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hram777 [196]
2 years ago
6

A roller coaster rider traveling in a straight line changes from a speed of 4 m/s to 16 m/s in 3 seconds. What is the accelerati

on of the rider?
A. 1.33 m/s2
B. 3 m/s2
C. 5.33 m/s2
D. 4 m/s2
Physics
1 answer:
frez [133]2 years ago
3 0

Answer:27

Explanation:

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A hypothesis is an educated guess. It's your own opinion!
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Which of the following characteristics do all unicellular organisms share?
Fed [463]

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Asexual production they can be eukaryotes or prokaryotes

Explanation:

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Pretty sure it's C) condensation because all of the others required heat to be added
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An organ pipe open at both ends is 1.5 m long. A second organ pipe that is closed at one end and open at the other is 0.75 m lon
telo118 [61]

Answer:

A. 110Hz,220Hz, 330 Hz

Explanation:

for organ open at open both ends;

the length of the organ for the fundamental frequency, L = A---->N + N----->A

A---->N  = λ /4 and N----->A = λ /4

L = λ /4 + λ /4 = λ /2

L = \frac{\lambda}{2} \\\\\lambda = 2L

λ  = 2 x 1.5m = 3.0 m

Wave equation is given by;

V = Fλ

Where;

V is the speed of sound

F is the frequency of the wave

F = V/ λ

F₀ = V / 2L

Where;

F₀  is the fundamental frequency

F₀ = 330 / 2(1.5)

F₀ = 330 / 3

F₀ = 110 Hz

the length of the organ for the first overtone, L = A---->N + N----->A + A----->N +  N----->A

L = 4λ /4

L = λ

λ = 1.5 m

F₁ = 330 / 1.5

F₁ = 220 Hz

Thus, F₁ = 2F₀

For open organ at one end

the length of the organ for the fundamental frequency, L = N------A

L = λ /4

λ = 4L

F₀ = V/4L

F₀ = 330 / (4 x 0.75)

F₀ = 110 Hz

the length of the organ for the first overtone, L = N-----N + N-----A

L = λ/2 + λ / 4

L = 3λ /4

F₁ = 3F₀

F₁ = 3 x 110

F₁ = 330 Hz

Thus the fundamental frequency for both organs is 110 Hz,

The first overtone for the organ open at both ends is 220 Hz

The first overtone for the organ open at one end is 330 Hz

The correct option is "A. 110Hz,220Hz, 330 Hz"

6 0
2 years ago
A bungee jumper of mass m jumps off a bridge. Assume that the bungee cord behaves like am ideal spring of spring constant k. Whe
DanielleElmas [232]

Answer:

b) √[(kx²/m) - 2gx]

Explanation:

The energy at the lowest point is equal to:

E_{elas}=\frac{1}{2} *k*x^{2}

where:

Eelas = elastic energy [J]

k = spring constant [N/m]

x = extension of the spring [m]

We consider the lowest point, as the point where the potential energy is zero. At the moment when the person goes back through the point of the normal length of the elastic cord, it is this point that the person will have potential energy and kinetic energy.

E_{elas}=E_{pot}+E_{kine}\

\frac{1}{2}*k*x^{2}=m*g*x +\frac{1}{2} *m*v^{2}  \\v^{2} = \frac{k*x^{2} }{m}-2*g*x\\ v=\sqrt{\frac{k*x^{2} }{m}-2*g*x}

8 0
2 years ago
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