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gayaneshka [121]
2 years ago
11

An isotope undergoes radioactive decay by emitting radiation that has no mass. What other characteristic does the radiation have

?
some shielding required
positive or negative charge
low penetrating power
no charge
Physics
1 answer:
vitfil [10]2 years ago
4 0

Other characteristic of radiation is the presence of positive and negative charge.

<h3 /><h3>What other characteristic does the radiation have?</h3>

Positive or negative charge are the other characteristic that is present in the radiation. Beta radiation have negative charge while on the other hand, alpha radiation have positive charge.

So we can conclude that other characteristic of radiation is the presence of positive and negative charge.

Learn more about radiation here: brainly.com/question/893656

#SPJ1

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The argument against your claim (what the other side would say if they disagreed with your claim.) is:
Vlad1618 [11]

Answer:

a counterclaim

Explanation:

authors purpose is what an author wrote somthing for

opinion is someones thoughts or "side" on a argument

an arguement is a battle of opinions if that makes sense

3 0
3 years ago
The magnetic field at 8 cm distance from a long straight wire, carrying is 0.2x10^-5 T. How much is the electric current in the
FrozenT [24]

Answer:

The electric current in the wire is 0.8 A

Explanation:

We solve this problem by applying the formula of the magnetic field generated at a distance by a long and straight conductor wire that carries electric current, as follows:

B=\frac{2\pi*a }{u*I}

B= Magnetic field due to a straight and long wire that carries current

u= Free space permeability

I= Electrical current passing through the wire

a  = Perpendicular distance from the wire to the point where the magnetic field is located

Magnetic Field Calculation

We cleared (I) of the formula (1):

I=\frac{2\pi*a*B }{u} Formula(2)

B=0.2*10^{-5}  T = 0.2*10^{-5} \frac{weber}{m^{2} }

a  =8cm=0.08m

u=4*\pi *10^{-7} \frac{Weber}{A*m}

We replace the known information in the formula (2)

I=\frac{2\pi*0.08*0.2*10^{-5}  }{4\pi *10x^{-7} }

I=0.8 A

Answer: The electric current in the wire is 0.8 A

4 0
3 years ago
A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−
Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m.

Explanation:

Given that,

The actual wavelength of the hydrogen atom, \lambda_a=5.13\times 10^{-7}\ m

A hydrogen atom in a galaxy moving with a speed of, v=6.65\times 10^6\ m/s

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}

\lambda_o is the observed wavelength

\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m

So, the observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m. Hence, this is the required solution.

8 0
3 years ago
3, A 4kg block is pushed 2m at an acceleration of 0.2m/s square up a vertical wall by a constant force F applied at an angle of
Andrews [41]

The work done by the applied force on the block against the frictional force is 15.75 J.

<h3>Work done by the applied force</h3>

The work done by the applied force is calculated as follows;

W = Fd

F - Ff = ma

where;

  • F is applied force
  • Ff is frictional force

Fcos(37) - μmgsin(37) = ma

Fcos(37) - (0.3)(4)(9.8)sin(37) = 4(0.2)

0.799F - 7.077 = 0.8

F = 9.86 N

W = Fdcosθ

W = 9.86 x 2 x cos(37)

W = 15.75 J

Thus, the work done by the applied force on the block against the frictional force is 15.75 J.

Learn more about work done here: brainly.com/question/25573309

#SPJ1

5 0
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
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