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3 years ago

Please help! Will mark best answer!!

Physics

2 answers:

MA_775_DIABLO [31]3 years ago

7 0

The answer is (A) hope it helps

ICE Princess25 [194]3 years ago

7 0

Hey there Chainsmokers!

A nuclear fusion would actually be known to be a (reaction) of a very low fuse. Which on this case, it would then carry alot of energy, why? This would happen because the fuse would then form heavier substances such as nucleas.

So, from looking at your options listed above, it would seem that the correct answer would be your option "a".

"the process of two or more atomic particles sticking together to form a heavier atomic nucleus.".

Your correct answer would be option a.)

Hope this helps you dear!

You might be interested in

two cars with initial speed 2v and v, lock their brakes and skid to a stop. what is the ratio of the distance travelled

denpristay [2]

Answer:

4:1

Explanation:

Given that the initial speed of the first car, u = 2v while the initial speed of the second car, u = v. To find the distance travelled, we are going to apply one of the equations of motion. The equation chosen is

v² = u² - 2as, where

s = the distance needed

a = acceleration due to gravity

u = initial velocity which is v & 2v

v = final velocity which is 0

For the first car with initial velocity, 2v, on substituting into the equation, we have

v² = u² - 2as(1)

0 = 4v - 2as(1)

4v = 2as(1)

2v = as(1), making s(1) subject of formula we have

s(1) = 2v/a

Taking the second car, we have u = v

v² = u² - 2as(2)

0 = v - 2as(2)

v = 2as(2), making s(2) subject of formula, we have

s(2) = v/2a

Not, ratio of s1 : s2 =

2v/a : v/2a

s1/s2 = 2v/a ÷ v/2a

s1/s2 = 2v/a * 2a/v

s1/s2 = 4av/av

s1/s2 = 4/1

Therefore, the ratio of the first car to the second car is 4:1

Please leave a like if it helped you

8 0

2 years ago

When a 360 nF air capacitor is connected to a power supply, the energy stored in the capacitor is 1.85 x 10-5 J. While the capac

GaryK [48]

Answer:

(a) Approximately $10.1\; {\rm V}$ .

Explanation:

Let $C$ denote the capacitance of a capacitor. Let $V$ be the potential difference (voltage) between the two plates of this capacitor. The energy $E$ stored in this capacitor would be:

$\displaystyle E = \frac{1}{2}\, C\, (V^{2})$ .

Rearrange this equation to find an expression for the potential difference $V$ in terms of capacitance $C$ and energy $E$ :

$\begin{aligned}V^{2} &= \frac{2\, E}{C} \end{aligned}$ .

$\begin{aligned}V &= \sqrt{\frac{2\, E}{C}} \end{aligned}$

The capacitance $C$ of this capacitor is given in nanofarads. Convert that unit to standard unit (farads):

$\begin{aligned}C &= 360\; {\rm nF} \\ &= 360\; {\rm nF} \times \frac{1\; {\rm F}}{10^{9}\; {\rm nF}} \\ &= 3.60 \times 10^{-7}\; {\rm F}\end{aligned}$ .

Given that the energy stored in this capacitor is $E = 1.85 \times 10^{-5}\; {\rm J}$ , the potential difference across the capacitor plates would be: