QuestionmULTIPLE CHOIGE
1. An object moves in a circle at a constant speed. Which of the following is not true of the object? A. Its centripetal acceleration points toward the center of the circle. B. Its tangential speed is constant. C. Its velocity is constant. D. A centripetal force acts on the object.
Use the passage below to answer questions 2-3. A car traveling at $15 \mathrm{~m} / \mathrm{s}$ on a flat surface turns in a circle with a radius of 25 m .
2. What is the centripetal acceleration of the car? F. $2.4 \times 10^{-2} \mathrm{~m} / \mathrm{s}^{2}$ G. $0.60 \mathrm{~m} / \mathrm{s}^{2}$ H. $9.0 \mathrm{~m} / \mathrm{s}^{2}$ J. zero
3. What is the most direct cause of the car's centripetal acceleration? A. the torque on the steering wheel B. the torque on the tires of the car C. the force of friction between the tires and the road D. the normal force between the tires and the road
4. Earth ( $m=5.97 \times 10^{24} \mathrm{~kg}$ ) orbits the sun ( $m=1.99 \times$ $10^{30} \mathrm{~kg}$ ) at a mean distance of $1.50 \times 10^{11} \mathrm{~m}$ . What is the gravitational force of the sun on Earth? ( $G=$ $6.673 \times 10^{-11} \mathrm{~N}^{-\mathrm{m}^{2}} / \mathrm{kg}^{2}$ ) F. $5.29 \times 10^{32} \mathrm{~N}$ G. $3.52 \times 10^{22} \mathrm{~N}$ H. $5.90 \times 10^{-2} \mathrm{~N}$ J. $1.77 \times 10^{-8} \mathrm{~N}$
5. Which of the following is a correct interpretation of the expression $a_{s}=g=G \frac{m_{E_{2}}}{r^{2}}$ ? A. Gravitational field strength changes with an object's distance from Earth. B. Free-fall acceleration changes with an object's distance from Earth. C. Free-fall acceleration is independent of the falling object's mass. D. All of the above are correct interpretations.
6. What data do you need to calculate the orbital speed of a satellite? F. mass of satellite, mass of planet, radius of orbit G. mass of satellite, radius of planet, area of orbit H. mass of satellite and radius of orbit only J. mass of planet and radius of orbit only
7. Which of the following choices correctly describes the orbital relationship between Earth and the sun? A. The sun orbits Earth in a perfect circle. B. Earth orbits the sun in a perfect circle. C. The sun orbits Earth in an ellipse, with Earth at one focus. D. Earth orbits the sun in an ellipse, with the sun at one focus.
Use the diagram below to answer questions 8-9.
8. The three forces acting on the wheel above have equal magnitudes. Which force will produce the greatest torque on the wheel? F. $\mathrm{F}_{1}$ G. $\mathrm{F}_{2}$ H. $\mathrm{F}_{3}$ J. Each force will produce the same torque. $8 \quad$ Chapter 7

Studdy Solution

STEP 1

What is this asking? We've got a mix of questions here, from circular motion and forces to gravity and orbits!
Let's tackle them one by one. Watch out! Remember the difference between velocity and speed: velocity has direction, speed doesn't.
Also, gravity depends on both masses and the distance between them, so keep track of those values.

STEP 2

1. Multiple Choice Conceptual Questions
2. Centripetal Acceleration Calculation
3. Cause of Centripetal Acceleration
4. Gravitational Force Calculation
5. Gravitational Field Strength Interpretation
6. Data for Orbital Speed
7. Earth-Sun Orbital Relationship
8. Torque on a Wheel

STEP 3

For question 1, an object moving in a circle at a constant speed has a constantly changing direction of its velocity.
Since velocity is a vector (magnitude and direction), its velocity isn't constant!

STEP 4

The answer to question 1 is C.

STEP 5

For question 2, the formula for centripetal acceleration is $a_c = \frac{v^2}{r}$ , where $v$ is the tangential speed ( $15 \text{ m/s}$ ) and $r$ is the radius ( $25 \text{ m}$ ).

STEP 6

Let's plug in the values: $a_c = \frac{(15 \text{ m/s})^2}{25 \text{ m}} = \frac{225 \text{ m}^2/\text{s}^2}{25 \text{ m}}$ .

STEP 7

Dividing gives us $a_c = 9 \text{ m/s}^2$ .

STEP 8

For question 3, the centripetal force, which causes the centripetal acceleration, is provided by the friction between the tires and the road.
This friction keeps the car moving in a circle.

