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Blizzard [7]
1 year ago
10

A superman cyclist rode a bike uphill at 20 miles/hour for two hours. To sustain this constant speed the cyclist was exerting 50

pounds. (a) How much work has the cyclist done during the trip?
Physics
1 answer:
NeX [460]1 year ago
5 0

Answer:

1.056 x 10⁷ lb-ft

Explanation:

v = Speed of the bike = 20 mph

t = time of travel = 2 h

d = distance traveled by cyclist

Distance traveled by cyclist is given as

d = v t

d = (20) (2)

d = 40 miles

We know that, 1 mile = 5280 ft

d = 40 (5280) ft

d = 211200 ft

F = force applied by cyclist = 50 lb

W = work done by cyclist

Work done by cyclist is given as

W = F d

W = (50) (211200)

W = 1.056 x 10⁷ lb-ft

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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
Delvig [45]

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}}\\

7 0
1 year ago
You are working as an assistant to an air-traffic controller at the local airport, from which small airplanes take off and land.
Alika [10]

Answer:

d = 2021.6 km

Explanation:

We can solve this distance exercise with vectors, the easiest method s to find the components of the position of each plane and then use the Pythagorean theorem to find distance between them

Airplane 1

Height   y₁ = 800m

Angle θ = 25°

           cos 25 = x / r

           sin 25 = z / r

           x₁ = r cos 20

           z₁ = r sin 25

          x₁ = 18 103 cos 25 = 16,314 103 m = 16314 m

          z₁ = 18 103 sin 25 = 7,607 103 m= 7607 m

2 plane

Height   y₂ = 1100 m

Angle θ = 20°

          x₂ = 20 103 cos 25 = 18.126 103 m = 18126 m

          z₂ = 20 103 without 25 = 8.452 103 m = 8452 m

The distance between the planes using the Pythagorean Theorem is

         d² = (x₂-x₁)² + (y₂-y₁)² + (z₂-z₁)²2

Let's calculate

        d² = (18126-16314)²  + (1100-800)² + (8452-7607)²

        d² = 3,283 106 +9 104 + 7,140 105

        d² = (328.3 + 9 + 71.40) 10⁴

        d = √(408.7 10⁴)

        d = 20,216 10² m

        d = 2021.6 km

7 0
2 years ago
Starting from equilibrium at point 0, what point on the pv diagram will describe the ideal gas after the following process? lock
Anna71 [15]
Since the product of P·Vis constant along an isotherm, an expansion to twice the volume implies a pressure reduction to half the original pressure. I hope my answer has come to your help. God bless and have a nice day ahead!<span>
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3 0
1 year ago
Read 2 more answers
On a guitar, the lowest toned string is usually strung to the E note, which produces sound at 82.4 82.4 Hz. The diameter of E gu
Vsevolod [243]

The complete and comprehensive solution is attached.

8 0
1 year ago
After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r
mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

2 n t = (m + \dfrac{1}{2})\lambda

At m = 0 for the minimum thickness, so

2\times 1.45 \times t = (0 + 0.5)\times 600

2.9 \times t =300

t =\dfrac{300}{2.9}

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0
2 years ago
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