Question 1

Uniform circular motion: A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 2.00 m making 7.00 revolutions every 10.0 seconds. What is the magnitude of the acceleration of the ball?

Accepted Answer

The angular velocity of the ball is given by:
ω = (2π*7)/10 = 4.4 rad/s
The centripetal acceleration of the ball is given by:
a = rω^2 = (2.00 m)(4.4 rad/s)^2 = 38.7 m/s^2.
Therefore, the magnitude of the acceleration of the ball is 38.7 m/s^2. Option B is the correct answer.

Question 2

Uniform circular motion: A satellite orbits the earth a distance of 1.50 × 10⁷ m above the planet's surface and takes 8.65 hours for each revolution about the earth. The earth's radius is 6.38 × 10⁶ m. The acceleration of this satellite is closest to

Accepted Answer

The acceleration can be calculated using the formula for centripetal acceleration: a = v²/r. First, find the orbital radius (r = Earth's radius + altitude) and the orbital speed (v = 2πr/T, where T is the period in seconds). Then, substitute these into the formula to find the acceleration.

Question 3

Uniform circular motion: An electrical motor spins at a constant  If the armature radius is  what is the acceleration of the outer edge of the armature?

Accepted Answer

The answer of Uniform circular motion: An electrical motor spins...

Question 4

Uniform circular motion: A disk-shaped space station 175 m in diameter spins uniformly about an axis perpendicular to the plane of the disk through its center. How many rpm (rev/min) must this disk make so that the acceleration of all points on its rim is g/2?

Accepted Answer

The answer of Uniform circular motion: A disk-shaped space station...

Question 5

Uniform circular motion: An object moves in a circle of radius R at constant speed with a period T. If you want to change only the period in order to cut the object's acceleration in half, the new period should be

Accepted Answer

The answer of Uniform circular motion: An object moves in...

Question 6

Relative velocity: A plane flies directly between two cities, A and 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  less than the trip from A to B. What is the airspeed of the plane, assuming it is the same in both directions?

Accepted Answer

The answer of Relative velocity: A plane flies directly between...

Question 7

Uniform circular motion: 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, which sets the wings in circular motion. If it takes it 35 s to complete the circle and the wingspan of the plane is 11 m, what is the acceleration of the wing tip?

Accepted Answer

The acceleration of the wing tip is given by:$$a = \frac{v^2}{r}$$where:* v is the speed of the wing tip* r is the radius of the circleThe speed of the wing tip is given by:$$v = \frac{2\pi r}{T}$$where:* T is the period of the circleThe period of the circle is given as 35 s. The radius of the circle is half of the wingspan, which is 11 m / 2 = 5.5 m.Substituting these values into the equation for the acceleration of the wing tip, we get:$$a = \frac{\left(\frac{2\pi (5.5 	ext{ m})}{35 	ext{ s}}ight)^2}{5.5 	ext{ m}} = 0.18 	ext{ m/s}^2$$Therefore, the acceleration of the wing tip is 0.18 m/s^2.

Question 8

Relative velocity: A cart is moving with a constant horizontal velocity of 5.00 m/s. A small pebble is launched from the front of the cart with a velocity of 8.00 m/s at 60.0° above the horizontal as measured relative to the cart (see figure) and experiences no significant air resistance. Just as the pebble returns to the level from which it was launched, its distance from the front of the cart is closest to

Accepted Answer

Since the cart is moving with a constant horizontal velocity and there is no vertical component to its motion, the pebble's motion can be analyzed independently in the horizontal and vertical directions.

In the horizontal direction, the pebble's velocity relative to the ground is the vector sum of its velocity relative to the cart and the cart's velocity relative to the ground. Since the cart is moving with a constant velocity and the pebble is launched horizontally from the front of the cart, the pebble's horizontal velocity relative to the ground is also 5.00 m/s.