STEP 9

For question 4, we'll use Newton's Law of Universal Gravitation: $F = G\frac{m_1 m_2}{r^2}$ .
Here, $G = 6.673 \times 10^{-11} \text{ N} \cdot \text{m}^2/\text{kg}^2$ , $m_1$ is the mass of the Sun ( $1.99 \times 10^{30} \text{ kg}$ ), $m_2$ is the mass of Earth ( $5.97 \times 10^{24} \text{ kg}$ ), and $r$ is the distance between them ( $1.50 \times 10^{11} \text{ m}$ ).

STEP 10

Plugging in the values: $F = (6.673 \times 10^{-11})\frac{(1.99 \times 10^{30})(5.97 \times 10^{24})}{(1.50 \times 10^{11})^2}$ .

STEP 11

Calculating the numerator: $(6.673 \times 10^{-11})(1.99 \times 10^{30})(5.97 \times 10^{24}) = 7.926 \times 10^{44}$ .

STEP 12

Calculating the denominator: $(1.50 \times 10^{11})^2 = 2.25 \times 10^{22}$ .

STEP 13

Dividing gives us $F = \frac{7.926 \times 10^{44}}{2.25 \times 10^{22}} = 3.52 \times 10^{22} \text{ N}$ .

STEP 14

For question 5, the given expression shows that both gravitational field strength and free-fall acceleration depend on distance from Earth and are independent of the falling object's mass.

STEP 15

For question 6, to calculate the orbital speed of a satellite, you need the mass of the planet and the radius of the orbit.

STEP 16

For question 7, Earth orbits the Sun in an ellipse, with the Sun at one focus.

STEP 17

For question 8, $F_1$ will produce the greatest torque because it has the largest perpendicular lever arm (distance from the pivot point to the line of action of the force).

STEP 18

1. C
2. H
3. C
4. G
5. D
6. J
7. D
8. F

Was this helpful?

Studdy Solution

STEP 1

STEP 2

1. Multiple Choice Conceptual Questions
2. Centripetal Acceleration Calculation
3. Cause of Centripetal Acceleration
4. Gravitational Force Calculation
5. Gravitational Field Strength Interpretation
6. Data for Orbital Speed
7. Earth-Sun Orbital Relationship
8. Torque on a Wheel

STEP 3

For question 1, an object moving in a circle at a constant speed has a constantly changing direction of its velocity.
Since velocity is a vector (magnitude and direction), its velocity isn't constant!

STEP 4

The answer to question 1 is C.

STEP 5

For question 2, the formula for centripetal acceleration is $a_c = \frac{v^2}{r}$ , where $v$ is the tangential speed ( $15 \text{ m/s}$ ) and $r$ is the radius ( $25 \text{ m}$ ).

STEP 6

Let's plug in the values: $a_c = \frac{(15 \text{ m/s})^2}{25 \text{ m}} = \frac{225 \text{ m}^2/\text{s}^2}{25 \text{ m}}$ .

STEP 7

Dividing gives us $a_c = 9 \text{ m/s}^2$ .

STEP 8

For question 3, the centripetal force, which causes the centripetal acceleration, is provided by the friction between the tires and the road.
This friction keeps the car moving in a circle.

STEP 9

STEP 10

Plugging in the values: $F = (6.673 \times 10^{-11})\frac{(1.99 \times 10^{30})(5.97 \times 10^{24})}{(1.50 \times 10^{11})^2}$ .

STEP 11

Calculating the numerator: $(6.673 \times 10^{-11})(1.99 \times 10^{30})(5.97 \times 10^{24}) = 7.926 \times 10^{44}$ .

STEP 12

Calculating the denominator: $(1.50 \times 10^{11})^2 = 2.25 \times 10^{22}$ .

STEP 13

Dividing gives us $F = \frac{7.926 \times 10^{44}}{2.25 \times 10^{22}} = 3.52 \times 10^{22} \text{ N}$ .

STEP 14

For question 5, the given expression shows that both gravitational field strength and free-fall acceleration depend on distance from Earth and are independent of the falling object's mass.

STEP 15

For question 6, to calculate the orbital speed of a satellite, you need the mass of the planet and the radius of the orbit.

STEP 16

For question 7, Earth orbits the Sun in an ellipse, with the Sun at one focus.

STEP 17

For question 8, $F_1$ will produce the greatest torque because it has the largest perpendicular lever arm (distance from the pivot point to the line of action of the force).

STEP 18

1. C
2. H
3. C
4. G
5. D
6. J
7. D
8. F

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Multiple Choice Conceptual Questions2. Centripetal Acceleration Calculation3. Cause of Centripetal Acceleration4. Gravitational Force Calculation5. Gravitational Field Strength Interpretation6. Data for Orbital Speed7. Earth-Sun Orbital Relationship8. Torque on a Wheel

STEP 3

For question 1, an object moving in a circle at a constant speed has a constantly changing *direction* of its velocity.Since velocity is a vector (magnitude and direction), its velocity isn't constant!

