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Studentka2010 [4]
3 years ago
15

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
Leftover planetesimals that formed in the region of the solar system now occupied by the jovian planets are called
rewona [7]

Answer:

Comets

Explanation:

Comets are planetary celestial bodies consisting of ice and dust, sometimes rocky particles  formed in the region of the solar system. Long-period comets propagate towards the Sun by gravitational perturbations caused by passing stars. Some comets usually hyberbolic comets, move through the inner Solar System  prior to entering the interstellar region. Short period comet lies beyond the orbit of the Neptune.

The Jovian planets include Jupiter, Saturn, Uranus, and Neptune.

Therefore, leftovers of comets (planetesimal bodies) formed in the region of the solar system that are now occupied by the Jovian planets is due to the dusty particles  associated with the comets.

7 0
3 years ago
A mass of 3.0 kg rests on a smooth surface inclined 34° above the horizontal. It is kept from sliding down the plane by a spring
azamat

Answer:

The spring stretched by x = 13.7 cm

Explanation:

Given data

Mass = 3 kg

k = 120 \frac{N}{m}

Angle \theta = 34°

From the free body diagram

Force acting on the box = mg sin\theta

⇒ F = 3 × 9.81 × \sin34

⇒ F = 16.45 N ------- (1)

Since box is attached with the spring so a spring force also acts on the box.

F_{sp} = k x

F_{sp} = 120 x -------- (2)

The net force acting on the body is given by

F_{net} = ma

Since acceleration of the box is zero so

F_{net} = 0

F - F_{sp} = 0

F = F_{sp}

Put the values from equation (1) & (2) we get

16.45 = 120x

x = 0.137 m

x = 13.7 cm

Therefore the spring stretched by x = 13.7 cm

3 0
3 years ago
If you kicked your mom <br><br> would she be mad?
vfiekz [6]

Answer:

Yes !

Explanation:

8 0
3 years ago
In a thin film experiment, a wedge of air is used between two glass plates. If the wavelength of the incident light in air is 48
cluponka [151]

Answer:

The thickness is  \Delta y =  2.4 *10^{-6} \  m

Explanation:

From the question we are told that

   The wavelength is  \lambda  = 480 \ nm = 480*10^{-9} \  m

    The first order of the dark  fringe is  m_1 =  16

     The second order of dark fringe considered is  m_2 = 6

Generally the condition for destructive interference is mathematically represented as

        y = \frac{m \lambda}{2}

Here y is the path difference between the central maxima(i.e the origin) and any dark fringe

So  the path difference between the 16th dark fringe and the 6th dark fringe is mathematically represented as

      y_1 - y_2 = \Delta y =  \frac{m_1 \lambda}{2} -  \frac{m_2 \lambda}{2}

=>  y_1 - y_2 = \Delta y =  \frac{16 *480*10^{-9}}{2} -  \frac{6 *480*10^{-9}}{2}

=>  y_1 - y_2 = \Delta y =  5 (480*10^{-9})

=>  \Delta y =  2.4 *10^{-6} \  m

8 0
3 years ago
Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1
lesantik [10]

Answer:

a)\lambda=6.63\times10^{-31}m

b)\lambda=6.63\times10^{-39}m

c)\lambda=9.97\times10^{-11}m

d)\lambda=4.03\times10^{-36}m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

\lambda=\frac{h}{P}=\frac{h}{mv}

with h the Plank's constant (6.63\times10^{-34}\frac{m^{2}kg}{s}) and P the momentum of the object that is mass (m) times velocity (v).

a)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}

\lambda=6.63\times10^{-31}m

b)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}

\lambda=6.63\times10^{-39}m

c)\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}

\lambda=9.97\times10^{-11}m

d)\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}

\lambda=4.03\times10^{-36}m

e) \lambda=\frac{6.63\times10^{-34}}{(74*0)}

λ=∞

6 0
3 years ago
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