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RoseWind [281]
3 years ago
9

1) Based on this graph of force vs. extension for a certain spring, what should be the Hooke's law constant (or spring constant)

?
Show the full calculation using data from the figure, and report the answer in N/m.

2) Suppose a 250 g mass was hung from the spring in the previous question. What would be the extension of the spring when it is in equilibrium and at rest?

Write the equation needed to answer this question and the solve for the extension, recording the answer in either m or cm.

3) Now the 250 g mass hanging from the spring in the previous question is pulled down an extra 10 cm and released. What would be the period and the counting frequency of the oscillations?

Once again, begin with an equation and show the steps needed to find the answers.

Physics
1 answer:
madam [21]3 years ago
3 0
Oh junk I need help with this
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A plane traveling horizontally at 120 ​m/s over flat ground at an elevation of 3610 m must drop an emergency packet on a target
antoniya [11.8K]

Answer:

Explanation:

Horizontal displacement

x = 120 t

Vertical position

y = 3610 - 4.9 t²

y = 0 for the ground

0 = 3610 - 4.9 t²

t = 27.14 s

This is the time it will take to reach the ground .

During this period , horizontal displacement

x = 120 x 27.14 m

= 3256.8 m

So packet should be released 3256.8 m before the target.

8 0
3 years ago
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A very long stick ruled with meter markings is placed in empty space. A spaceship of rest length L = 100 m runs lengthwise along
alexira [117]

Answer:

Meter marks are on cut-off portion of stick is 100 - 60 = 40 m

Explanation:

Given data:

Spaceship  length of L = 100 m

Relative velocity between the ship and stick is given as

v = \frac{4}{5} c

The observed length observed by the outside observer is

L' = L\sqrt{1 -\frac{v^2}{c^2}}

putting all value to get observe length

putv = \frac{4}{5} c

L ' = 100\times \sqrt{1 - \frac{(4/5)^2c^2}{c^2}

L' = 60 m

Meter marks are on cut-off portion of stick is 100 - 60 = 40 m

8 0
3 years ago
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Julli [10]

I can't see the answers clearly, but I can see the question. So, I'll just give you a clue/hint.

A stars brightness actually depends on how far it is from your location. If it's far away, it will be dimmer than its counterpart that is closer. To summarize it, if two stars have the same brightness level and one is farther away than the other, the one farther away will appear dimmer than its closer counterpart.

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3 years ago
Physics questions , will give brainliest
creativ13 [48]

The applicable relationship here is

... acceleration = rate of change of velocity

4. The slope of the velocity curve is constant, so the acceleration is constant. That slope is 3 m/s in 5 s, or (3 m/s)/(5 s) = (3/5) m/s² = 0.6 m/s²

7. You're looking for a point on the velocity curve where its slope is -2 m/s². That will be somewhere between t=0 and t=4, because slope is positive for t>4. The only available choice in that region is t = 2 s.

8. At 6.00 m/s² for 3 seconds, velocity will change (3 s)×(6.00 m/s²) = 18.00 m/s. That would get you from an initial speed of 3.44 m/s to 21.44 m/s, so clearly 3 s is almost right, but a little too long for the given change in velocity. The best choice is 2.91 s.

If you want to actually figure it out, the change in velocity is 20.9 -3.44 = 17.46 m/s. Using the relation a = ∆v/∆t, we can rearrange it to ∆t = ∆v/a, or

... ∆t = (17.46 m/s)/(6 m/s²) = 2.91 s

9. This problem combines the determination of acceleration with a units conversion problem. Numbers in the problem are given in kph and seconds, and answers are given in m/s². At some point, you need to convert from km/h to m/s. The multiplier for that is (1000 m/km)/(3600 s/h) = 1/3.6 (m·h)/(km·s).

Your change in speed is -24.6 km/h = (1/3.6)·(-24.6) m/s ≈ -6.8333 m/s. When that change in speed occurs over 3.56 seconds, the acceleration is

... (-6.8333 m/s)/(3.56 s) ≈ -1.919 m/s² ≈ -1.92 m/s²

5. At 10 s, the velocity is about 14 m/s. Looking for grid points the curve goes through, we can use (11, 16) and (9, 12). That is, over the 2-second range from 9 s to 11 s, the velocity increases 4 m/s from 12 m/s to 16 m/s. The acceleration is

... ∆v/∆t = a = (4 m/s)/(2 s) = 2 m/s²

6. ∆v/∆t = a = (28 m/s - 0 m/s)/(4.22 s) ≈ 6.6351 m/s² ≈ 6.64 m/s²

10. No change in velocity means the acceleration is 0 m/s².

4 0
3 years ago
Static electric fields are caused by charged particles. What characterizes a positively charged particle?
sergeinik [125]

Static electric fields are caused by charged particles when the atoms which compose the particle will have the same number of electrons as protons.

Each point in space has an electric field associated with it when the charge of any kind is present. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field.

Both the electromagnetic wave produced by a radio broadcast monopole antenna and the field created in the dielectric of a parallel-plate capacitor are examples of electric fields (which create a time-varying field). Every place in space where a charge, regardless of its shape, is present is said to have an electric field, which is an electric attribute. The electric field's equation is given as E = F / Q.

To learn more about electric field please visit -
brainly.com/question/15800304
#SPJ1

8 0
2 years ago
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