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xenn [34]
3 years ago
7

Which of the following materials is necessary to stop an alpha particle? a. three feet of concrete c. single sheet of aluminum f

oil b. three inches of lead d. single sheet of paper
Physics
2 answers:
Nat2105 [25]3 years ago
8 0
A single sheet of paper can stop an alpha particle
aleksklad [387]3 years ago
3 0

Answer:

d) Single sheet of paper

Explanation:

Since alpha particles are heaviest particles that comes out from nucleus during the radioactive decay so out of all these are having least energy.

So here in this we can say that penetration power of alpha particles is least and it can be stopped by least thickness due to less penetration power and least energy

So here we can say that alpha particles can be stopped just by using the least thickness.

So out of all given options we can have

d) Single sheet of paper

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For a sound coming from a point source, the amplitude of the sound is inversely
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2 years ago
A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout
nika2105 [10]

Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

  • If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
  • So, we can write the following equation:

       E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)

  • As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
  • So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
  • So, on the outer surface of the shell there must be a charge that be the difference between them:

        Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C

  • Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

      E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C

  • As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
6 0
3 years ago
A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the
harina [27]

Answer:

R=2.78\ \Omega

Explanation:

Given that,

The current flowing in the circuit, I = 3 A

The power of the battery, P = 25 W

We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :

P=I^2R

Put all the values to find R.

R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega

So, the resistance is equal to 2.78\ \Omega.

7 0
2 years ago
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