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snow_tiger [21]
3 years ago
11

What is the total energy of a 850 kg sea plane flying at 38 m/s and 20 m above the ocean?

Physics
1 answer:
Nikolay [14]3 years ago
3 0

Answer: 7.8×10^5 J

Explanation:

Total energy = kinetic energy of plane + gravitational potential energy of plane.

Total energy = ((1/2)×m×v²) + (m×g×h)

Here, m is mass of plane, v is velocity of plane, g is acceleration due to gravity and h is height above the ocean.

Now,

Total energy = ((1/2)×m×v²) + (m×g×h)

Total energy = m((v²/2) + (g×h))

Total energy = 850((38²/2) + (9.81×20))

Total energy = 850(722 + 196.2)

Total energy = 7.8×10^5 J

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Suppose you design a new thermometer called the "x" thermometer. on the x scale, the boiling point of water is 130.0 ox and the
Hoochie [10]

You've told us:

-- 130°x  =  212°F

and

-- 10°x  =  32°F

Thank you.  Those are two points on a graph of °x vs °F .  With those, we can figure out the equation of the graph, and easily convert ANY temperature on one scale to the equivalent temperature on the other scale.

-- If our graph is going to have °x on the horizontal axis and °F on the vertical axis, then the two points we know are  (130, 212)  and  (10, 32) .

-- The slope of the line through these two points is

Slope = (32 - 212) / (10 - 130)

Slope = (-180) / (-120)

Slope = 1.5

So far, the equation of the graph is

F = 1.5 x + (F-intercept)

Plug one of the points into this equation.  I'll use the second point  (10, 32) just because the numbers are smaller:

32 = 1.5 (10) + F-intercept

32 = 15 + (F-intercept)

F-intercept = 17

So the equation of the conversion graph is

F = 1.5 x + 17

There you are !  Now you can plug ANY x temperature in there, and the F temperature jumps out at you.

The question is asking what temperature is the same on both scales. This seems tricky, but it's not too bad.  Whatever that temperature is, since it's the same on both scales, you can take the conversion equation, and write the same variable in BOTH places.

We can write [ x = 1.5x + 17 ], solve it for  x, and the solution will be the same temperature in  F  too.

or

We can write [ F = 1.5F + 17 ], solve it for  F, and the solution will be the same temperature in  x  too.

F = 1.5F + 17

Subtract  F  from each side:  0.5F + 17 = 0

Subtract 17 from each side:   0.5F = -17

Multiply each side by 2 :  F = -34

That should be the temperature that's the same number on both scales.

Let's check it out, using our handy-dandy conversion formula (the equation of our graph):

F = 1.5x + 17

Plug in -34 for  x:  

F = 1.5(-34) + 17

F = -51 + 17

<em>F = -34</em>

It works !  -34 on either scale converts to -34 on the other one too. If the temperature ever gets down to -34, and you take both thermometers outside, they'll both read the same number.

<em>yay !</em>

6 0
3 years ago
a) Calculate the magnitude of displacement of the car in 40 seconds. b) During which part of the journey was the car acceleratin
Zielflug [23.3K]

Answer:

a)  600 meters

b) between 0 and 10 seconds, and between 30 and 40 seconds.

c) the average of the magnitude of the velocity function is 15 m/s

Explanation:

a) In order to find the magnitude of the car's displacement in 40 seconds,we need to find the area under the curve (integral of the depicted velocity function) between 0 and 40 seconds. Since the area is that of a trapezoid, we can calculate it directly from geometry:

Area \,\,Trapezoid=(\left[B+b]\,(H/2)\\displacement= \left[(40-0)+(30-10)\right] \,(20/2)=600\,\,m

b) The car is accelerating when the velocity is changing, so we see that the velocity is changing (increasing) between 0 and 10 seconds, and we also see the velocity decreasing between 30 and 40 seconds.

Notice that between 10 and 30 seconds the velocity is constant (doesn't change)  of magnitude 20 m/s, so in this section of the trip there is NO acceleration.

c) To calculate the average of a function that is changing over time, we do it through calculus, using the formula for average of a function:

Average\,of\,f(x)=\frac{1}{b-a} \int\limits^b_a {f(x)} \, dx

Notice that the limits of integration for our case are 0 and 40 seconds, and that we have already calculated the area under the velocity function (the integral) in step a), so the average velocity becomes:

Avearage=\frac{600\,\,m}{40\,\,s}= 15\,\,\frac{m}s}

7 0
3 years ago
What is the goal of the scientific method
IgorLugansk [536]

Answer:

Regardless of how the steps are documented, the goal of scientific method is to gather data that will validate or invalidate a cause and effect relationship.

Hope this helped!!!

7 0
3 years ago
A ball is thrown from ground level so as to just clear wall 4m height at distance 4m from wall and falls 14 m from wall . Veloci
kolbaska11 [484]
Ok, this is a 2d kinematics problem,  the falls 14 m part is confusing, I think it means in the x direction, but you don't need it anyway.

If we know it goes 4m into the air, we know  d = 4m (height of wall), we also know the acceleration a=-9.8m/s^2 (because gravity) and that the vertical velocity when it just clears the wall will be 0 m/s, which we'll call our final velocity (Vf). Using Vf^2 = Vi^2 +2a*d, we can solve this for Vi and drop Vf because it's zero to get: Vi = sqrt(-2ad), plug in numbers (don't forget a is negative) and you get 8.85 m/s in the vertical direction. The x-direction velocity requires that we solve the y-direction for time, using Vf= Vi + at, we solve for t, getting t= -Vi/a, plug in numbers t= -8.85/-9.8 = 0.9 s. Now we can use the simple v = d/t (because x-direction has no acceleration (a=0)), and plug in the distance to the wall and the time it takes to get there v = (4/.9) = 4.444 m/s, this is the velocity in the x direction, we use Pythagoras' theorem to find the total velocity, Vtotal = sqrt(Vx^2 + Vy^2), so Vtotal = sqrt(8.85^2+4.444^2) = 9.9m/s. Yay physics!
8 0
3 years ago
Nearly all conventional piston engines have flywheels on them to smooth out engine vibrations caused by the thrust of individual
Artyom0805 [142]

Answer:

This is because, the flywheel has a very large moment of Inertia and hence sudden piston torques have negligible effect on the flywheel, but every piston combined has a significant torque. This smoothens out the vibrations.

Explanation:

7 0
3 years ago
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