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3 years ago

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An object of mass 0.93 kg is initially at rest. When a force acts on it for 2.9 ms it acquires a speed of 16.9 m/s. Find the mag

nitude of the average force acting on the object during the 2.9 ms time interval.

1 answer:

natulia [17]3 years ago

7 0

Use force equals mass times change in speed divided by time taken which is Newton’s second law.

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Un caballo parte del reposo y alcanza una velocidad de 15m/s en un tiempo de 8s. ¿ Cuál fue su aceleración y que distancia recor

Lemur [1.5K]

Answer: A 120 metros por segundo

Explanation: multiplicas la velocidad por el tiempo

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2 years ago

A 0.40 kg mass hangs on a spring with a spring constant of 12 N/m. The system oscillated with a constant amplitude of 12 cm. Wha

Vaselesa [24]

Answer:

The maximum acceleration of the system is 359.970 centimeters per square second.

Explanation:

The motion of the mass-spring system is represented by the following formula:

$x(t) = A\cdot \cos (\omega \cdot t + \phi)$

Where:

$x(t)$ - Position of the mass with respect to the equilibrium position, measured in centimeters.

$A$ - Amplitude of the mass-spring system, measured in centimeters.

$\omega$ - Angular frequency, measured in radians per second.

$t$ - Time, measured in seconds.

$\phi$ - Phase, measured in radians.

The acceleration experimented by the mass is obtained by deriving the position equation twice:

$a (t) = -\omega^{2}\cdot A \cdot \cos (\omega\cdot t + \phi)$

Where the maximum acceleration of the system is represented by $\omega^{2}\cdot A$ .

The natural frequency of the mass-spring system is:

$\omega = \sqrt{\frac{k}{m} }$

Where:

$k$ - Spring constant, measured in newtons per meter.

$m$ - Mass, measured in kilograms.

If $k = 12\,\frac{N}{m}$ and $m = 0.40\,kg$ , the natural frequency is:

$\omega = \sqrt{\frac{12\,\frac{N}{m} }{0.40\,kg} }$

$\omega \approx 5.477\,\frac{rad}{s}$

Lastly, the maximum acceleration of the system is:

$a_{max} = \left(5.477\,\frac{rad}{s})^{2}\cdot (12\,cm)$

$a_{max} = 359.970\,\frac{cm}{s^{2}}$

The maximum acceleration of the system is 359.970 centimeters per square second.

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3 years ago

Two pianos each sound the same note simultaneously, but they are both out of tune. On a day when the speed of sound is 349 m/s,

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Answer:

Time period between the successive beats will be 0.1703 sec

Explanation:

We have given speed of the sound v = 349 m/sec

Wavelength of piano $A\lambda _A=0.766m$

Wavelength of piano $B\lambda _B=0.776m$

So frequency of piano A $f_1=\frac{v}{\lambda _1}=\frac{349}{0.766}=455.61Hz$

Frequency of piano B $f_2=\frac{v}{\lambda _1}=\frac{349}{0.776}=449.74Hz$

So beat frequency f = 455.61 - 449.74 = 5.87 Hz

So time period $T=\frac{1}{f}=\frac{1}{5.87}=0.1703sec$

So time period between the successive beats will be 0.1703 sec

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3 years ago

A 19kg block is being pulled with a constant horizontal force of 95 Newton’s while also experiencing a constant friction force o

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Answer:

A

⋅(19kg)=19Akg

95=19Akg

19=19Ak

Explanation:

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3 years ago

How will creat thunderstrom

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Answer:

the air has to be unstable as well as it needs to be moved upwards.

Explanation:

it needs to be moved upwards and also needs to have unstable air.

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2 years ago