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3 years ago

The cable of a crane is lifting a 950 kg girder. The girder increases its speed from 0.25 m/s to 0.75 m/sin a distance of 5.5m

Physics

1 answer:

Umnica [9.8K]3 years ago

7 0

Explanation:

a) How much work is done by gravity?

w = f x d
w = 950 x 10 x 5.5 = 52250j

b) How much work is done by tension?

v²=u²+2as
0.75²=0.25²+2a x5.5
0.56=0.06+2a x5.5
2a x5.5 = 0.56 - 0.06
2a x 5.5 =0.5
11a=0.5
a = 0.5/11 = 0.05m/s²

w = f x d

w = 950 x 0.05 x 5.5 = 261.25j

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Define limiting friction.

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The maximum friction that can be generated between two static surfaces in contact with each other. Once a force applied to the two surfaces exceeds the limiting friction, motion will occur. For two dry surfaces, the limiting friction is a product of the normal reaction force and the coefficient of limiting friction.

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A liquid of density 1270 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the f

FinnZ [79.3K]

Answer:

${P_2}-P_1=49.99\ KPa$

Explanation:

$v_1=9.43\ m/s$

$d_1=11.7\ cm/s$

$d_2=17.5\ cm/s$

From continuity equation

$A_1v_1=A_2v_2$

$v_1d_1^2=v_2d_2^2$

$v_2=\dfrac{9.43\times 11.7^2}{17.5^2}$

$v_2=4.21\ m/s$

$\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+y_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+y_2$

${P_1}+\rho\dfrac{v_1^2}{2}+\rho y_1g={P_2}+\rho\dfrac{v_2^2}{2}+\rho y_2g$

$\rho\dfrac{v_1^2}{2}+\rho y_1g -\rho\dfrac{v_2^2}{2}-\rho y_2g={P_2}-P_1$

$1270\times \dfrac{9.43^2}{2}+1270\times 0\times 10 -1270\times\dfrac{4.21^2}{2}-1270\times 0.175\times 10={P_2}-P_1$

${P_2}-P_1=49.99\ KPa$

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3 years ago

While visiting the Albert Michelson exhibit at Clark University, you notice that a chandelier (which looks remarkably like a sim

Fudgin [204]

Answer:

a) 0,1613 Hz

b) 1,01342 rad/sec

c) 9.5422 m

d) $9.4314 m/sec^2$

Explanation:

In the Albert Michelson exhibit at Clark University, we know the period of oscillation of the chandelier is T = 6.2 seconds

The chandelier will be modeled as a simple pendulum

a) Since the frequency is the reciprocal of the period, we have

$f=\frac{1}{T}=\frac{1}{6.2sec}=0,1613 Hz$

b) The angular frequency is computed as

$w=2\pi f=2\pi (0,1613) = 1,01342\ rad/sec$

c) The period of a simple pendulum is given by

$T=2\pi \sqrt{\frac{L}{g}}$

Where L is its length and g is the local acceleration of gravity (assumed 9.8 $m/sec^2$ for this part)

We want to know the length of the pendulum, so we isolate L

$L=\frac{T^2g}{4\pi^2}$

$L=\frac{(6.2)^2(9.8))}{4\pi^2}=9.5422 m$

d) While hanging out in J.J. Thompson’s House O’ Blues, the new period is 6.2+0.11=6.32 sec. Since the chandelier is the very same (same length), we can assume the gravity is slightly different. We use again the formula for T, but now we'll isolate g as follows

$g=\frac{4\pi^2L}{T^2}=\frac{4\pi^2(9.5422m))}{(6.32sec)^2}$

Which results

$g=9.4314\ m/sec^2$

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4 years ago

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Answer:

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Explanation:

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3 years ago

You want to photograph a circular diffraction pattern whose central maximum has a diameter of 1.0 cm . You have a helium-neon la

Nataly [62]

Answer:

71 cm

Explanation:

The formula to apply here is θ = 1.22λ/D,

diameter of central maximum =1.0 cm

wavelength, λ= 633 nm

Diameter of pinhole = 0.11 mm

where θ = arc length/distance = s/R

s/R = 1.22λ/D

(.005)/R = (1.22)(633 X 10^-9)/(1.1 X 10^-4)

R = .71 m which is 71 cm

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3 years ago