STEP 4

The answer to question 1 is C.

STEP 5

For question 2, the formula for centripetal acceleration is ac=v2ra_c = \frac{v^2}{r}ac​=rv2​, where vvv is the **tangential speed** (15 m/s15 \text{ m/s}15 m/s) and rrr is the **radius** (25 m25 \text{ m}25 m).

STEP 6

Let's plug in the values: ac=(15 m/s)225 m=225 m2/s225 ma_c = \frac{(15 \text{ m/s})^2}{25 \text{ m}} = \frac{225 \text{ m}^2/\text{s}^2}{25 \text{ m}}ac​=25 m(15 m/s)2​=25 m225 m2/s2​.

STEP 7

Dividing gives us ac=9 m/s2a_c = 9 \text{ m/s}^2ac​=9 m/s2.

STEP 8

For question 3, the centripetal force, which causes the centripetal acceleration, is provided by the *friction* between the tires and the road.This friction keeps the car moving in a circle.

STEP 9

STEP 10

Plugging in the values: F=(6.673×10−11)(1.99×1030)(5.97×1024)(1.50×1011)2F = (6.673 \times 10^{-11})\frac{(1.99 \times 10^{30})(5.97 \times 10^{24})}{(1.50 \times 10^{11})^2}F=(6.673×10−11)(1.50×1011)2(1.99×1030)(5.97×1024)​.

STEP 11

Calculating the numerator: (6.673×10−11)(1.99×1030)(5.97×1024)=7.926×1044(6.673 \times 10^{-11})(1.99 \times 10^{30})(5.97 \times 10^{24}) = 7.926 \times 10^{44}(6.673×10−11)(1.99×1030)(5.97×1024)=7.926×1044.

STEP 12

Calculating the denominator: (1.50×1011)2=2.25×1022(1.50 \times 10^{11})^2 = 2.25 \times 10^{22}(1.50×1011)2=2.25×1022.

STEP 13

Dividing gives us F=7.926×10442.25×1022=3.52×1022 NF = \frac{7.926 \times 10^{44}}{2.25 \times 10^{22}} = 3.52 \times 10^{22} \text{ N}F=2.25×10227.926×1044​=3.52×1022 N.

STEP 14

For question 5, the given expression shows that both gravitational field strength and free-fall acceleration depend on distance from Earth and are independent of the falling object's mass.

STEP 15

For question 6, to calculate the orbital speed of a satellite, you need the mass of the planet and the radius of the orbit.

STEP 16

For question 7, Earth orbits the Sun in an ellipse, with the Sun at one focus.

STEP 17

For question 8, F1F_1F1​ will produce the greatest torque because it has the largest perpendicular lever arm (distance from the pivot point to the line of action of the force).

STEP 18

1. C2. H3. C4. G5. D6. J7. D8. F

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Multiple Choice Conceptual Questions
2. Centripetal Acceleration Calculation
3. Cause of Centripetal Acceleration
4. Gravitational Force Calculation
5. Gravitational Field Strength Interpretation
6. Data for Orbital Speed
7. Earth-Sun Orbital Relationship
8. Torque on a Wheel

For question 1, an object moving in a circle at a constant speed has a constantly changing direction of its velocity.
Since velocity is a vector (magnitude and direction), its velocity isn't constant!

For question 2, the formula for centripetal acceleration is $a_c = \frac{v^2}{r}$ , where $v$ is the tangential speed ( $15 \text{ m/s}$ ) and $r$ is the radius ( $25 \text{ m}$ ).

Let's plug in the values: $a_c = \frac{(15 \text{ m/s})^2}{25 \text{ m}} = \frac{225 \text{ m}^2/\text{s}^2}{25 \text{ m}}$ .

Dividing gives us $a_c = 9 \text{ m/s}^2$ .

For question 3, the centripetal force, which causes the centripetal acceleration, is provided by the friction between the tires and the road.
This friction keeps the car moving in a circle.

Plugging in the values: $F = (6.673 \times 10^{-11})\frac{(1.99 \times 10^{30})(5.97 \times 10^{24})}{(1.50 \times 10^{11})^2}$ .

Calculating the numerator: $(6.673 \times 10^{-11})(1.99 \times 10^{30})(5.97 \times 10^{24}) = 7.926 \times 10^{44}$ .

Calculating the denominator: $(1.50 \times 10^{11})^2 = 2.25 \times 10^{22}$ .

Dividing gives us $F = \frac{7.926 \times 10^{44}}{2.25 \times 10^{22}} = 3.52 \times 10^{22} \text{ N}$ .

For question 8, $F_1$ will produce the greatest torque because it has the largest perpendicular lever arm (distance from the pivot point to the line of action of the force).

1. C
2. H
3. C
4. G
5. D
6. J
7. D
8. F