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3 years ago

Example of nuclear fission

Physics

1 answer:

goldenfox [79]3 years ago

7 0

Answer:

In nuclear power plants, energetic neutrons are directed into a sample of the isotope uranium-235. ... A common fission reaction produces barium-141 and krypton-92. In this particular reaction, one uranium nucleus breaks into a barium nucleus, a krypton nucleus, and two neutrons

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A 100 g ball collides elastically with a 300 g ball that is at rest. If the 100 g ball was traveling

sammy [17]

Answer:

The magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

Explanation:

We can find the velocity of the two balls after the collision by conservation of linear momentum and energy:

$P_{1} = P_{2}$

$m_{1}v_{1_{i}} + m_{2}v_{2_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Where:

m₁: is the mass of the ball 1 = 100 g = 0.1 kg

m₂: is the mass of the ball 2 = 300 g = 0.3 kg

$v_{1_{i}}$ : is the initial velocity of the ball 1 = 6.20 m/s

$v_{2_{i}}$ : is the initial velocity of the ball 2 = 0 (it is at rest)

$v_{1_{f}}$ : is the final velocity of the ball 1 =?

$v_{2_{f}}$ : is the initial velocity of the ball 2 =?

$m_{1}v_{1_{i}} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

$v_{1_{f}} = v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}}$ (1)

Now, by conservation of kinetic energy (since they collide elastically):

$\frac{1}{2}m_{1}v_{1_{i}}^{2} = \frac{1}{2}m_{1}v_{1_{f}}^{2} + \frac{1}{2}m_{2}v_{2_{f}}^{2}$

$m_{1}v_{1_{i}}^{2} = m_{1}v_{1_{f}}^{2} + m_{2}v_{2_{f}}^{2}$ (2)

By entering equation (1) into (2) we have:

$m_{1}v_{1_{i}}^{2} = m_{1}(v_{1_{i}} - \frac{m_{2}v_{2_{f}}}{m_{1}})^{2} + m_{2}v_{2_{f}}^{2}$

$0.1 kg*(6.20 m/s)^{2} = 0.1 kg*(6.2 m/s - \frac{0.3 kg*v_{2_{f}}}{0.1 kg})^{2} + 0.3 kg(v_{2_{f}})^{2}$

By solving the above equation for $v_{2_{f}}$ :

$v_{2_{f}} = 3.1 m/s$

Now, $v_{1_{f}}$ can be calculated with equation (1):

$v_{1_{f}} = 6.20 m/s - \frac{0.3 kg*3.1 m/s}{0.1 kg} = -3.1 m/s$

The minus sign of $v_{1_{f}}$ means that the ball 1 (100g) is moving in the negative x-direction after the collision.

Therefore, the magnitude of the velocities of the two balls after the collision is 3.1 m/s (each one).

I hope it helps you!

5 0

3 years ago

a plane travels 2.5 km at an angle of 35 degrees to the ground and then changes Direction and travels 5.2 km and an angle of 22

Bess [88]

We can solve for the resultant x and y components by using the sine and cosine functions.

resultant x = 2.5 cos 35 + 5.2 cos 22 = 6.87 km

resultant y = 2.5 sin 35 + 5.2 sin 22 = 3.38 km

The resultant displacement is calculated using hypotenuse equation:

displacement = sqrt (6.87^2 + 3.38^2)

displacement = 7.66 km

The resultant angle is:

θ = tan^-1 (3.38 / 6.87)

θ = 26.20°

Therefore the magnitude and direction is:

7.66 km, 26.20° to the ground

3 0

4 years ago

A projectile proton with a speed of 500 m/s collides elastically with a target proton initially at rest. the two protons thenmov

Kay [80]

Because the two paths are perpendicular, therefore the target proton's new path must be at 30 degrees from the original direction.

Using the law of conservation of momentum about the original direction:
m (400 m/s) = m (v1) cos(60) + m (v2) cos(30)
Cancelling m since the two protons have similar mass.
(v1)cos(60) + (v2)cos(30) = 500 m/s                         ---> 1

Now by using the law conservation of momentum perpendicular to the original direction:
m (0 m/s) = m (v1) sin(60) – m (v2) sin(30)
Which simplifies to:
(v1)sin(60) - (v2)sin(30) = 0 m/s
v2 = v1 * sin(60) / sin(30) = v1 * sqrt(3)                 ---> 2

Plugging equation 2 to equation 1:
(v1) (1/2) + (v1 * sqrt(3)) sqrt(3)/2 = 500 m/s
(1/2) (v1) + (3/2) (v1) = 500 m/s
2 (v1) = 500 m/s
v1 = 250 m/s

Thus, from equation 2:

v2 = v1*sqrt(3) = (250 m/s) sqrt(3) = 433.01 m/s

So,
A. The target proton's speed is about 433 m/s
B. The projectile proton's speed is 250 m/s

8 0

4 years ago

Why is charcoal black?

WARRIOR [948]

The answer is C.

I just took the test!

Hope this helps. :)

8 0

3 years ago

One person is in a pool and is diving to a depth of 2.3m, and another person is diving to a depth of 3m. What pressure does each

Vlad [161]

Explanation:

The static pressure is P = ρgh, where ρ is the density of the fluid and h is the depth.

For the first person:

P = (1000 kg/m³) (9.8 m/s²) (2.3 m)

P = 22,500 Pa

For the second person:

P = (1000 kg/m³) (9.8 m/s²) (3 m)

P = 29,400 Pa

6 0

3 years ago

Example of nuclear fission​

Example of nuclear fission