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2 years ago

Where does the energy released in a nuclear decay reaction come from

Physics

1 answer:

erastovalidia [21]2 years ago

8 0

The answer is electrons

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Representar las uniones interatomicas<br> a. kcl<br> b. ZnO<br> c. ZnBr2<br> d. BeI2

Rashid [163]

Answer:

Explanation:

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3 years ago

Suppose that an electromagnetic wave is traveling toward the east. At one instant at a given point, the electric field vector po

Mrrafil [7]

Answer: east

Explanation: A propagating electromagnetic wave in space has it electric and magnetic field perpendicular to each other.

For this question of ours, the wave is moving in the eastward direction with the electric field upward (north), the magnetic field must be perpendicular at all time to the electric field in the direction of the motion of the electromagnetic wave.

Hence, the magnetic field is due east

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3 years ago

Using local acceleration of gravity data from the Internet, de- termine the weight, in N, of a person whose mass is 80 kg living

Anna007 [38]

Answer

mass of a person is 80 kg

a) Mexico City, Mexico

acceleration due to gravity in mexico = 9.776 m/s²

weight = 80 × 9.776 = 782.08 N

b) Cape Town, South Africa

acceleration due to gravity in cape town = 9.796 m/s²

weight = 80 × 9.796 = 783.68 N

c) Tokyo, Japan

acceleration due to gravity in cape town = 9.798 m/s²

weight = 80 × 9.798 = 783.84 N

d) Chicago, IL

acceleration due to gravity in cape town = 9.803 m/s²

weight = 80 × 9.803 = 784.24 N

e) Copenhagen, Denmark

acceleration due to gravity in cape town = 9.815 m/s²

weight = 80 × 9.815 = 785.2 N

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3 years ago

Select the correct images.

Afina-wow [57]

It could be the third and the first one.

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3 years ago

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An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.275 T. If the kinetic energy of the electr

xxMikexx [17]

Answer:

Radius, $r=2.14\times 10^{-5}\ m$

Explanation:

It is given that,

Magnetic field, B = 0.275 T

Kinetic energy of the electron, $E=4.9\times 10^{-19}\ J$

Kinetic energy is given by :

$E=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2E}{m}}$

$v=\sqrt{\dfrac{2\times 4.9\times 10^{-19}}{9.1\times 10^{-31}}}$

v = 1037749.04 m/s

The centripetal force is balanced by the magnetic force as :

$qvB\ sin90=\dfrac{mv^2}{r}$

$r=\dfrac{mv}{qB}$

$r=\dfrac{9.1\times 10^{-31}\times 1037749.04}{1.6\times 10^{-19}\times 0.275 }$

$r=2.14\times 10^{-5}\ m$

So, the radius of the circular path is $2.14\times 10^{-5}\ m$ . Hence, this is the required solution.

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3 years ago

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