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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 metal has a strength of 414 MPa at its elastic limit and the strain at that point is 0.002. Assume the test specimen is 12.8-m
ser-zykov [4K]

To solve this problem, we will start by defining each of the variables given and proceed to find the modulus of elasticity of the object. We will calculate the deformation per unit of elastic volume and finally we will calculate the net energy of the system. Let's start defining the variables

Yield Strength of the metal specimen

S_{el} = 414Mpa

Yield Strain of the Specimen

\epsilon_{el} = 0.002

Diameter of the test-specimen

d_0 = 12.8mm

Gage length of the Specimen

L_0 = 50mm

Modulus of elasticity

E = \frac{S_{el}}{\epsilon_{el}}

E = \frac{414Mpa}{0.002}

E = 207Gpa

Strain energy per unit volume at the elastic limit is

U'_{el} = \frac{1}{2} S_{el} \cdot \epsilon_{el}

U'_{el} = \frac{1}{2} (414)(0.002)

U'_{el} = 414kN\cdot m/m^3

Considering that the net strain energy of the sample is

U_{el} = U_{el}' \cdot (\text{Volume of sample})

U_{el} =  U_{el}'(\frac{\pi d_0^2}{4})(L_0)

U_{el} = (414)(\frac{\pi*0.0128^2}{4}) (50*10^{-3})

U_{el} = 2.663N\cdot m

Therefore the net strain energy of the sample is 2.663N\codt m

6 0
1 year ago
Where is the steering nozzle located on a pwc?
Dvinal [7]
At the rear.

PWC stands for personal watercraft, and it is a small powerboat. The main components of a PWC are the hull (body of the boat), deck (surface where people walk/stand), throttle (controls speed), steering nozzle and water intake.
3 0
2 years ago
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Which title best reflects the main idea of the passage? The Role of Convection in the Distribution of Earth's Energy The Role of
Leto [7]

Answer:

The Role of Heat Transfer Methods in the Distribution of Earth's Energy

Explanation:

8 0
2 years ago
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Plug variables expressed in SI units in the kinematic equation given in article: a = -v0^2/(2sg). What value of g you get as exp
Elanso [62]

Answer:

1) acceleration is increased by a factor of four 4X

2) the acceleration increases a factor of 2X

3) the correct answer of 400g

Explanation:

This is a kinematics exercise, where you use the velocity equation to obtain the acceleration, with the final velocity equal to zero.

           v² = v₀² + 2 a x

           0 = v₀² + 2 a x

           a = - v₀² / 2 x

           

In the case of wanting to give the acceleration as a function of g, we can find the relationship between the two quantities

         a / g = - v₀² / (2 x g)

Let's answer the different questions about this equation

1. The initial velocity is doubled, how much the acceleration is worth

           

       a/g = - (2v₀) 2 / 2xg

       a = 4 (-v₀² / 2xg) g

acceleration is increased by a factor of four 4X

2. if the stopping distance is reduced by 2, that is, x = x₀ / 2

we substitute

        a/g = (- v₀² / 2g) 2/x

         

        a =2  (-v₀² / 2x₀g)  g

       

therefore the acceleration increases a factor of 2X

3. the initial velocity of the hockey player is v₀ = 20 m / s and the stopping distance is

x = 5cm = 0.05m

we calculate the acceleration

        a / g = - 20² / (2 0.05)

        a / g = - 4000 / g

        a / g = - 4000 / 9.8 = 408

        a = 408 g

the correct answer of 400g, the value matches exactly if g = 10 m / s2 is taken

6 0
2 years ago
Oceanographers use submerged sonar systems, towed by a cable from a ship, to map the ocean floor. In addition to their downward
KATRIN_1 [288]

Answer:

Tension in the cable is T = 16653.32 N

Explanation:

Give data:

Cross section Area A = 1.3 m^2

Drag coefficient CD = 1.2

Velocity V = 4.3 m/s

Angle made by cable with horizontal  =30 degree

Density \rho \ of\  water= 1000 kg/m3

 Drag force FD is given as

F_{D} = \fracP{1}{2} \rho v^{2} C_{D} A

        = 0.5\times 1000\times 4.32\times  1.2\times 1.3

Drag force = 14422.2 N acting opposite to the motion

As cable made angle  of 30 degree with horizontal  thus horizontal component is take into action to calculate drag force

TCos30 = F_D

T = \frac{F_D}{cos30}

T =\frac{ 14422.2}{cos 30}

T = 16653.32 N

7 0
2 years ago
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