Lecture 1 (Examples & tests) Dr. Rabab Bahabry

Physics 110 Measurement Problems + (TB)

Lecture 1 (Examples & tests)

Dr. Rabab Bahabry

Problems (Chapter1)

r= 6.3710⁶ m.

Conversion factor (1m1km)= 1m

1m = 10⁻³km 1m r= 6.3710⁶ m 10⁻³km

1m = 6.37 10³ km

Circumference of earth(s)=2πr=2 3.14 6.37×103km =4.0×104km Area(A)=4πr2=4 3.14 6.37 ×103 km 2

=5.10×108km2

(a) 1Km=?m

Conversion factor (1km1m)=^1𝐾𝑚

1𝐾𝑚 = ^{103×106𝜇𝑚}

1𝑘𝑚 = ^109𝜇𝑚

1𝑘𝑚

1km=10⁹m.

(b) We Know that 1m=10^-6m

Conversion factor(1m1cm)=^1𝑚

1𝑚 = ^102𝑐𝑚

1𝑚

1m=10^-6m ^102𝑐𝑚

1𝑚 = 10^-4 cm

To convert the density ρ unit into Kg/m³ Conversion factor (1g1kg)=^1𝑔

1𝑔 = ^{10−3𝑘𝑔}

1𝑔

Conversion factor ( ¹

𝑐𝑚 ¹

𝑚)=^1𝑐𝑚

1𝑐𝑚 = ^1𝑐𝑚

10⁻²𝑚

ρ=7.87g/cm³ ^{10−3𝑘𝑔}

1𝑔  ^1𝑐𝑚

10⁻²𝑚

3= 7.87kg/cm³ ^1𝑐𝑚3

10⁻⁶𝑚³=7.87kg ¹⁰⁻³

10⁻⁶𝑚³ = 7.87kg  ¹

10⁻³𝑚³= 7.87  10³kg/𝑚³

If we ignore the empty spaces between the close-packed spheres

The density of the individual iron atom will be the same as the same as the density of an iron atom

𝜌 = 𝑚

𝑤ℎ𝑒𝑟𝑒 𝑚 𝑖𝑠 𝑡ℎ𝑒 𝑚𝑎𝑠𝑠 𝑎𝑛𝑑 𝑉 𝑖𝑠 𝑡ℎ𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑎𝑛 𝑎𝑡𝑜𝑚𝑉

𝑉 = 𝑚

𝜌 = 9.27 × 10⁻²⁶𝑘𝑔 7.870 × 10³ 𝑘𝑔

𝑚³

= 1.18 × 10⁻²⁹𝑚³

1- We can write the speed of light (c = 299,000,000 m/s) using the scientific notation as:

(a) 2.99 x 10⁸ (b) 29.9 x 10⁸ (c) 0.299 x 10⁸ (d) 299 x 10⁸ Note:

(,) comma used for every 3 decimal places

(.) Decimal point a represent units in a decimal fraction

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2- A car moving with a speed of 100 km/h, what is its speed in m/s?

(a) 27.8 m/s (b) 16.7 m/s (c) 277.8 m/s (d) 167.7 m/s Answer: Step#1 conversation factor

conversation factor (1km1 m)= ¹km

1km = 10³m

1km conversation factor ( 1

1h → 1

1s)= 1h

1h = 1h 3600s

Step#2

Result= Origin*Conversion factors= 100 km

h  10³m

1km  ¹h

3600s= 27.77m/s Correct answer is (a) 27.8 m/s.

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3. We can express the very small number ( 0.000 000 004 56 ) using the scientific notation as:

(a) 4.56 x 10^-8 (b) 4.56 x 10^-9 (c) 4.56 x 10^-10 (d) 4.56 x 10^-11 _________________________________________

4. The conversion factor to convert 3 min to seconds is (a) 3600s

3 min (b) 60s

3min (c) 3600s

1min (d) 60s 1min

Answer: conversion factor (1min1s) conversion factor= 1min

1min = 60s 1min The correct answer is (d) 60s

1min

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5. Which of the following is not a base quantity ?

(a) speed (b) mass (c) length (d) time _________________________________________

6. How many centimeters in 1 km?

(a) 10⁵ cm (b) 10² cm (c) 10 cm (d) 10⁴ cm

Answer:

Conversion factor is required to convert (1km1cm) Step#1Conversion factor(1km1m)=1km

1km= 10³m 1km 1Km 10³m

1km = 10³m

Step #2 (10³m 1cm)= 10³m 1m

1m= 10³m 10²cm

1m = 10⁵cm.

The correct answer is (a) 10⁵ cm

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7. The conversion factor to convert hours to seconds is:

(a) 1s

3600h (b) 3600h

1 s (c) 1 h

3600s (d) 3600s 1 h

Answer:

conversion factor (1 h1s)= 1h

1h = 3600s 1h Correct answer is (d) 3600s

1 h .

