Question 1

What is the maximum distance we can shoot a dart,provided our toy dart gun gives a maximum initial velocity of 2.78 m/s?
A)0.79 m
B)1.58 m
C)1.39 m
D)More information needed

Accepted Answer

The maximum distance a dart can be shot is given by the equation:d = v^2 * sin(2 * theta) / gwhere:d is the distancev is the initial velocitytheta is the angle of projectiong is the acceleration due to gravityIn this case, the initial velocity is 2.78 m/s and the angle of projection is 45 degrees. Plugging these values into the equation, we get:d = (2.78 m/s)^2 * sin(2 * 45 degrees) / 9.81 m/s^2d = 0.79 mTherefore, the maximum distance we can shoot a dart is 0.79 m.

Question 2

A child is sitting on the outer edge of a merry-go-round that is 18 m in diameter.If the merry-go-round makes 8.3 rev/min,what is the velocity of the child in m/s?
A)7.8 m/s
B)15.6 m/s
C)1.2 m/s
D)5.5 m/s

Accepted Answer

The velocity of the child can be calculated using the formula for the tangential velocity: v = ωr, where ω is the angular velocity in radians per second and r is the radius. First, convert revolutions per minute to radians per second: ω = 8.3 rev/min × (2π rad/rev) × (1 min/60 s) = 0.868 rad/s. The radius r is half the diameter, so r = 18 m / 2 = 9 m. Therefore, v = 0.868 rad/s × 9 m = 7.8 m/s.

Question 3

A plane flying at 70.0 m/s suddenly stalls.If the acceleration during the stall is 9.8 m/s² directly downward,the stall lasts 5.0 s,and the plane was originally climbing at 25° to the horizontal,what is the velocity after the stall? A)66 m/s at -17° B)66 m/s at +17° C)80 m/s at -37° D)80 m/s at +37°

Accepted Answer

The initial velocity components are 70 m/s * cos(25°) horizontally and 70 m/s * sin(25°) vertically. After 5 seconds of downward acceleration, the vertical velocity component changes by 9.8 m/s² * 5 s. Calculating the final velocity components and using trigonometry gives a resultant velocity of 66 m/s at an angle of -17° to the horizontal.

Question 4

A hobby rocket reaches a height of 72.3 m and lands 111 m from the launch point.What was the angle of launch?
A)69.0°
B)67.4°
C)22.6°
D)-44.8°

Accepted Answer

The correct answer is A because the angle of launch can be calculated using the formula:theta = arctan(2 * h / d)where h is the height reached by the rocket and d is the horizontal distance traveled by the rocket. Plugging in the given values, we get:theta = arctan(2 * 72.3 m / 111 m) = 69.0°

Question 5

A ship moving initially at 10.7 m/s due North,drifts due NE at the same speed.To correct the bearing,the captain accelerates the engine and turns the rudder NW.If the engines deliver a constant acceleration of 4.4 m/s²,how long will it take to assume the original course again? A)2.4 s B)3.4 s C)1.7 s D)1.2 s

Accepted Answer

The correct answer is A because the ship is moving at a constant speed of 10.7 m/s and the engines are delivering a constant acceleration of 4.4 m/s2. Therefore, the ship will take 2.4 seconds to assume its original course again.

Question 6

A plane flies directly from city A to city B,which are separated by 2300 mi.From A to B,the plane flies into a 65 mi/h headwind.On the return trip from B to A,the wind velocity is unchanged.The trip from B to A takes 65 min less than the trip from A to B.What is the airspeed (assumed constant)of the plane?
A)530 mi/h
B)400 mi/h
C)480 mi/h
D)610 mi/h

Accepted Answer

The answer of A plane flies directly from city A...

