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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:

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3 years ago
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Two objects, C &amp; D, have the same momentum. Object C has ½ the mass of object D. Find the value of the ratio of velocity C t
Savatey [412]
These are two questions and two answers.

Part 1. Fin the value of the ration of velocity C to velocity D.


Answer: 2

Explanation:

1) Formula: momentum = mass * velocity

2) momentum C = mass C * velocity C

3) momentum D = mass D * velocity D.

4) C and D have the same momentum =>

mass C * velocity C = mass D * velocity D

5) mass C = (1/2) mass D => mass C / mass C = 1/2

6) use in the equation stated in the point 4)

velocit C / velocity D = mass D / mass C

using the equation stated in point 5:

mass D / mass C = 1 / [ mass C / mass D] = 1 / [1/2] = 2

=>

7) velocity C / velocity D = mass D / mass C = 2

Part 2: <span>ratio of kinetic energy C to kinetic energy D.
</span>
Answer: 2

Explanation:

1) formula: kinetic energy KE = (1/2) mass * (velocity)^2

2) KE C = (1/2) mass C * (velocity C)^2

3) KE D = (1/2) mass D * (velocity D)^2

4) KE C / KE D =

(1/2) mass C * (velocity C)^2        mass C        (velocity C)^2
--------------------------------------- = --------------- * ---------------------- = (1/2) * (2)^2
(1/2) mass D *( velocity D)^2        mass D        v(velocity D)^2

= 4 / 2 = 2
3 0
3 years ago
What can you infer about the changes in the environment over time at the mystery fossil dig site? Record your inferences in the
vivado [14]

the fossil informed us about the climatic and geographical changes that occurs in the environment over time.

<h3>What the mystery fossil dig site informed us about environment?</h3>

We can infer about the climate and geographical changes in the environment over time at the mystery fossil dig site because the fossil provides us information about the climatic condition as well as the geography of land when this fossil organism were alive.

So we can conclude that the fossil informed us about the climatic and geographical changes that occurs in the environment over time.

Learn more about fossil here: brainly.com/question/11829803

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4 0
2 years ago
How do Newton's laws of motion explain why it is important to keep the ice smooth on a hockey rink so that players can
lord [1]

Answer:

I'm not sure..but please refer to your teacher later.

Answer: Based on Newton's First law of motion (where inertia is involved), smooth ice increases the forceused to accelerate the hockey puck.

Explanation;

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  • based on inertia theory ; the heavier the weight, the larger the inertia.. which explains it takes alot of force to move a heavier object than the lighter ones.. it also hard to *stop* the motion of heavier objects than the lighter ones.
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6 0
3 years ago
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A square plate of copper with 55.0 cm sides has no net charge and is placed in a region of uniform electric field of 82.0 kN/C d
Alex73 [517]

Answer

given,

Side of copper plate, L = 55 cm

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a) Charge density,σ = ?

  using expression of charge density

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ε₀ is Permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

now,

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 σ = 725.7 x 10⁻⁹ C/m²

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change density on the plates are 725.7 nC/m² and -725.7 nC/m²

b) Total change on each faces

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   Q = 725.7 x 10⁻⁹ x 0.55²

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Hence, charges on the faces of the plate are 219.52 nC and -219.52 nC

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