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pentagon [3]
3 years ago
7

What happens if we nuke a city?

Physics
2 answers:
nignag [31]3 years ago
3 0
The shadow would subsequently drop the temperature in the city, forcing warmer air to rush in and violently destroy countless buildings. There would also be “black rain,” which would be the radioactive ash and dust that would liquify and pour down on the city and we would all die.
irga5000 [103]3 years ago
3 0

Answer:

Those who survived the bomb may become deaf or blind, as well as suffer catastrophic burns and injuries. Even people who were not seriously injured could become trapped within a structure or unable to navigate through the wreckage.

Explanation:

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Does the shape of molecule affect the chemical bond?​
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There is a sharp distinction between ionic and covalent bonds when the geometric arrangements of atoms in compounds are considered. In essence, ionic bonding is nondirectional, whereas covalent bonding is directional.

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Jfdojbfkkjds vdf;ifdfdi
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hsjishhdhjs jdjyshskksndhu

6 0
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A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they
Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁  +  m₂u₂

P₁ = (1000 x 25)  +  (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v    is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

(1000 x 25)  +  (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

                                                         = 27.727(1000 + 1200)

                                                          = 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0
3 years ago
What net force would be required for a 50.5 kg wolf to accelerate at 5.0 m/s^2
Basile [38]

Answer:

\boxed {\boxed {\sf 252.5 \ Newtons }}

Explanation:

Force is the push or pull on object that can cause different things, like acceleration. According to Newton's 2nd Law of Motion, it is the product of mass and acceleration.

F=m*a

The mass of the wolf is 50.5 kilograms and the acceleration is 5.0 meters per square second. Therefore:

  • m= 50.5 \ kg
  • a= 5.0 \ m/s^2

Substitute the values into the formula.

F= 50.5 \ kg * 5.0 \ m/s^2

Multiply.

F=252.5 \ kg*m/s^2

  • 1 kilogram meter per square second is equal to 1 Newton.
  • Our answer of 252.5 kg*m/s² is equal to 252.5 Newtons.

F= 252.5 \ N

The force required is <u>252.5 Newtons</u>.

4 0
3 years ago
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