Question 7

A glider is tugged by an airplane at 81 m/s when it is released.If the original speed was along the horizontal and the glider is now under a constant acceleration of 2.4 m/s² at 1.1° below the horizontal due to air drag,how long will it take to reach the ground 5.7 km below? A)500 s B)250,000 s C)2.2 s D)4.8 s

Accepted Answer

The vertical component of acceleration is 2.4 * sin(1.1°). Using the equation of motion for vertical displacement, 5.7 km = 0.5 * a * t^2, solve for t.

Question 8

A fisherman casts his bait into the river at an angle of 25°.As the line unravels,he notices that the bait and hook reach a maximum height of 3.9 m.What was the initial velocity he launched the bait with?
A)21 m/s
B)9.2 m/s
C)10 m/s
D)8.4 m/s

Accepted Answer

The correct answer is A because the initial velocity is 21 m/s. This can be calculated using the formula v = sqrt(2gh), where v is the initial velocity, g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum height reached (3.9 m).

Question 9

The magnitude of a vector cannot be smaller than the magnitude of any of its components.

Accepted Answer

The magnitude of a vector is the square root of the sum of the squares of its components, which means it is always greater than or equal to the magnitude of any individual component.

Question 10

An aircraft performs a maneuver called an aileron roll.During this maneuver,the plane turns like a screw as it maintains a straight flight path,by using its ailerons to set the wings in circular motion.If it takes the plane 35 s to complete the circle and each wing length is 4.6 m,what is the acceleration of the wing tip? A)0.15 m/s² B)0.69 m/s² C)6.7 m/s² D)1.4 m/s²

Accepted Answer

The answer of An aircraft performs a maneuver called an...

Question 11

The magnitude of the displacement vector from A to B can never be less than the distance from A to B,but it can be greater than that distance.

Accepted Answer

The magnitude of the displacement vector from A to B is always equal to the straight-line distance between A and B and cannot be greater or less than that distance.

Question 12

The magnitude of the displacement vector of an object is equal to the distance that object has moved.

Accepted Answer

The magnitude of the displacement vector of an object is the straight-line distance between the initial and final positions of the object, which may be different from the actual distance traveled due to changes in direction or path taken.

Question 13

A boat crosses a river at a constant engine speed of 8.0 cm/s under the influence of a constant perpendicular current acceleration of 0.060 m/s².How far off a straight course does it drift in 2.3 min? A)570 m B)0.16 m C)4.1 m D)1700 m

Accepted Answer

The boat's velocity due to the current is:$$v_c=at=(0.060	ext{ m/s}^2)(2.3	ext{ min})(60	ext{ s/min})=8.28	ext{ m/s}$$The boat's displacement due to the current is:$$d_c=v_ct=(8.28	ext{ m/s})(2.3	ext{ min})(60	ext{ s/min})=570	ext{ m}$$

Question 14

Two particles,A and B,are in uniform circular motion about a common center.The acceleration of particle A is 4.7 times that of particle B.Particle B takes 2.4 times as long for a rotation as particle A.The ratio of the radius of the motion of particle A to that of particle B is closest to:
A)0.82
B)2.0
C)3.8
D)0.51
E)11

Accepted Answer

Let's denote the radius of the circular motion of particle A by r and that of particle B by s. Let the velocities of A and B be denoted by v and u, respectively. 
The acceleration of a particle in uniform circular motion is given by a = v^2/r.
We are given that a_A = 4.7a_B, which means that v_A^2/r = 4.7v_B^2/s. 
We are also given that B takes 2.4 times as long as A for one rotation. This implies that B travels around its circle 2.4 times for every time A travels around its circle once. Therefore, the ratio of their respective velocities is v/u = s/r * 2.4 (since they both cover the same angle in their respective circles in the same amount of time).
We can solve for v_A^2/v_B^2 and v/u in terms of r/s, and then equate the two expressions to get the ratio of r/s. After simplifying, we get:
r/s = (4.7 * 2.4)^1/3 ≈ 0.82, which is closest to choice A.

