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3 years ago

5

A uniform thin rod is hung vertically from one end and set into small amplitude oscillation. If the rod has a length of 2.6 m, t

his rod will have the same period as a simple pendulum of length ____ cm. Round your answer to the nearest whole number.

1 answer:

Alinara [238K]3 years ago

5 0

Answer:

Length of pendulum, l = 1.74 meters

Explanation:

The time period of simple pendulum is, $T=2\pi\sqrt{\dfrac{l}{g}}$

Where

l is the length of simple pendulum

The time period of uniform thin rod is hung vertically from one end is, $T=2\pi\sqrt{\dfrac{2l'}{3g}}$

l' is the length of uniform rod, l' = 2.6 m

It is given that the rod and pendulum have same time period. So,

$2\pi\sqrt{\dfrac{l}{g}}=2\pi\sqrt{\dfrac{2l'}{3g}}$

After solving above expression, the value of length of the pendulum is, l = 1.74 meters. Hence, this is the required solution.

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Which quantity may be calculated directly using Newton's second law of motion?

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Answer:

Weight

Explanation:

The second law of motion by Newton can be used to determine weight.

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2 years ago

Dario, a prep cook at an Italian restaurant, spins a salad spinner and observes that it rotates 20.0 times in 5.00 seconds and t

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Answer:

$-\frac{8\pi}{3}rad/s^2$

Explanation:

To solve this problem we need to apply the concept related to Angular Acceleration. We can find it through the equation

$\omega_f^2-\omega_i^2=2\alpha\theta$

Where for definition,

$\omega_i = \frac{\theta}{t}$

The number of revolution $(\theta)$ was given by 20 times, then

$\omega_i = \frac{20*2pi}{5}$

$\omega = 8\pi rad/s$

We know as well that the salad rotates 6 more times, therefore in angle measurements that is

$\theta = 6*2\pi rad = 12\pi rad$

The cook at the end stop to spin, then using our first equation,

$0-8\pi = 2\alpha (12\pi)$

re-arrange to solve $\alpha$ ,

$\alpha = \frac{-8\pi}{2*12\pi}$

$\alpha = -\frac{8\pi}{3}rad/s^2$

We can know find the required time,

$\omega_f-\omega_i = \alpha t$

Re-arrange to find t, and considering that $\omega_f=0$

$t= \frac{\omega_i}{\alpha}$

$t=\frac{-8\pi}{-8\pi/3}$

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3 years ago

Read 2 more answers

1. Determine the energy released per kilogram of fuel used.

yarga [219]

200 MeV of energy
E1/E2=7.61=8

U is equal to 1 kilogram or 1000 g.
There are 6.02310 23 atoms in one mole, or 235 g, of uranium. Therefore, 6.02310 23 atoms are present in 1000 g of 92/235 U.
It is understood that one atom releases 200 MeV of energy during its fission.

As a result, the energy released from the fission of one kilogram of 92/235 is given by E 2 = 6.02310 23 1000200/235 =5.10610 26 MeV E1/E2=7.61=8

In light of this, the energy released during the fusion of one kilogram of hydrogen is roughly eight times greater than the energy generated during the fission of one kilogram of uranium.

To learn more about Fission please visit -
brainly.com/question/27923750
#SPJ1

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2 years ago

Two lasers are shining on a double slit, with slit separation d. Laser 1 has a wavelength of d/20, whereas laser 2 has a wavelen

xxTIMURxx [149]

Answer:

A) first laser

B) 0.08m

C) 0.64m

Explanation:

To find the position of the maximum you use the following formula:

$y=\frac{m\lambda D}{d}$

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λ: wavelength

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d: distance between slits

A) for the first laser you use:

$y_1=\frac{(1)(d/20)(4.80m)}{d}=0.24m\\$

for the second laser:

$y_2=\frac{(1)(d/15)(4.80m)}{d}=0.32m$

hence, the first maximum of the first laser is closer to the central maximum.

B) The difference between the first maximum:

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hence, the distance between the first maximum is 0.08m

C) you calculate the second maximum of laser 1:

$y_{m=2}=\frac{(2)(d/20)(4.80m)}{d}=0.48m$

and for the third minimum of laser 2:

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Finally, you take the difference:

$1.12m-0.48m=0.64m$

hence, the distance is 0.64m

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3 years ago

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