3?<br> А. С.<br> B. d<br> C. a<br> D. b

worty [1.4K]

Answer:

B.d

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3 years ago

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Sam is walking at 2m/s down the hallway when he realized he will be late to class. He speeds up to 4m/s and makes it to English

Salsk061 [2.6K]

Answer:

.5 m/s²

Explanation:

a= (vf-vi)/t

a= (4-2)/4

a=2/4

a= .5 m/s²

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2 years ago

A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand

enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

$v=gt$

And the distance traveled downwards is:

$\displaystyle y=\frac{gt^2}{2}$

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

$\displaystyle t=\sqrt{\frac{2y}{g}}$

Replacing into the first equation:

$\displaystyle v=g\sqrt{\frac{2y}{g}}$

Rationalizing:

$\displaystyle v=\sqrt{2gy}$

Let's call v1 the final speed of the package dropped from a height H. Thus:

$\displaystyle v_1=\sqrt{2gH}$

Let v2 be the final speed of the package dropped from a height 4H. Thus:

$\displaystyle v_2=\sqrt{2g(4H)}$

Taking out the square root of 4:

$\displaystyle v_2=2\sqrt{2gH}$

Dividing v2/v1 we can compare the final speeds:

$\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}$

Simplifying:

$\displaystyle v_2/v_1=2$

The final speed of the second package is twice as much as the final speed of the first package.

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3 years ago