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2 years ago

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The nucleus of a large atom splits and two smaller atoms are formed. Energy is released in the process. What is the name of this

process?

2 answers:

Ganezh [65]2 years ago

4 0

That process is called "nuclear fission".

creativ13 [48]2 years ago

4 0

Answer: Nuclear Fission

Explanation:

Nuclear fission reactions are one of the nuclear reactions in which heavier nuclei splits into two or more lighter nuclei. In these reactions, neutrons are released with a large amount of energy.

General equation for nuclear fission reactions is given by:

$_Z^A\textrm{X}\rightarrow _{Z-U}^{A-V}\textrm{Y}+_U^V\textrm{W}+3_0^1n+\text{Energy}$

Nuclear fusion reactions are one of the nuclear reactions in which two or more lighter nuclei combines to form heavier nuclei. In these reactions, large amount of energy is released.

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What force is necessary to keep a mass of 0.8 kg revolving in a horizontal circle of radius 0.7 m with a period of 0.5 s? What i

inn [45]

Answer:

88.34 N directed towards the center of the circle

Explanation:

Applying,

F = mv²/r................... Equation 1

F = Force needed to keep the mass in a circle, m = mass of the mass, v = velocity of the mass, r = radius of the circle.

But,

v = 2πr/t................... Equation 2

Where t = time, π = pie

Substitute equation 2 into equation 1

F = m(2πr/t)²/r

F = 4π²r²m/t²r

F = 4π²rm/t²............. Equation 3

From the question,

Given: m = 0.8 kg, r = 0.7 m, t = 0.5 s

Constant: π = 3.14

Substitute these values into equation 3

F = 4(3.14²)(0.7)(0.8)/0.5²

F = 88.34 N directed towards the center of the circle

8 0

2 years ago

Which of the following BEST describes a deer?

ad-work [718]

If we go down the list,

It wouldn’t be Primary Producer because Deer don’t really produce anything

Can’t be Carnivore because deer don’t eat meat

Same with tertiary

They do eat plants. So it would have to be the last one

6 0

3 years ago

Read 2 more answers

An example of when total internal reflection occurs is when all the light passing from a region of higher index of refraction to

Amiraneli [1.4K]

Answer:

is reflected back into the region of higher index

Explanation:

Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.

According to Snell's law, refraction of ligth is described by the equation

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

n1 is the refractive index of the first medium

n2 is the refractive index of the second medium

$\theta_1$ is the angle of incidence (in the first medium)

$\theta_2$ is the angle of refraction (in the second medium)

Let's now consider a situation in which

$n_1 > n_2$

so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

Where $\frac{n_1}{n_2}$ is a number greater than 1. This means that above a certain value of the angle of incidence $\theta_1$ , the term on the right can become greater than 1. So this would mean

$sin \theta_2 > 1$

But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by

$\theta_c =sin^{-1}(\frac{n_2}{n_1})$

8 0

3 years ago

5. A body travels in a circle at a constant speed, it

nata0808 [166]

A the same acceleration

5 0

2 years ago

Read 2 more answers

two point charges of 5*10^-19 C and 20*10^-19C are separated by a distance of 2m. at which point on the line joining them will h

Aneli [31]

Answer:

On that line segment between the two charges, at approximately $0.7\; \rm m$ away from the smaller charge (the one with a magnitude of $5 \times 10^{-19}\; \rm C$ ,) and approximately $1.3\; \rm m$ from the larger charge (the one with a magnitude of $20 \times 10^{-19}\; \rm C$ .)

Explanation:

Each of the two point charges generate an electric field. These two fields overlap at all points in the space around the two point charges. At each point in that region, the actual electric field will be the sum of the field vectors of these two electric fields.

Let $k$ denote the Coulomb constant, and let $q$ denote the size of a point charge. At a distance of $r$ away from the charge, the electric field due to this point charge will be:

$\displaystyle E = \frac{k\, q}{r^2}$ .

At the point (or points) where the electric field is zero, the size of the net electrostatic force on any test charge should also be zero.

Consider a positive test charge placed on the line joining the two point charges in this question. Both of the two point charges here are positive. They will both repel the positive test charge regardless of the position of this test charge.

When the test charge is on the same side of both point charges, both point charges will push the test charge in the same direction. As a result, the two electric forces (due to the two point charges) will not balance each other, and the net electric force on the test charge will be non-zero.

On the other hand, when the test charge is between the two point charges, the electric forces due to the two point charges will counteract each other. This force should be zero at some point in that region.

Keep in mind that the electric field at a point is zero only if the electric force on any test charge at that position is zero. Therefore, among the three sections, the line segment between the two point charges is the only place where the electric field could be zero.

Let $q_1 = 5\times 10^{-19}\; \rm C$ and $q_2 = 20 \times 10^{-19}\; \rm C$ . Assume that the electric field is zero at $r$ meters to the right of the $5\times 10^{-19}\; \rm C$ point charge. That would be $(2 - r)$ meters to the left of the $20 \times 10^{-19}\; \rm C$ point charge. (Since this point should be between the two point charges, $0 < r < 2$ .)

The electric field due to $q_1 = 5\times 10^{-19}\; \rm C$ would have a magnitude of:

$\displaystyle | E_1 | = \frac{k\cdot q_1}{r^2}$ .

The electric field due to $q_2 = 20 \times 10^{-19}\; \rm C$ would have a magnitude of:

$\displaystyle | E_2 | = \frac{k\cdot q_2}{(2 - r)^2}$ .

Note that at all point in this section, the two electric fields $E_1$ and $E_2$ will be acting in opposite directions. At the point where the two electric fields balance each other precisely, $| E_1 | = | E_2 |$ . That's where the actual electric field is zero.

$| E_1 | = | E_2 |$ means that $\displaystyle \frac{k\cdot q_1}{r^2} = \frac{k\cdot q_2}{(2 - r)^2}$ .

Simplify this expression and solve for $r$ :

$\displaystyle q_1\, (2 - r)^2 - q_2 \, r^2 = 0$ .

$\displaystyle 5\times (2 - r)^2 - 20\, r^2 = 0$ .

Either $r = -2$ or $\displaystyle r = \frac{2}{3}\approx 0.67$ will satisfy this equation. However, since this point (the point where the actual electric field is zero) should be between the two point charges, $0 < r < 2$ . Therefore, $(-2)$ isn't a valid value for $r$ in this context.

As a result, the electric field is zero at the point approximately $0.67\; \rm m$ away the $5\times 10^{-19}\; \rm C$ charge, and approximately $2 - 0.67 \approx 1.3\; \rm m$ away from the $20 \times 10^{-19}\; \rm C$ charge.

8 0

3 years ago