In the vertical direction, the pebble is launched at an angle of 60.0° with a velocity of 8.00 m/s relative to the cart. Ignoring air resistance, the only force acting on the pebble is gravity, so it follows a parabolic trajectory. Since the pebble returns to the same level from which it was launched, its time of flight is the same for both the upward and downward parts of its motion. Using the vertical component of its initial velocity relative to the cart, the time of flight can be found:

vy = v0y + gt
0 = 8.00 sin 60.0° + (-9.80 m/s^2)t
t = 1.03 s

During this time, the cart moves a horizontal distance of:

x = vt = (5.00 m/s)(1.03 s) = 5.15 m

Therefore, the distance between the pebble and the front of the cart at the instant when the pebble returns to the same level from which it was launched is:

d = sqrt(x^2 + y^2) = sqrt((5.15 m)^2 + (d0 + 8.00 cos 60.0° - (1/2)(9.80 m/s^2)(1.03 s)^2)^2)

where d0 is the initial distance between the front of the cart and the pebble at the instant the pebble is launched. We want to find the minimum value of d, which occurs when d0 is also the distance traveled by the pebble in the horizontal direction during the time of flight:

d0 = v0x t = (8.00 cos 60.0°)(1.03 s) = 4.16 m

Substituting this value of d0 into the expression for d and using a calculator, we get:

d ≈ 5.66 m

Therefore, the closest answer choice is C: 5.66 m.

Question 9

Uniform circular motion: Two particles, A and B, are in uniform circular motion about a common center. The acceleration of particle A is 8.5 times that of particle B. The period of particle B is 2.0 times the period of particle A. The ratio of the radius of the motion of particle A to that of particle B is closest to

Accepted Answer

The acceleration in uniform circular motion is given by $a = \frac{v^2}{r}$ or $a = \omega^2r$, where $v$ is the linear velocity, $r$ is the radius of the circle, and $\omega$ is the angular velocity. The angular velocity $\omega$ is related to the period $T$ by $\omega = \frac{2\pi}{T}$. Given that the acceleration of A is 8.5 times that of B and the period of B is 2 times that of A, we can use the relation $a = \omega^2r$ to find the ratio of the radii. Since $T_B = 2T_A$, it follows that $\omega_A = 2\omega_B$, and thus $a_A = 8.5a_B$. Substituting the relation for acceleration in terms of angular velocity and radius, we get $4\omega_B^2r_A = 8.5\omega_B^2r_B$, leading to $r_A/r_B = 8.5/4 = 2.125$, which is closest to 2.1.

Question 10

Uniform circular motion: A wind farm generator uses a two-bladed propeller (see figure) mounted on a pylon at a height of 20 m. The length of each propeller blade is 12 m. A small piece from the tip of the propeller breaks off when the propeller is vertical. At that instant, the period of the motion of the propeller is 1.2 s. The fragment flies off horizontally, falls with negligible air resistance, and strikes the ground at P.
(a) How far is point P from the base of the pylon?
(b) At what angle with respect to the vertical is the fragment moving just as it strikes the ground at P?

Accepted Answer

The answer of Uniform circular motion: A wind farm generator...

Quiz 4: Newtons Laws of Motion

Uniform circular motion: A ball is tied to the end of a cable of negligible mass. The ball is spun in a circle with a radius 2.00 m making 7.00 revolutions every 10.0 seconds. What is the magnitude of the acceleration of the ball?

Uniform circular motion: A satellite orbits the earth a distance of 1.50 × 107 m above the planet's surface and takes 8.65 hours for each revolution about the earth. The earth's radius is 6.38 × 106 m. The acceleration of this satellite is closest to

Uniform circular motion: An electrical motor spins at a constant If the armature radius is what is the acceleration of the outer edge of the armature?

Uniform circular motion: A disk-shaped space station 175 m in diameter spins uniformly about an axis perpendicular to the plane of the disk through its center. How many rpm (rev/min) must this disk make so that the acceleration of all points on its rim is g/2?

Uniform circular motion: An object moves in a circle of radius R at constant speed with a period T. If you want to change only the period in order to cut the object's acceleration in half, the new period should be

Uniform circular motion: A satellite orbits the earth a distance of 1.50 × 10⁷ m above the planet's surface and takes 8.65 hours for each revolution about the earth. The earth's radius is 6.38 × 10⁶ m. The acceleration of this satellite is closest to