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m_a_m_a [10]
2 years ago
7

A disk of known radius and rotational inertia can rotate without friction in a horizontal plane around its fixed central axis. T

he disk has a cord of negligible mass wrapped around its edge. The disk is initially at rest, and the cord can be pulled to make the disk rotate. Which of the following procedures would best determine the relationship between applied torque and the resulting change in angular momentum of the disk?
A. Pulling on the cord, exerting a force of 15 N for 2 s and then 25 N for 3 s, and measuring the final angular velocity of the disk

B. For five forces of different time intervals, pulling on the cord for 5 s, exerting a force of 15 N, and then measuring the angle through which the disk rotates in each case

C. For five forces of different time intervals, pulling on the cord for 5 s, exerting a force of 15 N, and then measuring the final angular velocity of the disk

D. For five forces of different magnitude, pulling on the cord for 5 s, and then measuring the final angular velocity of the disk
Physics
1 answer:
klio [65]2 years ago
5 0
I think the answer is C
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(a) Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.500 - μC charge and flies due west at a sp
12345 [234]

(a) 2.64\cdot 10^{-8} N north

We can treat the aircraft as a single point charge moving in a magnetic field. In this case, the magnetic force exerted on the plane is

F=qvB sin \theta

where

q=0.500 \mu C = 0.500\cdot 10^{-6} C is the charge on the plane

v = 660 m/s is the velocity

B=8.00\cdot 10^{-5} T is the magnitude of the magnetic field

\theta=90^{\circ} is the angle between the direction of motion of the jet and of the magnetic field

Substituting,

F=(0.5\cdot 10^{-6})(660)(8.0\cdot 10^{-5})=2.64\cdot 10^{-8} N

The direction can be found by using Fleming's left hand rule. We have:

- index finger: magnetic field direction (straight up)

- middle finger: velocity of the plane (due west)

- force: thumb --> north

(b) Not negligible

As we can see from part (a), the magnitude of the force is not really big, so the effects are negligible.

For instance, we can compare this force with the weight of a plane. If we take a Boeing 737, its mass is about 80,000 kg, so its weight is

W=mg=(80000)(9.8)=784,000 N

As we can see, this is several orders of magnitude bigger than the magnetic force calculated at point (a), so the effects of the magnetic force are negligible.

8 0
2 years ago
An object on a number line moved from x = 15 cm to x = 165 cm and then
olasank [31]

Answer:

v_avg = 2.9 cm/s

Explanation:

The average velocity of the object is the sum of the distance of all its trajectories divided the time:

v_{avg}=\frac{x_{all}}{t}

x_all is the total distance traveled by the object. In this case you have that the object traveled in the first trajectory 165cm-15cm = 150cm, and in the second one, 165cm - 25cm = 140cm

Then, x_all = 150cm + 140cm = 290cm

The average velocity is, for t = 100s

v_{avg}=\frac{290cm}{100s}=2.9\frac{cm}{s}

hence, the average velocity of the object in the total trajectory traveled is 2.9 cm/s

3 0
2 years ago
Read 2 more answers
a 4357 kg roller coaster car starts from rest at the top of a 36.5 m high track. determine the speed of the car at the top of a
andrey2020 [161]
The correct answer is 17.24 m/s. You get the answer by subtracting the two heights of the tracks which are 36.5 and 10.8 m, and the answer is 25.7. Since you already know the height at which the kinetic energy will be coming from, you then divide the amount of weight the roller coaster has to the distance it needs to travel in order for you to determine the speed of the car. So that is, 4,357 kg and 25.7 m and the answer is 169 kg/m. Dividing it to the earth's gravity of 9.8 m/s you'll get 17.24 m/s.
4 0
2 years ago
An electrical conductor is an element with __________ electrons in its outer orbit.
Setler [38]
An electric conductor is an element with free electrons in its outer orbit
5 0
2 years ago
A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

4 0
2 years ago
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