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CHAPTER 1
MEASUREMENT AND PROBLEM SOLVING
Remind students that their answers to odd-numbered exercises may be slightly different from those given here
because of rounding. Refer to Problem-Solving Hint: The "Correct" Answer in Chapter 1.
1.

(c).

2.

(b).

3.

(b).

4.

(c).

5.

Because there are no more fundamental quantities and all other quantities can be derived fromthe fundamental
ones.

6.

Weight changes depending on gravity of locations .

7.

Mean solar day replaced the original second definition. No, atomic clocks are now used .

8.

One major difference is decimal vs. duodecimal bases .
Another difference is mks (meter, kilogram, second) vs. fps (foot, pounds, second) .

9.

(b).

10.

(b).

11.

(a).

12.

(c).

13.Since 1 in. = 2.54 cm, 3 cm = (3 cm) ×

14.

This is by definition. 1 L = 1000 mL and 1 L = 1000 cm3 .

1 in.
= 1.2 in. That would have been a huge lady bug. No .
2.54 cm
2.2 lb.
Since 1 kg is equivalent to 2.2 lb., 10 kg = (10 kg) ×
=22 lb. Yes , salmons are quite large.
1 kg

© 2007 Pearson Prentice Hall, Upper Saddle River, NJ. All rights reserved. This material is protected underall copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1

Chapter 1

15.

Measurement and Problem Solving

Metric ton is defined as the mass of 1 m3 of water. 1 m3 = 1000 L, and 1 L of water has a mass of 1 kg. So one
metric ton is equivalent to 1000 kg.

16.

Thedecimal system (base 10) has a dime worth 10¢ and a dollar worth 10 dimes, or 100¢. By analogy, a
duodecimal system would have a dime worth 12¢ and a dollar worth 12 “dimes,” or $1.44 in decimal dollars. Then
1
a penny would be 144 of a dollar.

17.

(a) Different ounces are used for volume and weight measurements. 16 oz = 1 pt is a volume measure and 16 oz = 1
lb is a weight measure.
(b)Two different pound units are used. Avoirdupois lb = 16 oz‚ troy lb = 12 oz.

18 .

That is because 1 nautical mile = 6076 ft = 1.15 mi. A nautical mile is larger than a (statute) mile.

19.

(d).

20.

(d).

21.

(a).

22 .

No , it only tells if the equation is dimensionally correct.

23.

No , it cannot tell if an equation is correct. Unit analysis can only tell if it isdimensionally correct.

24.

By putting in units and solving for those of the unknown quantity .

25.

(Length) = (Length) +

26.

(d).

27.

(Length)
× (Time) = (Length) + (Length) .
(Time)

m2 = (m)2 = m2.

28 .

29.

Yes , since [m3] = [m]3 = [m3].

No . V = 4π r /3 = 4π (8r )/24 = 4π (2r) /24 = π d /6. So it should be V = π d 3/6 .
3

3

3

3

© 2007Pearson Prentice Hall, Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they
currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

2

Chapter 1
1
2

mv2, the unit of K, joule, is equal to kg(m/s)2 = kg⋅m2/s2 .

30.

Since K =

31.

Since ax2 is inmeters, a =
Since bx is in meters, b =

32 .

Measurement and Problem Solving

m
2 = 1/m .
m
m
= dimensionless .
m

c is in m .

Since p = ρv2, the unit of pressure is (kg/m3)(m/s)2 = kg/(m⋅s2).
No , this does not prove that this relationship is physically correct, because there might be a coefficient in the
equation.

33.

m
3
Since ρ = V , the unit of mass is kg, and theunit of volume is m ; the unit of density is equal to
kg
3
3 = kg/m .
m

34.

Yes , because m2 = 1 m(m + m) = m2 + m2.
2

35.

The first student , because m/s =

36 .

m /s2 = m/s.

(a) Since F = ma, newton = (kg)(m/s2) = kg·m/s2 .
(b) Yes , because (kg) ×

37.

2

(m/s2)(m) =

m2/s2
2
2
m = kg·m/s (F = mv /r).

(a) The unit of angular momentum is (kg)(m/s)(m) =...
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