Maximum scatter radiation to the operator occurs when

When the x-ray tube is above the patient

Explanation:

Maximum scatter radiation to the operator occurs when the x-ray tube is above the patient.

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3.3 s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 12 m/s

Height of tower (h₁) = 15 m

Total time (T) to reach the ground =?

Next, we shall determine the time taken to reach the maximum height from the tower. This can be obtained as follow:

Initial velocity (u) = 12 m/s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Time (t₁) to reach the maximum height from tower =?

v = u – gt₁ (ball is going against gravity)

0 = 12 – 10t₁

Rearrange

10t₁ = 12

Divide both side by 10

t₁ = 12/ 10

t₁ = 1.2 s

Next, we shall determine the maximum height reached by the ball from the tower.

Initial velocity (u) = 12 m/s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 10 m/s²

Maximum height from the tower (h₂) =?

v² = u² – 2gh₂

0² = 12² – 2 × 10 × h₂

0 = 144 – 20h₂

Rearrange

20h₂ = 144

Divide both side by 20

h₂ = 144 / 20

h₂ = 7.2 m

Next, we shall determine the maximum height from the ground. This can be obtained as follow:

Height of tower (h₁) = 15 m

Maximum height from the tower (h₂) = 7.2 m

Maximum height from the ground (H) =?

H = h₁ + h₂

H = 15 + 7.2

H = 22.2 m

Next, we shall determine the time taken for ball to get to the ground from the maximum height. This can be obtained as follow:

Acceleration due to gravity (g) = 10 m/s²

Maximum height from the ground (H) = 22.2 m

Time (t₂) taken for ball to get to the ground from the maximum height =?

H = ½gt₂²

22.2 = ½ × 10 × t₂²

22.2 = 5 × t₂²

Divide both side by 5

t₂² = 22.2 / 5

t₂² = 4.44

Take the square root of both side

t₂ = √4.44

t₂ = 2.1 s

Finally, we shall determine the time taken for the ball to get to the ground. This can be obtained as follow:

Time (t₁) to reach the maximum height from tower = 1.2 s

Time (t₂) taken for ball to get to the ground from the maximum height = 2.1 s

Time (T) taken for the ball to get to the ground =?

T = t₁ + t₂

T = 1.2 + 2.1

T = 3.3 s

Therefore, it will take the ball 3.3 s to get to the ground.