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3 years ago

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A spring that has a spring constant of 1400 N/m is stretched to a length of 2.5 m. If the normal length of the spring is 1.0 m,

how much elastic potential energy is stored in the spring? 700 J 1050 J 1575 J 4375 J

1 answer:

Daniel [21]3 years ago

7 0

The answer is 1575, I just took the Review.<span />

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If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

stich3 [128]

I would think that you would have to do 42/2=21Hz, but I'm not sure...

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3 years ago

a) A unit of time sometimes used in microscopic physics is the shake. One shake equals 10–8 s. Are there more shakes in a second

lawyer [7]

Answer:

a) Yes, there are $10^8$ shakes in a second ( $1 s \frac{1 shake}{10^{-8}s}=10^8 shake$ ) in a year there are 31,536,000 seconds... that is 3.1536x $10^7$ ( $1year\frac{365 day}{1 year} \frac{24 h}{1 day} \frac{60 min}{1 h}\frac{60s}{1min}=3.1536 *10^7$ )

b) Note that the defined universe day will have 60*60*24=86400 universe seconds if the seconds are defined as normal seconds.

Also one universe day (or 86400 universe seconds) is equivalent to 1010 years. Now using a rule of three we can know the seconds that humans have existed.

$106 years * \frac{86,400 universe-seconds}{1,010 years}=9,067.728 s$

That is about 2 and half hours.

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3 years ago

The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

Cerrena [4.2K]

Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

$I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }$

$I = 19.799\,kg\cdot m^{2}$

b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

$\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}$

$\theta_{T} = 344.580\,rad$

The total number of revolutions of the wheel is:

$n_{T} = \frac{\theta_{T}}{2\pi}$

$n_{T} = \frac{344.580\,rad}{2\pi}$

$n_{T} \approx 54.842\,rev$

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3 years ago

Which of the following best describes the upper respiratory tract?

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A). It takes air in from outside the body.

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2 years ago

Your gender identity includes your __________. (Select all that apply.)

earnstyle [38]

The last one - biological attributes

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3 years ago