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patriot [66]
3 years ago
9

Tom has built a large slingshot, but it is not working quite right. He thinks he can model the slingshot like an ideal spring, w

ith a spring constant of 45.0 N/m. When he pulls the slingshot back 0.515 m from a non-stretched position, it just doesn't launch its payload as far as he wants. His physics professor "helps" by telling him to aim for an elastic potential energy of 14.5 Joules. Tom decides he just needs elastic bands with a higher spring constant. By what factor does Tom need to increase the spring constant to hit his potential energy goal?
During a followup conversation, Tom's physics professor suggests that he should leave the slingshot alone and try pulling the slingshot back further without changing the spring constant. How many times further than before must Tom pull the slingshot back to hit the potential energy goal with the original spring constant?
Physics
1 answer:
lora16 [44]3 years ago
4 0
I don’t know right now, but I’ll get back to you later this afternoon in comment
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minim
barrel
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Also included are metric units which are dekaliter onwards.
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Explain how streams can erode soil and transport materials?
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8 0
3 years ago
Which of the following statements correctly describes the position of the intake and exhaust valves during most of the power sta
I am Lyosha [343]

Both valves are closed during the power stroke.

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If either of the valves was open, even just a crack,
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5 0
3 years ago
Read 2 more answers
A body in simple harmonic motion has a displacement x that varies in time t according to the equation x = 5cos(π t + π/3) , wher
zhuklara [117]

Answer:

1/2 Hz

Explanation:

A simple harmonic motion has an equation in the form of

x(t) = Acos(\omega t - \phi)

where A is the amplitude, \omega = 2\pi f is the angular frequency and \phi is the initial phase.

Since our body has an equation of  x = 5cos(π t + π/3) we can equate \omega = \pi and solve for frequency f

2\pi f = \pi

f = 1/2 Hz

7 0
3 years ago
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Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in
postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is \rho = \frac{m}{V}, and the volume of a pyramid is (confusingly written on the proposal) V=\frac{1}{3} Ah, so we can write:

m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h

Where s is the side of the base, being s^2 the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that \frac{1m}{100cm}=1, and we can use this factor to convert anything written in cm to anything written in m. Example:

500cm=500cm\frac{1m}{100cm}=5m

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so <em>it's not doing anything but changing the units</em>.

We can use this tool like this:

2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3

Where we have used the fact that 1^3=1 (<u>we can elevate any conversion factor to any number and they still will be 1</u>) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg

And now we will convert to short tons using two conversion factors at the same time:

m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0
3 years ago
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