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2 years ago

A cable that weighs 4 lb/ft is used to lift 1000 lb of coal up a mine shaft 700 ft deep. Find the work done.

Physics

1 answer:

Shkiper50 [21]2 years ago

6 0

Answer:

Explanation:

Step one:

given data

mass of cable= 4lb/ft

mass of coal= 1000lb

dept of mine= 700ft

Step two:

Required

the work-done to lift the coal and the rope combined

Work-done to lift coal

Wc=1000*700= 700,000 lb-ft

Work-done to lift rope

$Wr=\int\limits^{700} _0 {4(700-y)} \, dx \\\\Wr=4(700y-\frac{1}{2}y^2 )\limits^{700}_0$

substitute y=700 we have, since y=0 will result to 0

$Wr=4(700*700-\frac{1}{2}*700^2 )\\\\Wr=4(490000-245000)\\\\Wr=4(245000)\\\\Wr=980000ft-lbs$

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3 years ago

The speed of sound in aluminum is 5200 m/s. Can you hear a sound with a

romanna [79]

Answer:

v = wavelength * frequency

frequency = 5200 m/s / .2 m = 26000 / sec

20,000 / sec is optimistic for the upper frequency of human hearing

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2 years ago

An airplane is heading south at a speed of 600km/h. If a wind begins blowing from the southwest at a speed of 100km/h, calculate

Karo-lina-s [1.5K]

Answer:

Here's what I got:

Let's assume that N and E are + directions while S and W are - directions.

Wind is blowing from SW; thus, it is blowing towards NE (or at 45 deg N of E).

Dividing the wind's speed into components:y-component: +70.71 km/h; x-component: +70.71 km/h

Dividing the airplane's speed into components:y-component: -600 km/h; x-component: 0 km/h

Adding the components to get the resulting components:y-component: -529.29 km/h; x-component: +70.71

Using the Pythagorean Theorem to find the resulting speed:v^2 = y^2 + x^2 so v = 533.99 km/h

To find the angle of direction, use arctan (y/x):arctan (529.29/70.71) = 82.39 deg

ANSWER: velocity = 533.99 km/h at 82.39 deg S of E

Explanation:

8 0

3 years ago

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .