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Morgarella [4.7K]
3 years ago
7

Why is evidence good for practice

Physics
2 answers:
posledela3 years ago
8 0

Answer:

HEYO BRO! (my bad for da caps)

Explanation:

Evidence-based practice (EBP) is the idea that occupational practices ought to be based on scientific evidence. That at first sight may seem to be obviously desirable, but the proposal has been controversial.

Happy to Help From, Adam

attashe74 [19]3 years ago
8 0
Bc U know what ur doing...?
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A disk with mass m = 9.5 kg and radius r = 0.3 m begins at rest and accelerates uniformly for t = 18.1 s, to a final angular spe
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Ahhh this going to be confusing sorry...
1. α = Δω / Δt = 28 rad/s / 19s = 1.47 rad/s²

2. Θ = ½αt² = ½ * 1.47rad/s² * (19s)² = 266 rads

3. I = ½mr² = ½ * 8.7kg * (0.33m)² = 0.47 kg·m²

4. ΔEk = ½Iω² = ½ * 0.47kg·m² * (28rad/s)² = 186 J

5. a = α r = 1.47rad/s² * 0.33m = 0.49 m/s²

6. a = ω² r = (14rad/s)² * 0.33m = 65 m/s²

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8 0
3 years ago
What is the tangential velocity of a record player which makes 11 revolutions in 20 seconds? Please help
Deffense [45]

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Explanation:

4 0
3 years ago
Question 5(Multiple Choice Worth 3 points) (05.01 LC) Complete the following statement. The universe is filled with millions of
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the answer is a: galaxies. galaxy clusters are groups of galaxies that like to hang out together out there in space. the milky way's own galaxy cluster is called the local group.

8 0
3 years ago
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Ten identical steel wires have equal lengths L and equal "spring constants" k. The wires are connected end to end so that the re
lapo4ka [179]

Answer:

K_{system} = \frac{k}{10}

Explanation:

When the springs are connected end to end, it means they are connected in series. When the springs are connected in series, the stress applied to the system gets applied to each of the springs without any change in magnitude while the strain of the system is the sum total of strains of each spring. The spring constant of the resultant system is given as,

\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{n}})

Here, n = 10

Spring constant of each spring = k

Thus,

\frac{1}{K_{system}} = (\frac{1}{K_{1}})+(\frac{1}{K_{2}})+(\frac{1}{K_{3}})+ (\frac{1}{K_{4}})+.....+(\frac{1}{K_{10}})

\frac{1}{K_{system}} = (\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})+(\frac{1}{k})

\frac{1}{K_{system}} = \frac{10}{k}

K_{system} = \frac{k}{10}

7 0
3 years ago
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