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8. (1 m = 3.281 ft) then 1.5 ft/h equals:

(a) 1.37 x 10^-3 m/s (b) 1.27 x 10^-4 m/s (c) 1645.8 m/s (d) 17717.4 m/s Answer:

conversion factor (1ft1m)= 1ft

1ft = (1/3.281)m

1ft = 1m 3.281ft conversion factor ( 1

1h → 1

1s)= 1h

1h = 1h 3600s

1.5 ft/h =1.5 ft/h  1m

3.281ft  1h

3600s= 1.2710^-4 m Correct answer is (b) 1.27 x 10^-4 m/s

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9. A square with an edge of 1 cm has an area of: ( area = edge² ) (a) 10² m² (b) 10⁴ m² (c) 10^-4 m² (d) 10^-6 m² Converstion factor (1cm1m)=1cm

1cm = 10⁻²m 1cm Area=(1cm)²=(1cm 10⁻²m

1cm )² Area=110^-4m²

Correct answer is (c) 10^-4 m²

10. 10³ gigawatts is:

(a) 10¹² watts (b) 10⁹ watts (c) 10^-6 watts (d) 10^-3 watts Conversion factor (1GW1W)=1GW

1GW = 10⁹W 1GW Then10³ GW=10³ GW 10⁹ W

1 GW=10¹² W.

Correct answer is (a) 10¹² watts

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11. The conversion factor to convert 10 kg to g is:

(a)10³g

1 Kg (b) 10³g

10 Kg (c) 1 kg

10³ g (d) 10 kg 10³ g conversion factor (1Kgg)= 1 Kg

1 Kg= 10³g 1 Kg Then the correct answer is (a)

12. Which prefix is true?

(a) milli = 10³ (b) micro = 10^-9 (c) mega = 10⁶ (d) pico = 10⁹

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13. 1 mm² =

(a) 10^{- 3} m² (b) 10^-6 m² (c) 10 ^{- 9} m² (d) 10^{- 12} m² Conversion factor (1mm1m)=^1𝑚𝑚

1𝑚𝑚 = ^10−3𝑚

1𝑚𝑚 . Then 1mm²= 1mm 10−3m

1mm

2

=10^-6 m^2.

The correct answer is (b).

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14. If the length, height, and width of a rectangular block are 3 cm, 4 cm, and 5 cm respectively, then the volume is

(a) 60 m³ (b) 60 cm³ (c) 60 m (d) 60 cm

Physics 110

Motion Along a Straight Line

Lecture 2- 1/28/2018 Dr. Rabab Bahabry

Dr. Arwa Kutbee

Motion

Motion : is the change in position of an object with respect to a frame of reference 1-Dimensional motion

(1D)

2-Dimensional motion (2D)

3-Dimensional motion (3D)

Definition Motion only in one out of the three coordinates

Motion of an object in two out of the three coordinates

Motion of the object when all the three

example Motion of car, free fall Projectile motion

(𝑥₁, 𝑦₁)

(𝑥₂, 𝑦₂)

𝑦₁

𝑦₂ 𝑥₁ 𝑥₂

(𝑥₂, 𝑦₂, 𝑧₂) (𝑥₁, 𝑦₁, 𝑧₁)

Watch https://www.youtube.com/watch?v=X6_wXmjxUdM

The position (vector quantity)

Position sign in 1-D Horizontal motion Vertical motion Positive The object is located to the

Right of the origin

The object is located above the origin

Negative The object is located to the Left of the origin

The object is located below the origin

• In order to locate an object, we need a reference point.

• Reference point or origin: is the “zero” point from which something is measured

• Position: is the location of an object with respect to a chosen reference point that that we can consider to be the (origin) of a coordinate system

Examples on position

• The position is a vector quantity: has magnitude and direction.

• Choosing the stop sign as the origin denoted as x=0 m

Position meaning of the sign

(Object location to the origin) X= 3m +ve direction from the origin X=-3m –ve direction from the origin

+ sign don’t need to be shown – sign must always to be shown

Displacement (vector quantity)

• The displacement ∆x of a particle is defined as its change in position in some time interval.

• As the particle moves from an initial position x_i to a final position x_f , its displacement is given by

∆𝒙 = 𝒙_𝒇 − 𝒙_𝒊 or ∆𝒙 = 𝒙_𝟐 − 𝒙_𝟏

• We use the capital Greek letter delta (∆) to denote the change in a quantity. From this definition, we see that

The displacement is a vector quantity: has magnitude and direction.Displacement meaning of the sign (Direction of motion )

∆X= + Object is moving to the right

∆X= - Object is moving to the left

𝐱₁

initial position

x₂

final position

𝜟𝒙 = 𝒙_𝟐 − 𝒙_𝟏 Displacement

Direction of motion

-3m (-)

5m (+)

Δx = 5 − −3 = +8 m (+)

Positive direction Object is moving to the right

-3m (-)

-7m (-)

Δx = −7 − −3 = −4 m (-)

negative direction Object is moving to the left

𝐱₁

initial position

x₂

final position

𝜟𝒙 = 𝒙_𝟐 − 𝒙_𝟏 Displacement

Direction of motion

7m (+)

-3m (-)

Δx = −3 − 7 = −10m (-)

negative direction

Object is moving to the left