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3 years ago

When U-235 splits, it usually emits

1 answer:

5 0

Splitting atoms. 'Fission' is another word for splitting. The process of splitting a nucleus is called nuclear fission. ... For fission to happen, the uranium-235 or plutonium-239 nucleus must first absorb a neutron.

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What voltage would produce a current of 100amps through an aluminum wire assuming that the re sistance of the wire is3.44×10-4 o

agasfer [191]

V = IR

I = current
R = resistance

Voltage = 100 * (3.44x 10^-4) = do the calculation

Hope this helps

8 0

2 years ago

What part of a thunderstorm kills the most people each year?

alexandr1967 [171]

Here is the answer. The part of a thunderstorm that kills the most people each year is the LIGHTNING. Thunder is only the sound created and will not hurt anyone, but it is the lightning that can kill anyone who will be struck by it. Hope this answers your question. Have a great day!

5 0

3 years ago

Read 2 more answers

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

2 years ago

You need to put a metal rod

Lubov Fominskaja [6]

-- Put the rod into the freezer for a while. As it cools,
it contracts (gets smaller) slightly.

-- Put the cylinder into hot hot water for a while. As it heats,
it expands (gets bigger) slightly.

-- Bring the rod and the cylinder togther quickly, before the
rod has a chance to warm up or the cylinder has a chance
to cool off.

-- I bet it'll fit now.

-- But be careful . . . get the rod exactly where you want it as fast
as you can. Once both pieces come back to the same temperature,
and the rod expands a little and the cylinder contracts a little, the fit
will be so tight that you'll probably never get them apart again, or even
move the rod.

8 0

3 years ago

Recall that the differential equation for the instantaneous charge q(t) on the capacitor in an lrc-series circuit is l d 2q dt 2

Aloiza [94]

DE which is the differential equation represents the LRC series circuit where
L d²q/dt² + Rdq/dt +I/Cq = E(t) = 150V.
Initial condition is q(t) = 0 and i(0) =0.
To find the charge q(t) by using Laplace transformation by
Substituting known values for DE
L×d²q/dt² +20 ×dq/dt + 1/0.005× q = 150
d²q/dt² +20dq/dt + 200q =150

5 0

3 years ago