How does nuclear energy work? Explain

1 answer:

6 0

There are many processes to get nuclear energy. Nuclear energy is basically energy from an atom. For example fission is where the nucleus of an atom ( typically radioactive atoms ) gets split then energy is released ( typically heat). And in radioactive decay radiation is released from an radioactive atom. Hope this helps

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A crate is pulled with a force of 165 N at an angle 30 ° northwest. What is the resultant horizontal force on the crate?

Ad libitum [116K]

Answer:

Resultant horizontal force = 143 N

Explanation:

Since the a gle is 30° northwest, then it means the resultant force will be horizontal and as such;

Resultant horizontal force = 165 * cos 30

Resultant horizontal force = 142.89

Approximating to a whole number gives;

Resultant horizontal force = 143 N

5 0

2 years ago

A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the angular

bagirrra123 [75]

To solve this problem we will use the concepts related to energy conservation. Both potential energy, such as rotational and linear kinetic energy, must be conserved, and the gain in kinetic energy must be proportional to the loss in potential energy and vice versa. This is mathematically

$PE = KE_{lineal} + KE_{rotational}$

$mgh = \frac{1}{2}mv^2 +\frac{1}{2} I\omega^2$

Where,

m = mass

v = Tangential Velocity

$\omega$ = Angular velocity

I = Moment of Inertia

g = Gravity

Replacing the value of Inertia in a Disk and rearranging to find h, we have

$mgh = \frac{1}{2}mv^2 +\frac{1}{2} I\omega^2$

$mgh = \frac{1}{2}mr^2\omega^2 + \frac{1}{2}(\frac{1}{2}mr^2)\omega^2 )$

$h = \frac{3}{4} \frac{r^2\omega^2}{g}$

Replacing,

$h = \frac{3}{4} \frac{(1.6)^2 (5.35)^2}{9.8}$

$h = 5.607m$

Therefore the height of the inclined plane is 5.6m

3 0

3 years ago

Find the electric field at a point midway between two charges of +40.0 x 10^-9 c and.+60.0 x 10^-9 c

PIT_PIT [208]

Missing part in the text: "...the charges are <span>separated by a distance of 30.0 cm."
</span>
Solution:
The point midway between the two charges is located 15.0 cm from one charge and 15.0 from the other charge. The electric field generated by each of the charges is
$E=k_e \frac{q}{r^2}$
where
ke is the Coulomb's constant
Q is the value of the charge
r is the distance of the point at which we calculate the field from the charge (so, in this problem, r=15.0 cm=0.15 m).

Let's calculate the electric field generated by the first charge:
$E_1 = (8.99 \cdot 10^9 Nm^2 C^{-2} ) \frac{+40.0 \cdot 10^{-9} C}{(0.15 m)^2}=1.6 \cdot 10^4 N/C$

While the electric field generated by the second charge is
$E_2 = (8.99 \cdot 10^9 N m^2 C^{-2} ) \frac{+60.0 \cdot 10^{-9} C}{(0.15 m)^2}=2.4 \cdot 10^4 N/C$

Both charges are positive, this means that both electric fields are directed toward the charge. Therefore, at the point midway between the two charges the two electric fields have opposite direction, so the total electric field at that point is given by the difference between the two fields:
$E=E_2 - E_1 = 2.4 \cdot 10^4 N/C - 1.6 \cdot 10^4 N/C = 8000 N/C$