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nika2105 [10]
3 years ago
6

A student performs an experiment that involves the motion of a pendulum. The student attaches one end of a string to an object o

f mass M and secures the other end of the string so that the object is at rest as it hangs from the string. When the student raises the object to a height above its lowest point and releases it from rest, the object undergoes simple harmonic motion. As the student collects data about the time it takes for the pendulum to undergo one oscillation, the student observes that the time for one swing significantly changes after each oscillation. The student wants to conduct the experiment a second time. Which two of the following procedures should the student consider when conducting the second experiment?
Physics
1 answer:
nadya68 [22]3 years ago
7 0

Answer:

-use the exact same equipment

-use the exact same measurements (dropping height)

Explanation:

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How to do this, i'm completely lost
vaieri [72.5K]
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.

You need to figure out t4 to know the tension in the string.

Since the whole thing is not moving t1 + t2 + t3 = t4.

torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)

t1 =3.2 * 44g 
t2 = 7 * 49g 
t3 = 3.5 * 24g 

t4 = t1 + t2 + t3 = 5570,118

The t4 also is given by:

t4 = r * T * sin Ф

r = 7
Ф = 32°
T: tension in the string

T = t4 / (r * sinФ)

T = t4 / (7 * sin(32°)) 

T = 1501,6 N

8 0
3 years ago
Please help its URGENT!!
Greeley [361]

Answer:

1 D ,2 B ,5 C ,3 A ,4 E i dont know why you gotta have 20 words to answer it but yeah

6 0
2 years ago
Read 2 more answers
A friend in your class tells you that she never uses hints when doing her Mastering homework. She says that she finds the hints
AnnZ [28]

Answer:

A, B, and C are good reasons for my friend not to worry

Explanation:

The following reasons are reason not to worry

A. The only way to lose additional partial credit on a hint is by using the "give up" button or entering incorrect answers. Leaving the question blank will not cost you any credit (Regardless of whether you open a link or not, you will lose credit if you enter a wrong answer or if you give up on a question by hitting the "give up" button. Even after opening a hint, you can leave the question blank if the hint does not provide relevant hints or if the hint brings up more question. Once the question is left blank, you do not lose additional partial credit)

B. As an incentive for thinking hard about the problem, your instructor may choose to apply a small hint penalty, but this penalty is the same whether the hint simply gives information or asks another question (In a situation where you decide to use a hint, the instructor may have put a penalty for using the hint, so whether it asks a question or help in the solution of the question, as long as the hint is consulted, the hint penalty still applies)

C. Getting the correct answer to the question in a hint actually gives you some partial credit, even if you still can't answer the original question (An advantage of using hint is that you get some partial credit for using it if you answer the hint question correctly and fails to answer the original question)

6 0
3 years ago
An green hoop with mass mh = 2.8 kg and radius rh = 0.17 m hangs from a string that goes over a blue solid disk pulley with mass
vladimir2022 [97]
The mass of the hoop is the only force which is computed by:F net = 2.8kg*9.81m/s^2 = 27.468 N 
the slow masses that must be quicker are the pulley, ring, and the rolling sphere. 
The mass correspondent of M the pulley is computed by torque τ = F*R = I*α = I*a/R F = M*a = I*a/R^2 --> M = I/R^2 = 21/2*m*R^2/R^2 = 1/2*m 
The mass equal of the rolling sphere is computed by: the sphere revolves around the contact point with the table. So using the proposition of parallel axes, the moment of inertia of the sphere is I = 2/5*mR^2 for spin about the midpoint of mass + mR^2 for the distance of the axis of rotation from the center of mass of the sphere. I = 7/5*mR^2 M = 7/5*m 
the acceleration is then a = F/m = 27.468/(2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2
6 0
3 years ago
The diameter of an atom is 1.1×10−10m and the diameter of its nucleus is 1.0×10−14m. Part A What percent of the atom's volume is
77julia77 [94]

To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,

\% Percent = \frac{V_{nucleus}}{V_{atom}}*100

\% Percent = \frac{4/3 \pi (d_{nucleus}/2)^3}{4/3 \pi (d_{atom}/2)^3}*100

\% Percent = \frac{(d_{nucleus}/2)^3}{ (d_{atom}/2)^3}*100

\% Percent =\frac{(1.0*10^{-14}/2)^3}{ (1.1*10^{-10}/2)^3}*100

\% Percent = 7.51*10^{-13}*100

\% Percent = 7.51*10^{-11}\%

Therefore the percent of the atom's volume is occupied by mass is 7.51*10^{-11}\%

3 0
3 years ago
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