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jekas [21]
3 years ago
8

With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons

of fuel in its tank? The slope is 5.7
Physics
1 answer:
Shtirlitz [24]3 years ago
3 0

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

m=\frac{Y-Y_{1}}{X-X_{1}}  (1)

Where:

m=5.7 is the slope of the line

Y_{1}=2390.7pounds is the airplane weight with  51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

X_{1}=51gallons is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

(X_{1},Y_{1})=(51,2390.7)

Rewritting (1):

Y=m(X-X_{1})+Y_{1}  (2)

As Y is a function of X:

Y=f_{(X)}=m(X-X_{1})+Y_{1}  (3)

Substituting the known values:

f_{(X)}=5.7(X-51)+2390.7  (4)

f_{(X)}=5.7X-290.7+2390.7  (5)

f_{(X)}=5.7X+2100  (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

f_{(81)}=5.7(81)+2100  (7)

f_{(81)}=2561.7  (8)   This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

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4 0
4 years ago
Two large metal plates of area 0.7 m2 face each other. They are 5.1 cm apart and have equal but opposite charges on their inner
german

Explanation:

The given data is as follows.

    Area (A) = 0.7 m^{2}.

  Electric field between the plates, (E) = 55 N/C

Since, electric field is related to surface charge density as follows.

          \sigma = \frac{Q}{A}

Also,    E = \frac{\sigma}{\epsilon_{o}}

or,        \sigma = \epsilon_{o} E

Therefore, charge will be calculated as follows.

          Q = \sigma \times A = \epsilon_{o} E \times A

             = 8.854 \times 10^{-12} \times 55 \times 0.7

             = 3.41 \times 10^{-10} C

             = 341 pC

Thus, we can conclude that magnitude of the charge on each plate is 341 pC.

8 0
4 years ago
If an athlete expends 3480. kJ/h, how long does she have to play to work off 1.00 lb of body fat? Note that the nutritional calo
Sunny_sXe [5.5K]

Answer:

The time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute

Explanation:

1 lb of body fat = 4.1 k cal

1 k cal = 4.184 Kilo joule

1 lb of body fat = 4.1 × 4.184 = 17.1544 Kilo joule

Athlete expends 3480 Kilo joule in one hour

⇒ Time required to expand 3480 Kilo joule  = 60 minute

⇒ Time required to expand 1 Kilo joule = \frac{60}{3480} \frac{min}{KJ}

⇒ Time required to expand 17.1544 Kilo joule = \frac{60}{3480} × 17.1544 = 0.296 min

Therefore the time required by the Athlete to work off 1.00 lb of body fat = 0.296 minute

6 0
3 years ago
If the amplitude of a sound wave is tripled, by what factor will the intensity increase?
Shtirlitz [24]

Answer:

By a factor 9

Explanation:

The intensity of a sound wave is proportional to the square of the amplitude of the wave:

I \propto A^2

where

I is the intensity

A is the amplitude of the wave

In this problem, the amplitude of the sound wave is increased by a factor 3:

A' = 3A

So the intensity would change by

I' \propto A'^2 = (3A)^2 = 9 A^2 = 9I

So, the intensity would increase by a factor 9.

5 0
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What is the density of a box measuring 100 grams and 10 ml?
Alisiya [41]

Answer:

In order to convert density to grams, you have to put the mass on one side of the equation, and the density and the volume on the other. Therefore, d * v = m. Multiply the density by the volume. Using the example in step 1, you would multiply 2 g/mL by 4mL.

Explanation:

ok

6 0
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