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3 years ago

Which best describes nuclear fusion?

Physics

2 answers:

MrMuchimi3 years ago

5 0

Answer:

The statement that best describes nuclear fusion is;

Nuclei combine to form a heavier nucleus, releasing energy

Explanation:

In nuclear fusion, we have the reaction of the nuclei of two or more atoms coming together (combining) to form heavier elements and subatomic particles such as protons and neutrons accompanied by the release or absorption in energy depending on the difference between the mass of the reactants and the products

Some nuclear fusion reaction require an input of energy and such reactions are therefore not spontaneous

The best option is nuclei (two or more nuclei) combine to form a heavier nucleus, releasing energy.

NeX [460]3 years ago

5 0

Answer:

The answer is the last option or D

Explanation:

Nuclei combine to form a heavier nucleus, releasing energy.

Have a great day and stay safe!

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On a planet where g = 10.0 m/s2 and air resistance is negligible, a sled is at rest on a rough inclined hill rising at 30°. The

Cloud [144]

Answer:

Explanation:

a is the acceleration

μ is the coefficient of friction

Acceleration of the object is given by

$a = g (\sin \theta -\mu \cos \theta)\\\\=10( \sin 30 - 0.4 \cos 30)\\\\=10(0.5-0.3464)\\\\=1.54m/s^2$

Velocity at the bottom

$v^2=u^2+2as\\\\u=0\\\\v^2=2as\\\\=2*1.54*8\\\\=24.576\\\\v=4.96m/s$

after travelling 4m , its velocity becomes 0

$a=\frac{v^2-u^2}{2s}$

$a=\frac{0-u^2}{2s}$

$a=\frac{-(-4.96)^2}{2*4} \\\\=-3.075m/s^2$

Coefficient of kinetic friction

μ = F/N

$=\frac{ma}{mg} \\\\=\frac{3.075}{10} \\\\=0.31$

Therefore, the Coefficient of kinetic friction is 0.31

8 0

3 years ago

Read 2 more answers

A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$

$-F_{f} + ma = 0$

$\mu mg = ma$

$\mu = \frac{a}{g}$

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}$

Where:

d: is the distance traveled = 46.1 m

$v_{f}$ : is the final speed of the truck = 0 (it stops)

$v_{0}$ : is the initial speed of the truck = 17.9 m/s

$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

$\mu = \frac{a}{g}$

$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

$\mu = 0.35$

Therefore, the minimum coefficient of friction required is 0.35.

I hope it helps you!

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A rowboat passenger uses an oar to push the boat off the dock by exerting a force of 40N for 3.0s. What impulse acts on the boat

jarptica [38.1K]

Answer:

120 Ns

Explanation:

The impulse exerted on an object is given by:

$I=Ft$

where

F is the force applied

t is the time taken

In this problem, we have:

F = 40 N

t = 3.0 s

So, the impulse acting on the boat is

$I=Ft=(40 N)(3.0 s)=120 Ns$

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4 years ago