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vovikov84 [41]
3 years ago
8

If an object on a horizontal, frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If

it is displaced 0.120 m from its equilibrium position and released with zero initial speed, then after 0.800 s its displacement is found to be 0.120 m on the opposite side, and it has passed the equilibrium position once during this interval. Find (a) the amplitude; (b) the period; (c) the frequency. 14.3 . The tip of a tuning fork go
Physics
1 answer:
Rama09 [41]3 years ago
3 0

Answer:

A) 0.120

B) 1.6s

C) 0.625 Hz

Explanation:

Here, X = 0.120 m

x = 0.120 m after t = 0.800 s

A- Amplitude = max displacement from equillibrium position = 0.120 m

B- Period = 2 * 0.800 = 1.6 s

C- Frequency = 1/peroid = 1/1.6

f = 0.625 Hz

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Calculate the average induced voltage between the tips of the wings of an airplane flying above East Lansing at a speed of 885 k
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Answer:

=0.855V

Explanation:

The induced voltage can be calculated using below expression

E =B x dA/dt

Where dA/dt = area

B= magnetic field = 6.90×10-5 T.

We were given speed of 885 km/h but we will need to convert to m/s for consistency of unit

speed = 885 km/h

speed = 885 x 10^3 m/hr

speed = 885 x 10^3/60 x60 m/s

speed = 245.8 m/s

If The aircraft wing sweep out" an area

at t= 50.4seconds then we have;

dA/dt = 50.4 x 245.8

= 123388.32m^2/s

Then from the expression above

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E = 6.90 x 10^-5 x 12388.32 V

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2 years ago
En una báscula hidráulica colocamos una persona de 75 kg sobre un émbolo y un camión de 7200 kg sobre una plataforma de 5 m de l
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3 years ago
How do i solve the ones in red? A student fires a cannonball horizontally with a speed of 24.0m/s from a height of 51.0m. Neglec
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Horizontal speed = 24.0 m/s

height of the cliff = 51.0 m

For the initial vertical speed will are considering the vertical component. Therefore,

Since the student fires the canonical ball at the maximum height of 51 m, the initial vertical velocity will be zero. This means

\text{ Initial vertical velocity = 0 m/s}

let's find how long the ball remained in the air.

\begin{gathered} 0=51-\frac{1}{2}(9.8)t^2 \\ 4.9t^2=51 \\ t^2=\frac{51}{4.9} \\ t^2=10.4081632653 \\ t=\sqrt[]{10.4081632653} \\ t=3.22 \\ t=3.22\text{ s} \end{gathered}

Finally, let's find the how far from the base of the building the ball landed(horizontal distance)

\begin{gathered} S_x=u_xt+\frac{1}{2}a_xt^2 \\ S_x=24\times3.23 \\ S_x=\text{horizontal distance=77.28 m} \\ \text{note} \\ a_x=0m/s^2 \end{gathered}

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