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masya89 [10]
3 years ago
7

HELP PLZ QUICK HELP PLZ QUICK HELP PLZ QUICK HELP PLZ QUICK

Physics
2 answers:
BlackZzzverrR [31]3 years ago
7 0

B. Subtract the smallest value from the largest value

Explanation:

The range of a data set is determined by subtracting the smallest value from the largest one.

 Range = Highest value - lowest value.

  • The range gives us the spread of a data point.

To find the range;

  • Sort the given data in ascending order.
  • Record the highest number
  • Record the lowest number
  • Subtract the lowest number from the highest one.

The answer gives the range of data point.

Learn more:

Range brainly.com/question/1466393

#learnwithBrainly

solniwko [45]3 years ago
6 0

Answer:B

Explanation:example 12, 14, 17, 21, and 24. 24-12=12. So 12 is your range.

plz mark branliest

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Answer: For 18 the inch is longer im not sure by how much, for 19 the the kilogram is 2.2 times heavier than a pound, QUART IS A LITTLE LESS THAN A LITER. ... An easy way to figure from liters to gallons, for example, is that a quart is a little less than a liter and 4 liters is a little more than 1 gallon. To be exact, 1 liter is 0.264 gallon (a little more than a quart), and 4 liters is 1.06 gallons,  30°C or 30°F? Correct answer: the table shows that 30°C is about 86°F, which is warmer than 30°F.   And sorry Im not sure how to do the rest

Explanation:

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A bar magnet is cut in half, as shown.
Feliz [49]

Answer:

NS/NS

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Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. Take T = 1.3 kN·m. De
My name is Ann [436]

Answer:

τ (bc. max) =25.37 MPa

Explanation:

From the question, T = 1.3Kn.m;

(G = 77.2 GPa) and from the image of this solid shaft system i attached;

d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m

So ΣT = 0 → Ta + Tc = 1.3Kn.m

So the system is statically indeterminate.

Let's check at the equation that makes it compatible ;

ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0

[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =

ΣT = 0 and T(bc) = T(c)

ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m

Now,

[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0

So, T(c) = 273374 Nmm = 273.374Nm

T(a) = 1300Nm - 273.374Nm = 1026. 63Nm

From the beginning we saw that;

T(bc) = T(c) = 273.374Nm

Now let's find the maximum shear stress in shaft BC;

τ (bc. max) = {τ (bc) x r(bc)} / J(bc)

τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707

= 25.37 MPa

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