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2 years ago

Assuming motion is on a straight path, what would result in two positive components of a vector?

2 answers:

3 0

Answer: option A) a train moving north of east at an angle of 25°


Explanation:

1) You need to choose your axis. In this case North is vertical and positive, South is vertical and negative, East is horizontal and positive, and West is horizontal and negative.

2) The vector with the two positive components is a vector in the first quadrant (North and East). That is what North of East 25° means.

3) Regarding the other options:

B) a bus moving North of East at an angle of 95°: since the angle is greater tnan 90° the vector is in the second quadrant: its horizontal component is negative.


 C) a boat moving South of West at an angle of 40°: this is in the third quadrant: the two components are negative.

 D) a car moving south of west at an angle of 10°: as in the option C), this is in the third quadrant: the two components are negative.

avanturin [10]2 years ago

3 0

Two positive components means that the vector should lie in the first quadrant and not be parallel to the x or y axis; the only option that satisfies this is A.

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At the top of the circle the centripetal force (mv²/R) comes from the car's weight (mg)

So, the net downward force from the car (Fn) = (weight - centripetal force) .. and by reaction this is the upward force provided by the road ..

Fn = mg - mv²/R
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(b)
When the car's speed is such that all the weight is needed for the centripetal force .. then the net downward force (Fn), and the reaction from the road, becomes zero.

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2 years ago

A 4.00-kg mass is attached to a very light ideal spring hanging vertically and hangs at rest in the equilibrium position. The sp

Ahat [919]

Answer:

|v| = 8.7 cm/s

Explanation:

given:

mass m = 4 kg

spring constant k = 1 N/cm = 100 N/m

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1. Finding Ф from the initial conditions:

$-0.02 = 0.02cos(0 + \phi) => \phi = \pi$

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What tangential speed, v, must the bob have so that it moves in a horizontal circle with the string always making an angle θ fro

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Refer to the diagram shown below.

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r = the radius of the horizontal circle.
T = tension in the string.
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F = centripetal force
l = the length of the string

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r = l sin θ

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T cosθ = mg (1)
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Divide (2) by (1).
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Answer: $v = \sqrt{gl \, sin \theta \, tan \theta}$

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