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charle [14.2K]
1 year ago
7

In the past, salmon would swim more than 1130 km (700 mi) to spawn at the headwaters of the Salmon River in central Idaho. The t

rip took about 22 days, and the fish consumed energy at a rate of 2.0 W for each kilogram of body mass. Part A What is the total energy used by a 2.0-kg salmon while making this 22-day trip?
Physics
1 answer:
8_murik_8 [283]1 year ago
6 0

Answer:

E_t_o_t_a_l=7.603MJ

Explanation:

The salmon's total energy use , which is equivalent to work done,W, by swimming upstream, is related to the specific mechanical power. Mechanical \ power per unit mass is defined by the equation:

\frac{p}{m}=\frac{1}{m}|\frac{dW}{dt}|=2W/kg\\\frac{1}{2}|\frac{W}{22\times 24 \times 60\times 60}|=2\\\\W=7.603MJ

Hence, the energy total for the 22-day trip is 7.603MJ

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If two waves with identical crests and troughs meet, what is happening? The wave is reflecting. Constructive interference is occ
Kazeer [188]
The correct answer would be that destructive interference is happening. In this interference, the crest of a wave meets a trough of another wave resulting to an amplitude that is lower. The opposite is called the constructive interference. Hope this answers the question.
7 0
1 year ago
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Kimonoski takes a 9-minute shower every day. The shower uses about 1.8 gal per minute of water. He also uses 23 gallons of hot w
ioda

Answer:

Q_{week} = 458884.6\, BTU

Explanation:

The weekly water consumption of Kimonoski is:

m_{bath,week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (1.8\,\frac{gal}{min} )\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (\frac{1\,min}{60\,s} )\cdot (9\,min)\cdot (\frac{60\,s}{1\,min} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)

m_{bath.week} = 948.205\,lbm

m_{others, week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (23\,gal)\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)

m_{others, week} = 1346.218\,lbm

m_{week} = m_{bath,week} + m_{others, week}

m_{week} = 2294.423\,lbm

The total energy required per week for hot water is:

Q_{week} = m_{week}\cdot c_{p,water}\cdot \Delta T

Q_{week} =(2294.423\,lbm)\cdot (1\,\frac{BTU}{lbm\cdot ^{\textdegree}F} )\cdot (50^{\textdegree}F)

Q_{week} = 458884.6\, BTU

3 0
1 year ago
Physics help, please? :)
Svetllana [295]

Answer:

Explanation:

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4. The force of Earth’s gravity on the Sun is weaker than the force of the Sun’s gravity on Earth. The Sun’s attraction  affects the motion of Earth more than the Earth’s attraction affects the Sun’s motion because according to  Newton’s second law, force has mass as one of its factors. The Sun has a significantly higher mass than Earth,  meaning that its force of gravity would also be significantly higher. Newton’s third law is why the Earth doesn’t get  marginally closer to the Sun, stating that every action has an equal and opposite reaction. As the Sun is pulling  Earth towards itself, Earth is pulling away from the Sun.

8 0
1 year ago
You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac
NemiM [27]

Answer:

The horizontal range of the projectile = 26.63 meters

Explanation:

Step 1: Data given

Distance above the planet's surface = 630 km = 630000

The ship's orbal speed = 4900 m/s

Radius of the planet = 4.48 *10^6 m

Initial speed of the projectile = 13.6 m/s

Angle = 30.8 °

Step 2: Calculate g

g= GM /R² = (v²*(R+h)) /(R²)

⇒ with v= the ship's orbal speed = 4900 m/S

⇒ with R = the radius of the planet = 4.48 *10^6 m

⇒ with h = the distance above the planet's surface = 630000 meter

g = (4900² * ( 4.48*10^6+ 630000)) / ((4.48*10^6)²)

g = 6.11 m/s²

<u>Step 3:</u> Describe the position of the projectile

Horizontal component: x(t) = v0*t *cos∅

Vertical component: y(t) = v0*t *sin∅ -1/2 gt² ( will be reduced to 0 in time )

⇒ with ∅ = 30.8 °

⇒ with v0 = 13.6 m/s

⇒ with t= v(sin∅)/g = 1.14 s

Horizontal range d = v0²/g *2sin∅cos∅  = v0²/g * sin2∅

Horizontal range d =(13.6²)/6.11 * sin(2*30.8)

Horizontal range d =26.63 m

The horizontal range of the projectile = 26.63 meters

6 0
1 year ago
Water is projected from two rubber pipes at the same speed from one at an angle of 30°and from the other at 60°.why are the rang
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Thats hard...im not sure
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