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nataly862011 [7]

2 years ago

8

At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the le

ft. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant.

1 answer:

Nana76 [90]2 years ago

6 0

Answer:

Acceleration=24.9ft^2/s^2

Angular acceleration=1.47rads/s

Explanation:

Note before the ladder is inclined at 30° to the horizontal with a length of 16ft

Hence angular velocity = 6/8=0.75rad/s

acceleration Ab=Aa +(Ab/a)+(Ab/a)t

4+0.75^2*16+a*16

0=0.75^2*16cos30°-a*16sin30°---1

Ab=0+0.75^2sin30°+a*16cos30°----2

Solving equation 1

(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s

Also from equation 2

Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2

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A 18.0−μF capacitor is placed across a 22.5−V battery for a few seconds and is then connected across a 12.0−mH inductor that has

Gelneren [198K]

Answer:

Part a)

$i = 10.4 mA$

Part b)

in this case the charge on the capacitor will become zero

Part c)

$t_1 = 0.73 ms$

Part d)

$t = 2.2 ms$

Explanation:

As we know that first capacitor is charged with the battery and then it is connected to the inductor

So here we will have

$Q = CV$

$Q = (18\mu F)(22.5 V)$

$Q = 405 \mu C$

Part a)

now since the total energy of capacitor is converted into the energy of inductor

so by energy conservation we can say

$\frac{Q^2}{2C} = \frac{1}{2}Li^2$

so maximum current is given as

$i = \sqrt{\frac{L}{C}}Q$

$i = \sqrt{\frac{12\times 10^{-3}}{18\times 10^{-6}}}(405\times 10^{-6})$

$i = 10.4 mA$

Part b)

When current is maximum then whole energy of capacitor is converted into magnetic energy of inductor

So in this case the charge on the capacitor will become zero

Part c)

Time period of oscillation of charge between the plates and inductor is given as

$T = 2\pi\sqrt{LC}$

$T = 2\pi\sqrt{(18\mu F)(12 mH)}$

$T = 2.92 ms$

now capacitor gets discharged first time after 1/4 of total time period

$t_1 = 0.73 ms$

Part d)

Since time period is T and capacitor gets discharged two times in one complete time period of the motion

so first it will discharges in T/4 time

then next T/4 it will get charged again

then next T/4 time it will again discharged

so total time taken

$t = 3T/4$

$t = 2.2 ms$

4 0

1 year ago

A steel cable 1.25 in. in diameter and 50 ft long is to lift a 20-ton load without permanently deforming. What is the length of

Over [174]

Answer:

50.0543248872 ft

Explanation:

F = Load = 20 ton = $20\times 2000\ lb$

d = Diameter = 1.25 in

$L_1$ = Initial length = 50 ft

$L_2$ = Final length

A = Area = $\dfrac{\pi}{4}d^2$

Y = Young's modulus = $30\times 10^6\ psi$

Young's modulus is given by

$Y=\dfrac{FL}{A\Delta L}\\\Rightarrow Y=\dfrac{FL_1}{\dfrac{\pi}{4}d^2(L_2-L_1)}\\\Rightarrow L_2=\dfrac{4FL_1}{Y\pi d^2}+L_1\\\Rightarrow L_2=\dfrac{4\times 40000\times 50}{30\times 10^6\times \pi\times 1.25^2}+50\\\Rightarrow L_2=50.0543248872\ ft$

The length during the lift is 50.0543248872 ft

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2 years ago

A piece of wood that floats on water has a mass of 0.0175 kg. A lead weight is tied to the wood, and the apparent mass with the

crimeas [40]

Answer:

Specific gravity is 0.56

Explanation:

We know that

mass of water displaced by the wood is, m1( apparent mass when wood in air and lead is submerged in water) - m2(the apparent mass when wood and lead both are submerged in water)

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So the specific gravity of the wood is, = mass of wood / mass of water displaced by the wood

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=0.56

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2 years ago

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sweet [91]

Answer:

Speed of the electron will be $v=5.896\times 10^7m/sec$

Explanation:

We have given that charge on electron $e=1.6\times 10^{-19}C$

Mass of electron $m=9.11\times 10^{-31}kg$

Potential difference = $V=9.9KV=9.9\times 10^3volt$

Now according to energy conservation $eV=\frac{1}{2}mv^2$

$1.6\times 10^{-19}\times 9.9\times 10^3=\frac{1}{2}\times 9.11\times 10^{-31}v^2$

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