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tangare [24]
3 years ago
13

Car A is 16 kilometers to the west of car B. At noon, Car A begins driving east at a constant speed 4 kilometers/hour, and car B

begins driving north at a constant speed of 3 kilometers/hour. (a) (5 points) Let x be the distance that car A has traveled since noon, and let y be the distance that car B has traveled since noon. Let z be the distance between the two cars. Find an equation that relates x to x and y.
Physics
1 answer:
Vladimir [108]3 years ago
3 0

Answer:

The distance traveled can be found by kinematics equations

x = vt

x(t) = v_At - 16= 4t - 16\\y(t) = v_Bt = 3t\\z(t) = \sqrt{x(t)^2 + y(t)^2} = \sqrt{16t^2 - 128t + 256 + 9t^2} = \sqrt{25t^2 - 128t + 256}

Explanation:

The initial position of Car B is denoted as origin. Car A started at -16 km, and moving right. Car B started at origin and moving up. z is the magnitude of the vector with components x and y.

The relation can be found by pythagorean theorem.

For checking the solution, we can find the positions at t = 4h. Car A is at the origin, and Car B is 12 km north of origin. z = 12.

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Answer:

1.70 J

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With <em>m</em> = 0.175 kg,

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7 0
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You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner
omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

Q_1 = Q_2

V_1A_1=V_2A_2

Our values are given as,

A_1=\frac{1}{2}^2*\pi=0.785 in^2

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Re-arrange the equation to find the first ratio of rates we have:

\frac{V_1}{V_2}=\frac{A_2}{A_1}

\frac{V_1}{V_2}=\frac{1.227}{0.785}

\frac{V_1}{V_2}=1.56

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\frac{V2}{V1}=\frac{A_1}{A2}

\frac{V2}{V1}=\frac{0.785}{1.227}

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3 0
3 years ago
Two blocks with different temperatures had entropies of 10 J/K and 30 J/K before they were brought in contact. What can you say
True [87]

Answer:

a. Ssystem  > 40 J/K

Explanation:

Given that

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Therefore the answer is a.

Ssystem  > 40 J/K

6 0
2 years ago
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