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Ronch [10]
2 years ago
5

Uranium-235 Fission

Physics
1 answer:
ExtremeBDS [4]2 years ago
4 0

In the balanced nuclear reaction of uranium 235, we have thorium 231, helium atom and energy released at the right of the equation

<h3>Balanced nuclear reaction of Uranium</h3>

The balanced nuclear reaction of uranium is determined as follows;

^{235}_{92}U\ --- > \ ^4_2He\ + \ ^{231}_{90}Th\ + \ energy

Thus, in the balanced nuclear reaction of uranium 235, we have thorium 231, helium atom and energy released at the right of the equation while uranium 235 is on the left hand side of the nuclear reaction equation.

Learn more about nuclear reaction here:

brainly.com/question/25387647

#SPJ1

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An object can have forces acting upon it, but might not accelerate<br><br> True<br> or<br> False?
vivado [14]

Answer:

true

Explanation:

if you apply force to the top of a square it will not move

8 0
3 years ago
220V is applied to two different conductors made of the same material. One conductor is twice as long and twice as thick as the
ollegr [7]

Answer:

Explanation:

For calculating resistance of a conductor , the formula is

R = ρ l / A , ρ is specific resistance , l is length and A is cross sectional area of wire.

For first wire length is l₁ , area is A₁ resistance is R₁, for second resistance is R₂ , length is l₂ and area is A₂

Given , l₁ = 2l₂ , A₁ = 4A₂ , area is proportional to square of thickness.

R₁ / R₂ = I₁A₂ / I₂A₁

= 2l₂ x A₁ / 4 I₂A₁

= 1 / 2

2R₁ = R₂

Power = V² / R

Ratio of power = (V² / R₁) x (R₂ / V²)

= R₂ / R₁

= 2 .

7 0
3 years ago
Determine the thrust that a boat with a volume of 1.2m³ receives when it is stranded at sea. The density of seawater is 1020kg /
puteri [66]

Answer:

The maximum possible up-thrust on the boat is 11,995.2 N

Explanation:

According to Archimedes' principle, the thrust received by an object immersed a fluid is equal to the weight of the fluid displaced;

The given parameter of the boat in sea water are;

The volume of the boat = 1.2 m³

The density of seawater = 1020 kg/m³

Density = Mass/Volume

Therefore, Mass = Density × Volume

The maximum volume of water that the boat displaces = 1.2 m³

The mass of the water displaced by the boat = (Density of seawater) × (Volume of seawater displaced)

∴ The maximum possible mass of the water displaced by the boat = 1.2 m³ × 1020 kg/m³ = 1224 kg

The maximum possible mass of the water displaced by the boat, m = 1224 kg

Weight = Mass, m × g

Where;

g = The acceleration due to gravity = 9.8 m/s²

The up-thrust on the boat = The weight of the seawater displaced

∴ The maximum possible up-thrust on the boat = m × g = 1224 kg × 9.8 m/s² = 11,995.2 N

The maximum possible up-thrust on the boat = 11,995.2 N.

3 0
2 years ago
when an object falls to the ground, only the object moves down but the earth's motion is not noticeable​.why?
vlada-n [284]
Both the object and earth pulls each other towards itself but since the mass and pulling force of objects are very small the pulling force of objects are negligible.
7 0
3 years ago
On a straight road, a car speeds up at a constant rate from rest to 20 m/s over a 5 second interval and a truck slows at a const
IceJOKER [234]

Answer:

a)

Explanation:

  • Since the car speeds up at a constant rate, we can use the kinematic equation for distance (assuming that the initial position is x=0, and choosing t₀ =0), as follows:

        x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}   (1)

  • Since the car starts from rest, v₀ =0.
  • We know the value of t = 5 sec., but we need to find the value of a.
  • Applying the definition of acceleration, as the rate of change of velocity with respect to time, and remembering that v₀ = 0 and t₀ =0, we can solve for a, as follows:

       a_{c} =\frac{v_{fc}}{t} = \frac{20m/s}{5s} = 4 m/s2  (2)

  • Replacing a and t in (1):

       x_{fc} = v_{o}*t + \frac{1}{2}*a*t^{2}  = \frac{1}{2}*a*t^{2} = \frac{1}{2}* 4 m/s2*(5s)^{2} = 50.0 m.  (3)

  • Now, if the truck slows down at a constant rate also, we can use (1) again, noting that v₀ is not equal to zero anymore.
  • Since we have the values of vf (it's zero because the truck stops), v₀, and t, we can find the new value of a, as follows:

       a_{t} =\frac{-v_{to}}{t} = \frac{-20m/s}{10s} = -2 m/s2  (4)

  • Replacing v₀, at and t in (1), we have:

       x_{ft} = 20m/s*10.0s + \frac{1}{2}*(-2 m/s2)*(10.0s)^{2} = 200m -100m = 100.0m   (5)

  • Therefore, as the truck travels twice as far as the car, the right answer is a).
7 0
3 years ago
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