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sergey [27]
3 years ago
11

Identify one similarity between fission and fusion

Physics
2 answers:
Zina [86]3 years ago
4 0

Answer:

Fusion and fission are similar in that they both release large amounts of energy. Nuclear fusion is a process in which two nuclei join to form a larger nucleus. Nuclear fission is a process in which a nucleus splits into two smaller nuclei.

Explanation:

muminat3 years ago
3 0

Answer:

Both fission and fusion release large amounts of energy.

Explanation:

One or more atoms of a different weight to the reactant atom is formed in both fission and fusion reactions. There is a change in weight between the reactant (s) and product (s) in both fission and fusion reactions.

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The answer is A. Thermal energy is transferred.

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A 7.30 kg sign hangs from two wires. The
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Answer:

28.0\; {\rm N} to the right.

Explanation:

Since the sign is not moving, the net force on this sign should be 0\; {\rm N}. For that, the horizontal component (x-component) of external forces on this sign should be 0\; {\rm N}.

Sources of external forces on this sign include tension from the wires, as well as gravitational pull (weight) from the earth. The gravitational pull from the earth is entirely vertical (y-component,) with a magnitude of 0\; {\rm N} in the horizontal direction. Thus, the only external forces on this sign in the x-component would be from the two wires.

The question states that the x-component of the force from the first wire is 28.0\; {\rm N} to the left. Thus, for the net force in the x-direction to be 0\; {\rm N}, the force from the other wire in the x\!-component needs to be 28.0\; {\rm N}\! to the right (same magnitude but opposite direction.)

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As longitudional waves travel, particles in the medium are pushed together and then pulled apart. We call this
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Compression and rarefaction.
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4 years ago
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zanele has a mass of 40kg and is sitting inside a 20kg cart .zanele's friends pull a cart with a force of 500N at an angle of 20
Dmitriy789 [7]

The net horizontal force acting on the cart is 169.85 N.

The change in the carts momentum is 5,000 Ns.

The net horizontal force on Zanele is same as the net horizontal force on the cart.

If angle between the 500N force and the horizontal is decreased, the final velocity of the cart increases.

The given parameters:

  • Mass of zanele, m1 = 40 kg
  • Mass of the cart, m2 = 20 kg
  • Applied force, F = 500 N, at angle 20 degrees
  • Frictional force on the cart, Ff = 300 N
  • Time, t = 10 s

<h3>Net horizontal force on the cart;</h3>
  • The net horizontal force acting on the cart is calculated as follows;

F_{net} = F - F_f\\\\F_{net} = 500 \times cos(20) - 300\\\\F_{net} = 169.85 \ N

<h3>Change in momentum;</h3>
  • The change in the carts momentum at the given time of applied force is calculated as follows;

\Delta mv = Ft  \\\\\Delta mv = 500 \times 10= 5,000 \ Ns

The net horizontal force on Zanele is same as the net horizontal force on the cart.

<h3>The final velocity of the cart;</h3>
  • When the angle decreases the cart's final velocity would be affected  as follows;

F_{net} = F - F_f\\\\F_{net} = \frac{mv}{t} \\\\\frac{mv}{t} =  F - F_f\\\\  \frac{mv}{t} =  Fcos(\theta) - F_f\\\\let \ \theta  = 0^0\\\\\frac{mv}{t} =500 \times cos(0) - 300\\\\\frac{mv}{t} = 200\\\\v = \frac{200 t}{m} \\\\when \ \theta = 20^0\\\\\frac{mv}{t} =500 \times cos(20) - 300\\\\\frac{mv}{t} = 169.85\\\\v = \frac{169.85t}{m}

Thus, if angle between the 500N force and the horizontal is decreased, the final velocity of the cart increases.

Learn more about net horizontal force here: brainly.com/question/21684583

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Answer:

protons

Explanation:

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