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3 years ago

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Two billiard balls (each with mass equal to 170 g) collide head-on along the same line. Billiard ball A originally traveled east

ward at 8 m/s while billiard ball B originally traveled westward at 2 m/s. Calculate the speed and direction of each ball after the collision.

1 answer:

barxatty [35]3 years ago

5 0

Answer:

lucky mauld mauldgomary was an british poet...

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The one one of the times

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The highest waterfall in the world is the Salto Angel in Venezuela. Its longest single falls has a height of 807 m. If water at

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Answer:

Temperature at the bottom will be 19.18°C

Explanation:

We have given height h = 807 m

Temperature at the top $=17.3^{\circ}C$

Specific heat of water c = 4200 $J/kg/^{\circ}C$

From energy conservation

Kinetic energy at the bottom = potential energy at the top

So $mc\Delta T=mgh$

$\Delta T=\frac{gh}{c}=\frac{9.8\times 807}{4200}=1.88^{\circ}C$

So temperature at the bottom = 17.3+1.88 = 19.18°C

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How fast must a bug swim to keep up with the waves it produces? How fast must it move to produce a bow wave?

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Answer:

A bug must swim as fast as the wave speed to keep up with the waves it produces. Moreso, a boat must be moving faster than the waves it creates to produce a bow wave.

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3 years ago

A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.

SVETLANKA909090 [29]

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

$W=mg$ : weight of the ladder, with m = 20 kg (mass) and $g=9.8 m/s^2$ (acceleration of gravity)

$W_M=Mg$ : weight of the person, with M = 54 kg (mass)

$N_1$ : normal reaction exerted by the wall on the ladder

$N_2$ : normal reaction exerted by the floor on the ladder

$F_f = \mu N_2$ : force of friction between the floor and the ladder, with $\mu$ (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

$W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}$

where

$Wsin 21^{\circ}$ is the component of the weight of the ladder perpendicular to the ladder

$W_M sin 21^{\circ}$ is the component of the weight of the man perpendicular to the ladder

$N_1 sin 69^{\circ}$ is the component of the normal force perpendicular to the ladder

And solving for $N_1$ , we find the force exerted by the wall on the ladder:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N$

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of $N_2$ .

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

$\sum F_y = 0\\N_2 - W - W_M =0$

And substituting and solving for N2, we find:

$N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N$

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

$\sum F_x = 0\\F_f - N_1 = 0$

And re-writing the equation,

$\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257$

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

$\mu = \frac{N_1}{N_2}$

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}$

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

$N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)$

And substituting, we get

$N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N$

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

$\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332$

Learn more about torques and equilibrium:

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3 years ago

Air, considered an ideal gas, is contained in an insulated piston-cylinder assembly outfitted with a paddle wheel. It is initial

Maru [420]

Our data are,

State 1:

$P_1= 10psi=68.95kPa\\V_1 = 1ft^3=0.02831m^3\\T_1 = 100\°F = 310.93K$

State 2:

$P_2 =5psi=34.474kPa\\V_2 = 3ft^3=0.0899m^3$

We know as well that $3BTU=3.16kJ/K$

To find the mass we apply the ideal gas formula, which is given by

$P_1V_1=mRT_1$

Re-arrange for m,

$m= \frac{P_1V_1}{RT_1}\\m= \frac{68.95*0.02831}{(0.287)310.9}\\m=0.021893kg=0.04806lbm\\$

Because of the pressure, temperature and volume ratio of state 1 and 2, we have to

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Replacing,

$T_2 = \frac{P_2V_2}{P_1V_1}T_1\\T_2 =\frac{34.474*0.0844}{68.95*0.02831}*310.93\\T_2 = 464.217K=375.5\°F$

For conservative energy we have, (Cv = 0.718)

$W = m C_v = 0.718 \Delta T +dw\\dw = W - mv\Delta T\\dw = 3.16-(0.0218*0.718)(454.127-310.93)\\dw = 0.765kJ=0.72BTU$

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3 years ago