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2 years ago

Which of the following is a typical product of nuclear fission?

2 answers:

6 0

The correct answer for the question that is being presented above is this one: "c. radiation." The typical product of nuclear fission is radiation. Nuclear fission's another product is that<span> two different atoms that are both less massive than the original atom </span>

krok68 [10]2 years ago

6 0

On edge.nuity the answer is C- radiation. Hope this helps you out :)

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What is the overall charge of 1.5 x 10^10 electrons?

Troyanec [42]

Answer: Charge = -2.4x10^-9 Coulombs

Explanation:

The charge of one electron is e = -1.6x10^-19 C

Then, the charge of 1.5 x 10^10 electrons is equal to 1.5 x 10^10 times the charge of one electron:

Here i will use the relation (a^b)*(a^c) = a^(b + c)

Charge = ( 1.5 x 10^10)*( -1.6x10^-19 C) = -2.4x10^(10 - 19) C

Charge = -2.4x10^-9 C

7 0

2 years ago

What is the kinetic energy of an object that has a mass of 12 kilograms and moves with a velocity of 10 m/s?

Alik [6]

1/2mv^2

1/2x12x10^2=600J

The kinetic energy is 600J

3 0

2 years ago

Read 2 more answers

Describe mechanical energy in your own words

Vlad [161]

Answer:

motion......

movement.....

3 0

2 years ago

Read 2 more answers

An object is dropped from a bridge. A second object is thrown downward 1.48 s later. They both reach the water 48.1 m below at t

pashok25 [27]

To solve this problem we will apply the linear motion kinematic equations. With the data provided we will calculate the time of the first object to fall. Later we will get the time difference between the two. This difference will allow us to find the free fall distance. Through the distance we will find the initial velocity, that is,

$x = v_0 t +\frac{1}{2}at^2$

$48.1 = 0*t + \frac{1}{2} (9.8)t^2$

$t = 3.13s$

The second object is thrown downward at one second later and it meets the first object at the water is

$t' = 3.13 -1.48$

$t' = 1.65s$

The distance of the object will travel due to free fall acceleration is

$x = v_0 t+\frac{1}{2} at^2$

$x = 0*(1.65) +\frac{1}{2}(9.8)(1.65)^2$

$x = 13.34m$

The distance of the object will travel due to its initial velocity is

$v_0 = \frac{d_0}{t}$

$d_0 = v_0 t$

$48.1-13.34 = v_0 (1.65)$

$v_0 = 21.06m/s$

Therefore the initial speed of the second object is 21.06m/s

8 0

3 years ago

Pulsars "blink" because they _____.

rosijanka [135]

A pulsar, or a pulsing star, is a highly magnetized neutron star that emits a beam of electromagnetic radiation. So they blink when they are rotating because the beam of radiation they emit can only be seen when it is facing the Earth.
Hope this helps.

4 0

2 years ago