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elena-s [515]
2 years ago
8

The first law of thermodynamics states that ___. when a process converts energy from one form to another, some energy converted

into heat, an unusable form of energy energy can neither be created nor destroyed, but it can change from one form to another energy flows through ecosystems without being recycled most energy in ecosystems comes from the Sun
Physics
2 answers:
barxatty [35]2 years ago
7 0

The first law of thermodynamics states that energy cannot be created or destroyed, but it can change from one form to another

ratelena [41]2 years ago
5 0

The answer is: energy cannot be created or destroyed under normal conditions

The first law of thermodynamics was introduced by  Rudolf Clausius and  William Rankine in the year of 1950s. It stated that The energy cannot be destroyed or created, but it can be changed/altered. 'Normal' conditions in this context refers to the condition where the energy is not being intervened by external factors in any way.

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A goose with a mass of 2.0 kg strikes a commercial airliner with a mass of 160,000 kg head-on. Before the collision, the goose w
Goryan [66]

Answer:

The change in momentum of the goose during this interaction is 33.334 m/s

Explanation:

Given;

mass of goose, m₁ = 2.0 kg

mass of commercial airliner, m₂ = 160,000 kg

initial velocity of the bird, u₁ = 60 km/hr  = 16.667 m/s

initial velocity of the airliner, u₂ = 870 km/hr = 241.667 m/s

Change in momentum is given as;

ΔP = mv - mu

where;

u is the initial velocity of the bird

v is the final velocity of the bird

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the final velocity of bird and airliner after collision;

(2 x 16.667) + (160,000 x 241.667) = v (2 + 160,000)

38,666,753.334 = 160,002v

v = 38,666,753.334 / 160,002

v = 241.664 m/s

Thus, the final velocity of the bird is negligible compared to final  velocity of the airliner.

ΔP = mv - mu

ΔP = m(v - u)

ΔP = 2(0 - 16.667)

ΔP = -33.334 m/s

The negative sign implies a deceleration of the bird after the impact.

Therefore, the change in momentum of the goose during this interaction is 33.334 m/s

4 0
2 years ago
One student did an experiment with sand, rocks, pebbles, and water. The steps of the experiment are shown below.
Lubov Fominskaja [6]
The intended observation in this experiment is to check the flow of water through the cotton cloth when pouring the water through different material (sand, rocks and pebbles).
We will observe that the rate of water flow changes depending on which material it is flowing through.
The used materials in this experiment (sand, rocks and pebbles) are the same as those forming different layers of the earth.

Therefore, based on the above explanation, we can infer that the best answer is:
<span>D. groundwater moves at different rates through different layers of the Earth</span>
3 0
2 years ago
Using energy considerations and assuming negligible air resistance, show that a rock thrown from a bridge 20.0 m above water wit
melamori03 [73]

Answer:

Explanation:

Given that,

Height of the bridge is 20m

Initial before he throws the rock

The height is hi = 20 m

Then, final height hitting the water

hf = 0 m

Initial speed the rock is throw

Vi = 15m/s

The final speed at which the rock hits the water

Vf = 24.8 m/s

Using conservation of energy given by the question hint

Ki + Ui = Kf + Uf

Where

Ki is initial kinetic energy

Ui is initial potential energy

Kf is final kinetic energy

Uf is final potential energy

Then,

Ki + Ui = Kf + Uf

Where

Ei = Ki + Ui

Where Ei is initial energy

Ei = ½mVi² + m•g•hi

Ei = ½m × 15² + m × 9.8 × 20

Ei = 112.5m + 196m

Ei = 308.5m J

Now,

Ef = Kf + Uf

Ef = ½mVf² + m•g•hf

Ef = ½m × 24.8² + m × 9.8 × 0

Ef = 307.52m + 0

Ef = 307.52m J

Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw

7 0
2 years ago
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
2 years ago
A snowboarder travels 150 m down a mountain slope that is 65 degrees above horizontal. What is his vertical displacement?
choli [55]
This can be answered using trigonometric analysis. This sloped path that is 150 m long is the hypotenuse of the triangle. The adjacent angle would then be 65 degrees. Given these:

sin 65 = h / 150

Where: h = vertical displacement = 150 (sin 65)
h = 135.95 meters
3 0
2 years ago
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