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Tom [10]
3 years ago
7

Please help, my homework is due in a couple of days and I'm stuck. A plane is travelling 10.0 meters per second (this means the

plane would travel ten meters in one second, twenty meters in two seconds, thirty meters in three seconds, and so on) north, but then a strong wind of about 15.0 meters per second begins to blow east. (a) Is the speed of the plane equal to, greater than, or less than 25.0 meters per second? (Do not make any calculations.) (b) Explain your reasoning for your answer above. (c) Is the direction of the plane north, east, or somewhere between? (d) [!] If you said somewhere between, write (as an algebraic inequality, if you can) for the angle at which the plane might be travelling once the wind begins to blow.

Physics
1 answer:
ss7ja [257]3 years ago
4 0

Answer:

  • < 25 m/s
  • triangle inequality
  • between north and east
  • 45° < angle < 60°

Explanation:

(a) Just as one-dimensional numbers add on a number line by putting them end-to-end, so two-dimensional numbers add on a coordinate plane the same way.

Here, we choose to let the positive y-axis represent North, and the positive x-axis, East. This is the way a map is conventionally oriented. The velocity of the plane is represented by a vector pointing north (up). Its length represents the magnitude of the velocity. Likewise, the wind is represented by a vector of length 15 pointing east (right). The sum of these is the hypotenuse of the triangle they form.

The magnitude of the sum can be found here using the Pythagorean theorem, but for the purpose of this question, you're not asked to find that.

Instead, you're asked to estimate whether it is more or less than 25 (m/s).

Your knowledge of the triangle inequality will tell you that the hypotenuse (resultant) must be shorter than the sum of the lengths of the sides of the triangle, hence must be less than 10+15 = 25.

__

(b) The triangle inequality says the resultant is less than the sum of the other two sides of the triangle.

__

(c) Since the wind is blowing the plane toward the east, but the plane is traveling toward the north, the resulting direction is somewhere between north and east.

__

(d) "Somewhere between north and east" can be expressed as the inequality ...

  0° < angle < 90°

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An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:
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Heat added to the gas = Q = 743 Joules

Work done on the gas = W = -743 Joules

\texttt{ }

<h3>Further explanation</h3>

The Ideal Gas Law that needs to be recalled is:

\large {\boxed {PV = nRT} }

<em>P = Pressure (Pa)</em>

<em>V = Volume (m³)</em>

<em>n = number of moles (moles)</em>

<em>R = Gas Constant (8.314 J/mol K)</em>

<em>T = Absolute Temperature (K)</em>

Let us now tackle the problem !

\texttt{ }

<u>Given:</u>

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

<u>Unknown:</u>

Work done on the gas = W = ?

Heat added to the gas = Q = ?

<u>Solution:</u>

<h3>Step A:</h3>

<em>Ideal gas is allowed to expand isothermally:</em>

P_1V_1 = P_2V_2

5.00 \times 2.00 = 3.00 \times V_2

V_2 = 10 \div 3

V_2 = 3\frac{1}{3} \texttt{ L}

\texttt{ }

<em>Next we will calculate the work done on the gas:</em>

W_A = -P_2(V_2 - V_1)

W_A = -3.00(3\frac{1}{3} - 2.00)

W_A = \boxed{-4 \texttt{ L.atm}}

\texttt{ }

<h3>Step B:</h3>

<em>Using the same method as above:</em>

P_2V_2 = P_3V_3

3.00 \times 3\frac{1}{3} = 2.00 \times V_3

V_3 = 10 \div 2

V_3 = 5 \texttt{ L}

\texttt{ }

<em>Next we will calculate the work done on the gas:</em>

W_B = -P_3(V_3 - V_2)

W_B = -2.00(5 - 3\frac{1}{3})

W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}

\texttt{ }

<em>Finally we could calculate the total work done and heat added as follows:</em>

W = W_A + W_B

W = -4 + (-3\frac{1}{3})

W = -7\frac{1}{3} \texttt{ L.atm}

W = -7\frac{1}{3} \times 101.33 \texttt{ J}

\boxed{W \approx -743 \textt{ J}}

\texttt{ }

\Delta U = Q + W

0 = Q + (-743)

\boxed{Q = 743 \texttt{ J}}

\texttt{ }

<h3>Learn more</h3>
  • Minimum Coefficient of Static Friction : brainly.com/question/5884009
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\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Pressure

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Answer:

the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

Explanation:

Given the data in the question,

near point = 20 cm

far point = 2 m = 200 cm

Now, for an object that is infinitely far away, the image is at is its far point.

so using the following expression, we can determine the focal length

1/f = 1/i + 1/o

where f is the focal length, i is the image distance and o is the object distance.

here, far point i = 2 m = 200 cm  and v is ∞

so we substitute

1/f = 1/(-200 cm)  +  1/∞

f = -200 cm

Also, for object at its closest point, the image appear at near point,

so

1/f = 1/i + 1/o

we make o the subject of formula

o = ( i × f ) / ( i - f )

given that near point i = 20 cm

we substitute

o = ( -20 × -200 ) / ( -20 - (-200) )

o = 4000 / 180

o = 22.2 cm

Therefore, the closest distance that she can see an object clearly when she wears her contacts is 22.2 cm

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3 years ago
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Anastasy [175]

Answer:

<em>d. 268 s</em>

Explanation:

<u>Constant Speed Motion</u>

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

\displaystyle v=\frac{d}{t}

Where  

v = Speed of the object

d = Distance traveled

t  = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

And we can also solve it for t:

\displaystyle t=\frac{d}{v}

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The time it will take for them to meet is:

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t = 0.0746 hours

Converting to seconds: 0.0746*3600 = 268.56

The closest answer is d. 268 s

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TiliK225 [7]

The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N

<h3>Data obtained from the question </h3>
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F = m(v + u) / t

F = [2.5(7.5 + 8.5)]/ 0.25

F = 40 / 0.25

F = 160 N

Thus, the force exerted on the ball is 160 N

Learn more about momentum:

brainly.com/question/250648

#SPJ1

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