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Lemur [1.5K]
3 years ago
11

A moving roller coaster speeds up with constant acceleration for 2.3\,\text{s}2.3s2, point, 3, start text, s, end text until it

reaches a velocity of 35\,\dfrac{\text{m}}{\text{s}}35 s m ​ 35, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction. During this time, the roller coaster travels 41\,\text{m}41m41, start text, m, end text. We want to find the initial velocity of the roller coaster before it starts to accelerate. Which kinematic formula would be most useful to solve for the target unknown?
Physics
1 answer:
Ad libitum [116K]3 years ago
7 0

Answer:

Δx=(v+v0/2)t

Explanation:

We can figure out which kinematic formula to use by choosing the formula that includes the known variables, plus the target unknown.

In this problem, the target unknown is the initial velocity v_0v  

0

​  

v, start subscript, 0, end subscript of the roller coaster.

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The wheels of a car have radius 12 in. and are rotating at 600 rpm. Find the speed of the car in mi/h.
guajiro [1.7K]

Answer:

21.4 mph

Explanation:

Circumference of tire in FEET   = pi * d =  pi * 1 ft = pi  feet

pi feet  x  600 rot/min  *  60 min /hr  *  1 mile / 5280 feet = 21.4 mph

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10 months ago
A beam of sunlight falling on a prism refracts and forms seven color bands. This illustrates that
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That isn"t the right answer the correct answer is B.

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3 years ago
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Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat
kvasek [131]

Answer:

a) a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

b) k = \frac{9.8}{9.74}=1.006

Explanation:

Part a

For this case we can begin finding the period like this:

T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s

Then we know that the centripetal acceleration is given by:

a= \frac{v^2}{r}

And the velocity is given by:

v=\frac{2\pi r}{T}

If we replace this into the acceleration we got:

a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}

And we can replace the values and we got:

a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}

Part b

For this case we want to find a value of k such that:

a= k 9.8

Where a = 9.74, so then we can solve for k like this:

k = \frac{9.8}{9.74}=1.006

8 0
3 years ago
A 545-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth's mean radius. (a) Find the satelli
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Answer

given,

mass of satellite = 545 Kg

R = 6.4 x 10⁶ m

H = 2 x 6.4 x 10⁶ m

Mass of earth = 5.972 x 10²⁴ Kg

height above earth is equal to earth's mean radius

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     gravitational force = \dfrac{GMm}{r^2}

    equating both the above equation

    \dfrac{mv^2}{r} = \dfrac{GMm}{r^2}

      v = \sqrt{\dfrac{GM}{r}}

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b) T= \dfrac{2\pi\ r}{v}

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   T= \dfrac{2\pi\times 2\times 6.4 \times 10^6}{5578.5}

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c) gravitational force acting

  F = \dfrac{GMm}{r^2}

  F = \dfrac{6.67 \times 10^{-11}\times 545 \times 5.972 \times 10^{24} }{(6.46 \times 10^6)^2}

     F = 5202 N

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The one that would explain why the two groups of scientists got different results is : 
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