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3 years ago

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An amusement park ride moves a rider at a constant speed of 14 meters per second in a horizontal circular path of radius 10. met

ers. What is the rider's centripetal acceleration in terms of g, the acceleration due to gravity?
A) 1g
B) 2g
C) 3g
D) 0g

1 answer:

ASHA 777 [7]3 years ago

4 0

Answer:

B) 2g

Explanation:

<u>Given the following data;</u>

Velocity, v = 14m/s

Radius, r = 10m

To find the centripetal acceleration;

$Acceleration, a = \frac {v^{2}}{r}$

Substituting into the equation, we have;

$Acceleration, a = \frac {14^{2}}{10}$

$Acceleration, a = \frac {196}{10}$

Acceleration, a = 19.6m/s²

In terms of acceleration due to gravity, g = 9.8m/s²

We would divide by g;

Acceleration, a = 19.6/9.8 = 2

Hence, centripetal acceleration = 2g

Therefore, the rider's centripetal acceleration in terms of g, the acceleration due to gravity is 2g.

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just olya [345]

Answer:

$a=2.5\ m/s^2$

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly over time.

The equation that describes the change of velocities is:

$v_f=v_o+at$

Where:

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Solving the equation for a:

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The ball starts at rest (vo=0) and rolls down an inclined plane that makes it reach a speed of vf=7.5 m/s in t=3 seconds.

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3 years ago

What is the differential equation governing the growth of current in the circuit as a function of time after t=0? express the ri

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Answer:

$v_{b}=ir+L\frac{di}{dt}$

Explanation:

A differential equation that contain a term with di(t)/dt is in a RL circuit. Here we have

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I hope this is useful for you

regards

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Which of these best describes how the kinetic energy of carbon dioxide molecules change as the carbon dioxide is heated?

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B.
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4 years ago

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Use the graph for both answers.

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1. A. 6.00 sec

The graph shows the velocity of an object (y-axis) versus the time (x-axis). In order to find when the magnitude of the velocity reaches 36.00 km/h, we should find the time t (x-coordinate) at which the velocity (y-coordinate) is 36.

By looking at the graph, we see that this occurs when t=6.00 s.

2. A. positive acceleration

In a velocity-time graph like this one, the slope of the curve corresponds to the acceleration of the object. In fact, acceleration is defined as:

$a=\frac{\Delta v}{\Delta t}$

where $\Delta v$ is the variation of velocity and $\Delta t$ is the variation of time. We see that this quantity corresponds to the slope of the curve in the graph (in fact, $\Delta v$ represents the increment of the y coordinate, while $\Delta t$ represents the increment of the x coordinate). So, a positive slope means a positive acceleration: in this case, the slope is positive, so the acceleration is also positive.

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Baseball player a bunts the ball by hitting it in such a way that it acquires an initial velocity of 2.4 m/s parallel to the gro

LUCKY_DIMON [66]

Let $\mathbf r_A$ denote the position vector of the ball hit by player A. Then this vector has components

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where $g=9.8\,\dfrac{\mathrm m}{\mathrm s^2}$ is the magnitude of the acceleration due to gravity. Use the vertical component $r_{Ay}$ to find the time at which ball A reaches the ground:

$1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s$

The horizontal position of the ball after 0.49 seconds is

$\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m$

So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector $\mathbf r_B$ of the ball hit by player B has

$\begin{cases}r_{Bx}=v_0t\\r_{By}=1.6\,\mathrm m-\frac12gt^2\end{cases}$

Again, we solve for the time it takes the ball to reach the ground:

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