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What nursing action will significantly decrease the risk of a serious complication for a child in Bryant's traction?

A) Ensuring the weights hang freely off the end of the bed.
B) Increasing the weight used in the traction system.
C) Securing the traction strips tightly to the skin.
D) Allowing the child to freely move around the bed.

User Eyalsh
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1 Answer

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Final answer:

In physics, the child and basket's mass is 5.61 kg when the scale reads 55 N. The tension T₁ is 55 N, and T₂, which includes the scale's mass, is about 59.91 N.

Step-by-step explanation:

To resolve this problem, we'll use the principle that the tension in a rope holding a stationary object (in this case, the child and the basket) is equal to the weight of the object due to gravity. This is derived from Newton's second law of motion, which states that the net force on the object is zero if it is not accelerating.

Part (a): Mass of the child and basket

The scale reading is 55 N, which is the force due to gravity on the child and basket. Since weight (W) equals mass (m) times the acceleration due to gravity (g), which is approximately 9.81 m/s2 on Earth, we can calculate the mass.

W = m × g
m = W / g = 55 N / 9.81 m/s2 = 5.61 kg
Part (b): Tension T₁

The tension T₁ in the cord attaching the baby to the scale is equal to the weight of the child and basket, so T₁ = 55 N.Part (c): Tension T₂

To find T₂, we add the weight of the scale to T₁. The scale has a mass of 0.500 kg, so its weight is 0.500 kg × 9.81 m/s2 = 4.905 N. Thus, T₂ = T₁ + weight of the scale = 55 N + 4.905 N = 59.905 N, which we can approximately round to 59.91 N.

Finally, a sketch could be drawn showing the baby and basket at the bottom, the scale above it, followed by the cord T₁, then the scale, and above that the cord T₂ extending to the ceiling with arrows representing the direction and magnitude of the forces.