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vodomira [7]
2 years ago
15

What is the sources of error and suggestion on how to overcome it in the hooke's law experiment?

Physics
1 answer:
goblinko [34]2 years ago
6 0
In physics, Hooke's law is written in equation as:

F = kx

It states that the force F exerted on the spring is directly proportional to the displacement x by a constant called spring constant k.

In the laboratory, this is done in an experiment through the apparatus shown in the attached figure. The object experimented here is the spring, and you are to find the spring constant. A known mass of object is attached below the spring. That object carries a force in the form of gravitational pull in terms of weight. When the spring stretches, the displacement is measured with the use of the ruler.

There are a number of sources of error for this experiment. First, the reading from the ruler by the reader may be inaccurate. That's why digital balances are much more reliable because it minimizes human error. Reading the measurement on the ruler is subjective especially when you don't read it on eye level. Second, the force of the object might also be inaccurate if you use an unreliable weighing scale. Lastly, the apparatus might not be properly calibrated.

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The graph indicates Linda’s walk.
Sedaia [141]
I think the right answer is the first one. If she stops moving her Position does not change any more-and the Graph Shows that after 6 seconds she stays at the Position of 5 m. If she Went Back to the start point the Graph would have Developed Back to 0m(decreased).
3 0
2 years ago
Read 2 more answers
While traveling from Boston to Hartford, Person A drives at a constant speed of 55 mph for the entire trip. Person B drives at 6
Ne4ueva [31]

Answer:

B will take 1.034 times the time of A from Boston to Hartford.

Explanation:

Let the distance from Boston to Hartford be S.

Person A drives at a constant speed of 55 mph for the entire trip,

Time taken by person A

             t_A=\frac{S}{55}

Person B drives at 65 mph for half the distance and then drives 45 mph for the second half of the distance.

Time taken by person B

            t_B=\frac{\frac{S}{2}}{65}+\frac{\frac{S}{2}}{45}=\frac{S}{130}+\frac{S}{90}=\frac{220S}{130\times 90}=\frac{11S}{585}

Ratio of time of arrival of B to A

                      \frac{t_B}{t_A}=\frac{\frac{11S}{585}}{\frac{S}{55}}=\frac{121}{117}=1.034

B will take 1.034 times the time of A from Boston to Hartford.

8 0
2 years ago
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
2 years ago
3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing P
KIM [24]

Answer:

From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot  find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

E=\frac{F}{Q}

But the force F

F= \frac{kQ1Q2}{r^2}

But the electric field intensity due to a point charge Q at a distance r meters away is given by

E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }

<em>From the relation above we can conclude that the  as the distance between the two plate increases the electric field strength decreases</em>

6 0
2 years ago
How many calories are equal to one BTU? (One calorie = 4.186 J, one BTU = 1 054 J.)
I am Lyosha [343]
<h2>Option C is the correct answer.</h2>

Explanation:

We need to find how many calories is 1 BTU.

Given

          1 BTU = 1054 J

          1 calorie = 4.186 J

So we have

          1 BTU = 4.186 x 251.79 J

          1 BTU =251.79 calorie

          1 BTU = 252 calorie.

Option C is the correct answer.

3 0
2 years ago
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