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2 years ago

A crane uses a block and tackle to lift a 2200N flagstone to a height of 25m

Physics

1 answer:

Cloud [144]2 years ago

6 0

Remember the headline: ENERGY IS NEVER CREATED OR DESTROYED

The amount of energy before and after are always equal. All we ever do with energy is move it around from one place to another.

a). A crane can't create energy. Lifting the same rock in 20 different ways always takes the same amount of work. It doesn't matter whether one person picks the rock straight up, or 50 people get around it and lift it, or roll it up a ramp, or lift it with 16 pulleys and a mile of rope, or use a giant steam crane.

You want to lift a 2200N weight up 25m, you're going to have to supply

(2200N) x (25m) = 55,000 Joules of work.

c). YOU put out 55,000 Joules of energy. It had to GO someplace. Where is it now ? ===> It's the potential energy the rock has now, from being 25m higher than it was before. That 55,000 Joules is NOW the potential energy of the rock.

No energy was created or destroyed. It just got moved around.

55,000 Joules of energy began as nuclear energy in the core of the sun. Solar radiation carried it to the Earth. Plants absorbed it, and stored it as chemical energy. You ... or a cow that you ate later ... ate the plants and took the chemical energy. One way or the other, the chemical energy got stored in your blood and fat. When you needed to put it out somewhere, you moved it into your muscles, and they converted it into mechanical energy. Then you used the mechanical energy to exert forces. Today, you used the original 55,000 joules to lift the flagstone, and NOW that energy is in the flagstone, 25 meters up off the ground !

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The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This sp

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Answer:

Explanation:

Here is the full question and answer,

The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.

When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.

Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.

Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?

Answer

A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.

Calculate the speed of the archer fish.

The time of the flight of spitted water is,

$t = \frac{{2v\sin \theta }}{g}$

Substitute $9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for g and $60^\circ$ for $\theta$ in above equation.

$t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\$

Spitted water will travel $0.80{\rm{ m}}$ horizontally.

Displacement of water in this time period is

$x = vt\cos \theta$

Substitute $\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}$ for $t\rm 60^\circ[tex] for [tex]\theta$ and $0.80{\rm{ m}}$ for x in above equation.

$\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.

Calculate the height of insect above the surface.

Vertical component of the velocity is,

${v_v} = v\sin \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for v and $60^\circ$ for $\theta$ in above equation.

$\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

Horizontal component of the velocity is,

${v_h} = v\cos \theta$

Substitute $3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}$ for v and $60^\circ$ for $\theta$ in above equation.

$\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\$

When horizontal $({0.60\;{\rm{m}}}$ distance away from the fish.

The time of flight for distance (d) is ,

$t = \frac{d}{{{v_h}}}$

Substitute $0.60\;{\rm{m}}$ for d and $1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${v_h}$ in equation $t = \frac{d}{{{v_h}}}$

$\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\$

Distance of the insect above the surface is,

$s = {v_v}t + \frac{1}{2}g{t^2}$

Substitute $2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}$ for ${v_v}$ and $0.3987{\rm{ s}}$ for t and $- 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}$ for g in above equation.

$\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\$

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2 years ago