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Andre45 [30]
2 years ago
8

1) Compute the travel time ratio of the other objects to the hollow cylinder. 2) Discuss how the travel time ratio of solid cyli

nder to hollow cylinder have confirmed or contradicted the theoretical value. 3) Determine whether the billiard ball/racquetball is solid or hollow. 4) Discuss the object with the largest SEOM. 5) Discuss possible systematic errors in the measurement.
Physics
1 answer:
Illusion [34]2 years ago
6 0

Answer:

(1) Use the data in  Table R2 and error propagation rule to compute the travel time ratio (with  errors) of the other objects to the hollow cylinder  ? ℎ? . Fill in  Table R5 below. [6]  Table R5   Solid cylinder  Billiard ball  Racquetball  ? ? ℎ? ± ± ± (2) Discuss how the ratio of solid cylinder to hollow cylinder have confirmed or contradicted the  theoretical ratio in Eq. (8) in  Manual . Compute the % error and make discussions. [4]  Answer:  (3) From the travel time ratio, please determine (i) whether the billiard ball is solid or hollow,  and (ii) whether the racquetball is solid or hollow. Write down your reasons. (The answer  can be uncertain if your measurements do not show enough information.) [4]

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PHYS2125 Physics Laboratory I ©2018 Kuei Sun The University of Texas at Dallas 5 Answer: (4) Which object in Table R2has the largest SEOM? Please discuss the reason for this relatively largeSEOM and how to improveit. [3]  (5) Please discuss  TWO possible systematic errors in the measurement. [3]  Answer:  **Attach your calculation details. Use as many paper sheets as you need. Calculations that  demonstrate your understandings may gain you partial credit.  **Make sure your work-area and equipment are the same as they were before you leave.

Explanation:

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Two uniform, solid cylinders of radius R and total mass M are connected along their common axis by a short, light rod and rest o
sveta [45]

Explanation:

A) To prove the motion of the center of mass of the cylinders is simple harmonic:

System diagram for given situation is shown in attached Fig. 1

We can prove the motion of the center of mass of the cylinders is simple harmonic if

a_{x} = -\omega^{2}  x

where aₓ is acceleration when attached cylinders move in horizontal direction:

<h3>PROOF:</h3>

rotational inertia for cylinders  is given as:

                                  I=\frac{1}{2}MR^{2} -----(1)

Newton's second law for angular motion is:

                                             ∑τ = Iα ------(2)

For linear motion in horizontal direction it is:

                                             ∑Fₓ = Maₓ ------ (3)

By definition of torque:

                                               τ  = RF --------(4)        

Put (4) and (1) in (2)

                                       RF=\frac{1}{2}MR^{2}\alpha

                                       RF=\frac{1}{2}MR^{2}\alpha

from Fig 3 it can be seen that fs is force by which the cylinders roll without slipping as they oscillate

So above equation becomes

                                   f_{s}=\frac{1}{2}MR\alpha------ (5)

As angular acceleration is related to linear by:

                                          a= R\alpha

Eq (5) becomes

                                    f_{s}=\frac{1}{2}Ma_{x}---- (6)

aₓ shows displacement in horizontal direction

From (3)

                                              ∑Fₓ = Maₓ

Fₓ is sum of fs and restoring force that spring exerts:

                                  \sum F_{x} = f_{s} - kx ----(7)

Put (7) in (3)

                                  f_{s} - kx  = Ma_{x}[/tex] -----(8)

Using (6) in (8)

                               \frac{1}{2}Ma_{x} - kx =Ma_{x}

                                     a_{x} = \frac{2k}{3M} x --- (9)

For spring mass system

                                  a= -\omega^{2} x ----- (10)

Equating (9) and (10)

                                  \omega^{2} = \frac{2k}{3M}

\omega = \sqrt{ \frac{2k}{3M}}

then (9) becomes

                                a_{x} = - \omega^{2}x

(The minus sign says that x and  aₓ  have opposite directions as shown in fig 3)

This proves that the motion of the center of mass of the cylinders is simple harmonic.

<h3 /><h3>B) Time Period</h3>

Time period is related to angular frequency as:

                                   T=\frac{2\pi }{\omega}

                                  T = 2\pi \sqrt{\frac{3M}{2k}

                           

 

5 0
2 years ago
A particle is in uniform circular motion. Assume a standard rtz coordinate system. If you deconstruct the net force acting on th
Kitty [74]

Answer:

a) One

Explanation:

In a uniform circular motion there must be a force acting to keep it in the circular track. This force can either be centripetal or a centrifugal force.

According to the Newton's first law of motion a particle continues to be in state of rest or in uniform motion until acted upon by an external force.

Here the term uniform motion need to be understood which refers to the uniform velocity of the particle in accordance to the vector laws.

3 0
2 years ago
1) My 14V car battery could be used to charge my laptop, but I need to use an inverter to first convert it to a standard 120V. T
LenaWriter [7]

Answer:

1) Charge chord resistance is 75 Ω

2) Charge chord resistance is 6.33 Ω

Explanation:

1) To answer the question, we note that the the formula voltage is found as follows;

V = IR

Therefore,

R = \frac{V}{I} =  \frac{120}{1.6} = 75  \, \Omega

2) Where the voltage, V = 19.5 V and the current, I = 3.33 A, we have;

Initial resistance R₁ = 19.5 V/(3.33 A) = 5.86 Ω

However, to reduce the current to 1.6 A, we have;

R_T = \frac{19.5}{1.6} = 12.1875 \ \Omega

Therefore, where the resistance is found by the sum of the total resistance we have;

R_T = R₁ + Charge chord resistance

∴ 12.1875 = 5.86 + Charge chord resistance

Hence, charge chord resistance = 6.33 Ω

8 0
2 years ago
Which statement is not a good practice when working inside a computer case?
lozanna [386]

Answer:

a. be sure to hold expansion cards by the edge connectors

Explanation:

Removal of loose jewelry is a good safety practice. Also not touching a microchip with a magnetized screwdriver is also a good practice.

But holding expansion cards by the edge connectors is not a good practice, so it is the odd one in the question. Therefore answer option a provides the correct and best answer to the question

3 0
2 years ago
You start with spring that's already been stretched an unknown amount from equilibrium. After stretching it an additional 2.0 cm
maxonik [38]

Answer: 35*10^3 N/m

Explanation: In order to explain this problem we know that the potential energy for spring is given by:

Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.

We  also know that with additional streching of 2 cm of teh spring,  the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.

Then the difference of energy for both cases is 7 J so:

ΔUp= 1/2*k* (0.02)^2 then

k=2*7/(0.02)^2=35000 N/m

7 0
2 years ago
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