Question 15

An aircraft performs a maneuver called an aileron roll.During this maneuver,the plane turns like a screw as it maintains a straight flight path,by using its ailerons to set the wings in circular motion.Due to structural strength,an airplane can only withstand an acceleration of 6.1 m/s² on its wing tips.If the wing span is 6.0 m,what is the shortest time it can make a full turn when it performs an aileron roll? A)4.4 s B)6.0 s C)0.70 s D)6.2 s

Accepted Answer

The centripetal acceleration of the wing tips is given by:$$a_c = v^2/r$$where:* $a_c$ is the centripetal acceleration* $v$ is the speed of the wing tips* $r$ is the radius of the circleThe radius of the circle is half the wingspan, or 3.0 m. The maximum centripetal acceleration that the airplane can withstand is 6.1 m/s^2. Therefore, the speed of the wing tips is:$$v = \sqrt{a_c r} = \sqrt{6.1 \text{ m/s}^2 \cdot 3.0 \text{ m}} = 4.4 \text{ m/s}$$The time it takes to make a full turn is:$$t = 2\pi r/v = 2\pi \cdot 3.0 \text{ m} / 4.4 \text{ m/s} = 4.4 \text{ s}$$

Question 16

A small boat is moving at a velocity of 3.35 m/s when it is accelerated by a river current perpendicular to the initial direction of motion.If the current acceleration is 0.750 m/s²,what will the new velocity of the boat be after 33.5 s? A)25.3 m/s at 82.4° from initial direction of motion B)25.3 m/s at 7.59° from initial direction of motion C)640.1 m/s at 82.4° from initial direction of motion D)640.1 m/s at 7.59° from initial direction of motion

Accepted Answer

The boat's velocity components are 3.35 m/s in the initial direction and 0.750 m/s² * 33.5 s = 25.125 m/s perpendicular to it. The resultant velocity is √(3.35² + 25.125²) ≈ 25.3 m/s, and the angle is tan⁻¹(25.125/3.35) ≈ 82.4°.

Question 17

The product of a vector by a scalar is a vector having the same direction of the original vector but a different magnitude.

Accepted Answer

The answer of The product of a vector by a...

Question 18

An electron moves with a horizontal velocity of 3.0 × 10⁶ m/s as it enters a deflector inside a TV tube.The electron strikes the screen after traveling 17.0 cm horizontally and 40.0 cm vertically upward.What is the constant vertical acceleration provided by the deflector? A)2.5 × 10¹⁴ m/s² B)8.3 × 10² m/s² C)1.4 × 10⁴ m/s² D)1.2 × 10¹⁴ m/s²

Accepted Answer

The correct answer is A because the vertical acceleration can be calculated using the formula: a = (vf^2 - vi^2) / 2d, where vf is the final vertical velocity, vi is the initial vertical velocity, and d is the vertical displacement. Plugging in the given values, we get: a = (0^2 - (3.0 × 10^6 m/s)^2) / 2(0.40 m) = 2.5 × 10^14 m/s^2.

Question 19

You are taking a turn at 23.0 m/s on a ramp of radius 39.0 m.What is your acceleration? A)13.6 m/s² B)0.590 m/s² C)66.1 m/s² D)1.70 m/s²

Accepted Answer

The centripetal acceleration is given by a = v^2/r, where v is the velocity and r is the radius of the circular path. Plugging in the given values, we get a = (23.0 m/s)^2/39.0 m = 13.6 m/s^2. Therefore, the best choice is A.

Question 20

The magnitude of the resultant of two vectors must be greater than or equal to the magnitude of each of these vectors.

Accepted Answer

The answer of The magnitude of the resultant of two...

Quiz 3: Vectors and Motion in Two Dimensions

What is the maximum distance we can shoot a dart,provided our toy dart gun gives a maximum initial velocity of 2.78 m/s?

A child is sitting on the outer edge of a merry-go-round that is 18 m in diameter.If the merry-go-round makes 8.3 rev/min,what is the velocity of the child in m/s?

