FlipIt for College Physics (Algebra Version - Six Months Access)
FlipIt for College Physics (Algebra Version - Six Months Access)
17th Edition
ISBN: 9781319032432
Author: Todd Ruskell
Publisher: W.H. Freeman & Co
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Chapter 1, Problem 32QAP
To determine

(A)

The value of 238 feet into meter.

Expert Solution
Check Mark

Answer to Problem 32QAP

The numerical value for 238 ft is  72.5424 m

Explanation of Solution

Given data:

The given value is 238 ft

Formula used:

For numerical value convert ft into m,
  (ft×0.3048)m

Calculation:

Conversion of ft into m,
  238×0.3048 m72.5424 m

Conclusion:

Thus, numerical value for 238 ft is  72.5424 m

To determine

(B)

The value of 772 in into cm.

Expert Solution
Check Mark

Answer to Problem 32QAP

The numerical value for 772 in is  1960.88 cm

Explanation of Solution

Given data:

The given value is 772 in

Formula used:

For numerical value convert in into cm,
  (in×2.54)cm

Calculation:

Conversion of in into cm,
  772×2.54 cm1960.88 cm

Conclusion:

Thus, numerical value for 772 in is 196 0.88 cm

To determine

(C)

The value of 1220 in2 into cm2.

Expert Solution
Check Mark

Answer to Problem 32QAP

The numerical value for 1220 in2 is  7870.952 cm2

Explanation of Solution

Given data:

The given value is 1220 in2

Formula used:

For numerical value convert  in2 into cm2,
  (in2×2.542) cm2

Calculation:

Conversion of  in2 into cm2,
  1220×2.542 cm27870.952 cm2

Conclusion:

Thus, numerical value for 1220 in2 is  7870.952 cm2

To determine

(D)

The value of 559 oz into L.

Expert Solution
Check Mark

Answer to Problem 32QAP

The numerical value for 559 oz is 16.5316 L

Explanation of Solution

Given data:

The given value is 559 oz.

Formula used:

For numerical value convert oz into L,
  (oz×0.02957) L

Calculation:

Conversion of oz into L,
  559×0.02957 L16.5316 L

Conclusion:

Thus, numerical value for 559 oz is 16.5316 L

To determine

(E)

The value of  973lb to g

Expert Solution
Check Mark

Answer to Problem 32QAP

The numerical value for  973lb is  441345g.

Explanation of Solution

Given data:

The given value is  973lb

Formula used:

For numerical value convert  lb into  g,
  1lb=453.592g

Calculation:

Conversion of  973lb into g,
  973×453.592441345g

Conclusion:

Thus, numerical value for  973lb is  441345g.

To determine

(f)

The value of  122 ft3 in  m3.

Expert Solution
Check Mark

Answer to Problem 32QAP

The numerical value for  122 ft3 is  3.45m3

Explanation of Solution

Given data:

The given value is  122 ft3

Formula used:

For numerical value convert  ft3 into  m3,
  1ft3=0.0283m3

Calculation:

Conversion of  ft3 into  m3,
  122×0.02833.45m3

Conclusion:

The numerical value for  122 ft3 is  3.45m3

To determine

(g)

The value of  1.28mi2 in  km2.

Expert Solution
Check Mark

Answer to Problem 32QAP

The numerical value for  1.28mi2 is 3.315km2

Explanation of Solution

Given data:

The given value is  1.28mi2

Formula used:

For numerical value convert mi2 into km2,
  1mi2=2.5899km2

Calculation:

Conversion of mi2 into km2,
  1.28×2.58993.3155km2

Conclusion:

The numerical value for  1.28mi2 is 3.315km2

To determine

(h)

The value of  442in3 into  cm3.

Expert Solution
Check Mark

Answer to Problem 32QAP

The numerical value for  442in3 is  7243.08 cm3

Explanation of Solution

Given data:

The given value is  442in3

Formula used:

For numerical value convert  in3 into cm3,
  1in3=16.3871cm3

Calculation:

Conversion of  in3 into cm3,
  442×16.3817243.08cm3

Conclusion:

The numerical value for  442in3 is  7243.08 cm3

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************* ********************************* Q.1) Classify the following statements as a true or false statements: a. If M is a module, then every proper submodule of M is contained in a maximal submodule of M. b. The sum of a finite family of small submodules of a module M is small in M. c. Zz is directly indecomposable. d. An epimorphism a: M→ N is called solit iff Ker(a) is a direct summand in M. e. The Z-module has two composition series. Z 6Z f. Zz does not have a composition series. g. Any finitely generated module is a free module. h. If O→A MW→ 0 is short exact sequence then f is epimorphism. i. If f is a homomorphism then f-1 is also a homomorphism. Maximal C≤A if and only if is simple. Sup Q.4) Give an example and explain your claim in each case: Monomorphism not split. b) A finite free module. c) Semisimple module. d) A small submodule A of a module N and a homomorphism op: MN, but (A) is not small in M.
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