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2 years ago

If a 1-megaton nuclear weapon is exploded

Physics

1 answer:

Elenna [48]2 years ago

4 0

Answer:

1,780,000 N

Explanation:

0.2 atm × (1.013×10⁵ Pa/atm) = 20,260 Pa

Force = pressure × area

F = 20,260 Pa × (3.89 m × 22.6 m)

F = 1,780,000 N

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Newton's cradle consists of an aligned row of identical elastic balls suspended by strings so they barely touch one another. whe

Ilia_Sergeevich [38]

Initial momentum = 2 * m v

so if 1 ball popped out with 2 v momentum would be conserved

but the initial kinetic energy of the two balls is

1/2 m v^2 + 1/2 m v^2 = m v^2

But the kinetic energy of one ball with twice the speed would be

KE = 1/2 m (2 v)^2 = 2 m v^2 thus violating conservation of energy

8 0

3 years ago

A tube of water is open on one end to the environment while the other end is closed. The height of the water relative to the bas

Andreas93 [3]

Answer:

1.06 atm

Explanation:

On the open end of the tube, the pressure will be the sum of atmospheric pressure and the pressure due to the height of water

The pressure due to a height of water = ρgh

where ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the height of the water column

The height of water column on the open end = 100 cm = 1 m

pressure on this end = ρgh = 1000 x 9.81 x 1 = 9810 Pa

Atmospheric pressure = 101325 Pa

The total pressure on the open end = 101325 Pa + 9810 Pa = 111135 Pa

The pressure due to the water column on the closed end = ρgh

The height of the water in the closed end = 40 cm = 0.4 m

The pressure due to this column of water = 1000 x 9.81 x 0.4 = 3924 Pa

The resultant pressure on the water on the top of the closed end of the tube = 111135 Pa - 3924 Pa = 107211 Pa

In atm unit, this pressure = 107211/101325 = 1.06 atm

7 0

3 years ago

A 400-n block is dragged along a horizontal surface by an applied force as shown. the coefficient of kinetic friction is uk = 0.

gulaghasi [49]

The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.

We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force $F_f$ acting against the motion. Since their resultant must be zero, we have:
$F-F_f = 0$
The frictional force is
$F_f = \mu mg$
where
$\mu=0.4$ is the coefficient of kinetic friction
$mg=400 N$ is the weight of the block.

Substituting these values, we find the magnitude of the force F:
$F=F_f = \mu mg=(0.4 )(400 N)=160 N$

4 0

3 years ago

A 4 kg block is pulled with 4.1 N of force to the right. The block experiences 1.0 N of

Evgesh-ka [11]

Answer:

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