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levacccp [35]
3 years ago
13

The gauge pressure in your car tires is 2.50×105N/m2 at a temperature of 35.0ºC when you drive it onto a ferry boat to Alaska. W

hat is their gauge pressure later, when their temperature has dropped to –40.0ºC
Physics
1 answer:
vlada-n [284]3 years ago
8 0

Answer:

1.645 x 10^5 Pa

Explanation:

Let Po be the atmospheric pressure. Po = 1.01 x 10^5 Pa

P1 = 2.5 x 10^5 + Po = 2.5 x 10^5 + 1.01 x 10^5 = 3.51 x 10^5 Pa

T1 = 35 degree C = 35 + 273 = 308 k

T2 = - 40 degree C = - 40 + 273 = 233 k

Let the gauge pressure be P

So, P2 = P + Po

use

P1 / T1 = P2 / T2

3.51 x 10^5 / 308 = P2 / 233

P2 = 2.655 x 10^5

P + Po = 2.655 x 10^5

P = 2.655 x 10^5 - 1.01 x 10^5 = 1.645 x 10^5 Pa

Thus, the gauge pressure is 1.645 x 10^5 Pa.

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The density of the material would be 4.1 g/cm³.

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3 years ago
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A man starts walking from home and walks 2 miles at 20° north of west, then 4 miles at 10° west of south, then 3 miles at 15° no
Rzqust [24]

Answer:

a)  R = 2.5 mi   b)  To return to your case you must walk in the opposite direction or θ = 98º

This is 8º north west

Explanation:

This is a distance exercise with vectors the best way to work these is to decompose the vectors and perform the sum on each axis separately

To use the Cartesian system all angles must be measured from the positive side of the x-axis or the signs of the components must be assigned manually depending on the quadrant where they are.

First vector A = 2 to 20º north west

Measured from the positive x axis is θ = 180 -20 = 160º

We use trigonometry to find the components

     Cos 20 = Aₓ / A

     sin 20 = A_{y} / A

    Aₓ = A cos 160 = 2 cos 160

    A_{y}  = A sin160 = 2 sin160

    Aₓ = -1,879 mi

    A_{y}  = 0.684 mi

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Angle θ = 270 - 10 = 260º

    cos 2600 = Bₓ / B

    sin 260 = B_{y} / B

    Bₓ = B cos 260

     B_{y}  = B sin 260

    Bₓ = 4 cos 260

     B_{y}  = 4 sin 260

     Bₓ = -0.6946mi

     B_{y}  = - 3,939 mi

Third vector C = 3 mi to 15 north east

     cos 15 = Cₓ / C

     sin15 = C_{y} / C

     Cₓ = C cos 15

     C_{y} = C sin15

     Cₓ = 3 cos 15

    C_{y} = 3 sin 15

     Cₓ = 2,898 mi

    C_{y} = 0.7765 mi

Now we can find the final position of the person

    X = Aₓ + Bₓ + Cₓ

    X = -1.879 -0.6949 + 2.898

    X = 0.3241 mi

    Y = A_{y} +  B_{y} + C_{y}

    Y = 0.684 - 3.939 +0.7765

    Y = -2.4785 mi

a) We use Pythagoras' theorem

     R = √ (x2 + y2)

     R = √ (0.3241 2 + (-2.4785) 2)

     R = 2.4996 mi

     R = 2.5 mi

b) let's use trigonometry

     Tan θ = y / x

     Tanθ  = -2.4785 / 0.3241

     θ = tan⁻¹ (-7,647)

     θ = -82

Measured from the positive side of the x axis is Te = 360 - 82 = 278º

(90-82) south east

To return to your case you must walk in the opposite direction or Te = 98º

This is 8º north west

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