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3 years ago

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Two bricks are resting on the edge of the lab table. Shirley Sheshort stands on her toes and spots the two bricks. She acquires

an intense desire to know which of the two bricks are most massive. Since Shirley is vertically challenged, she is unable to reach high enough and lift the bricks; she can however reach high enough to give the bricks a push. Discuss how the process of pushing the bricks will allow Shirley to determine which of the two bricks is most massive. What difference will Shirley observe and how can this observation lead to the necessary conclusion?

1 answer:

PSYCHO15rus [73]3 years ago

4 0

Explanation:

since force =mass×acceleration and acceleration being g which is = 10m/s^2
the brick with the larger mass will reach the ground first since it has a greater force to resist resistance forces such as air resistance

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A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres

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Answer:

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putting all value

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$150\times 0.08^{1.4} = 80 \times V_2^{1.4}$

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Read 2 more answers

A mass free to vibrate on a level, frictionless surface at the end of a horizontal spring is pulled 35 cm from its equilibrium p

saul85 [17]

Answer:

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The displacement, $x$ , of an SHM is given by

$x = A\cos(\omega t)$

A is the amplitude and $\omega$ is the angular frequency.

We could use a sine function, in which case we will include a phase angle, to indicate that the oscillation began from a non-equilibrium point. We are using the cosine function for this particular case because the oscillation began from an extreme end, which is one-quarter of a single oscillation, when measured from the equilibrium point. One-quarter of an oscillation corresponds to a phase angle of 90° or $\frac{\pi}{4}$ radian.

From trigonometry, $\sin A =\cos B$ if A and B are complementary.

At $t = 0$ , $x = 3.5$

$3.5 = A\cos(\omega \times0)$

$A =3.5$

So

$x = 3.5\cos(\omega t)$

At $t = 0.12$ , $x = 1.5$

$1.5 = 3.5\cos(0.12\omega)$

$\cos(0.12\omega)=\dfrac{1.5}{3.5}=0.4286$

$0.12\omega =\cos^{-1}0.4286$

$0.12\omega = 1.13$

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The period, $T$ , is related to $\omega$ by

$T = \dfrac{2\pi}{\omega} = \dfrac{2\times3.14}{9.4}=0.67$

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