A plane flying at 70.0 m/s suddenly stalls.If the acceleration during the stall is 9.8 m/s2 directly downward,the stall lasts 5.0 s,and the plane was originally climbing at 25° to the horizontal,what is the velocity after the stall?

A hobby rocket reaches a height of 72.3 m and lands 111 m from the launch point.What was the angle of launch?

A ship moving initially at 10.7 m/s due North,drifts due NE at the same speed.To correct the bearing,the captain accelerates the engine and turns the rudder NW.If the engines deliver a constant acceleration of 4.4 m/s2,how long will it take to assume the original course again?

A glider is tugged by an airplane at 81 m/s when it is released.If the original speed was along the horizontal and the glider is now under a constant acceleration of 2.4 m/s2 at 1.1° below the horizontal due to air drag,how long will it take to reach the ground 5.7 km below?

A fisherman casts his bait into the river at an angle of 25°.As the line unravels,he notices that the bait and hook reach a maximum height of 3.9 m.What was the initial velocity he launched the bait with?

The magnitude of a vector cannot be smaller than the magnitude of any of its components.

The magnitude of the displacement vector from A to B can never be less than the distance from A to B,but it can be greater than that distance.

The magnitude of the displacement vector of an object is equal to the distance that object has moved.

A boat crosses a river at a constant engine speed of 8.0 cm/s under the influence of a constant perpendicular current acceleration of 0.060 m/s2.How far off a straight course does it drift in 2.3 min?

Two particles,A and B,are in uniform circular motion about a common center.The acceleration of particle A is 4.7 times that of particle B.Particle B takes 2.4 times as long for a rotation as particle A.The ratio of the radius of the motion of particle A to that of particle B is closest to:

A small boat is moving at a velocity of 3.35 m/s when it is accelerated by a river current perpendicular to the initial direction of motion.If the current acceleration is 0.750 m/s2,what will the new velocity of the boat be after 33.5 s?

The product of a vector by a scalar is a vector having the same direction of the original vector but a different magnitude.

An electron moves with a horizontal velocity of 3.0 × 106 m/s as it enters a deflector inside a TV tube.The electron strikes the screen after traveling 17.0 cm horizontally and 40.0 cm vertically upward.What is the constant vertical acceleration provided by the deflector?

You are taking a turn at 23.0 m/s on a ramp of radius 39.0 m.What is your acceleration?

The magnitude of the resultant of two vectors must be greater than or equal to the magnitude of each of these vectors.

A plane flying at 70.0 m/s suddenly stalls.If the acceleration during the stall is 9.8 m/s² directly downward,the stall lasts 5.0 s,and the plane was originally climbing at 25° to the horizontal,what is the velocity after the stall?

A ship moving initially at 10.7 m/s due North,drifts due NE at the same speed.To correct the bearing,the captain accelerates the engine and turns the rudder NW.If the engines deliver a constant acceleration of 4.4 m/s²,how long will it take to assume the original course again?

A glider is tugged by an airplane at 81 m/s when it is released.If the original speed was along the horizontal and the glider is now under a constant acceleration of 2.4 m/s² at 1.1° below the horizontal due to air drag,how long will it take to reach the ground 5.7 km below?

A boat crosses a river at a constant engine speed of 8.0 cm/s under the influence of a constant perpendicular current acceleration of 0.060 m/s².How far off a straight course does it drift in 2.3 min?

A small boat is moving at a velocity of 3.35 m/s when it is accelerated by a river current perpendicular to the initial direction of motion.If the current acceleration is 0.750 m/s²,what will the new velocity of the boat be after 33.5 s?

An electron moves with a horizontal velocity of 3.0 × 10⁶ m/s as it enters a deflector inside a TV tube.The electron strikes the screen after traveling 17.0 cm horizontally and 40.0 cm vertically upward.What is the constant vertical acceleration provided by the deflector?