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Anna71 [15]
3 years ago
12

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a sp

eed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.
Physics
2 answers:
Alex787 [66]3 years ago
7 0

Answer:

Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.

Explanation:

i want the answer i don't know

ipn [44]3 years ago
6 0

Answer:

Explanation:want answers

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A wave with a period of 0.008 second has a frequency of
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If you swim from one point at the edge of the pool to another, along a straight line, what is the longest distance d you can swi
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Long straight distance that a person can swim is 5.64 m.

<h3>What is the Long straight distance?</h3>

The line that runs form one end of the circle to another is called the diameter of the circle. The pool is a circle according to the question and the long straight distance that a person can swim is the same of the diameter of the circular pool.

Now we have;

A =  πr^2

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Missing parts;

An ad for an above-ground pool states that it is 25 m2. From the ad, you can tell that the pool is a circle. If you swim from one point at the edge of the pool to another, along a straight line, what is the longest distance d you can swim? Express your answer in three significant figures.

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2 years ago
A wheel in the shape of a flat, heavy, uniform, solid disk is initially at rest at the top of an inclined plane of height 2.00 m
olasank [31]

Answer:

Explanation:

If friction is neglected, the wheel cannot roll and can only slide frictionlessly and will have the same velocity at the bottom of the ramp as if it had been in free fall as it has converted the same amount of potential energy.

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v = √(2gh) = √(2(9.81)(2.00)) = 6.26418... = 6.26 m/s

However if we do not ignore all friction and the wheel rolls without slipping down the slope, the potential energy becomes linear and rotational kinetic energy

mgh = ½mv² + ½Iω²

mgh = ½mv² + ½(½mR²)(v/R)²

2gh = v² + ½v²

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v = √(4gh/3) =√(4(9.81)(2.00)/3) = 5.11468... = 5.11 m/s

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