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love history [14]
3 years ago
10

José is pinned against the walls of the Rotor, a ride with a radius of 3.00 meters that spins so fast that the floor can be remo

ved without the riders falling. When the ride accelerates to a speed of 17.0 m/s, his angular momentum is 3,570 kg m2/s. What is his mass?
a. 23.3 kg
b. 70.0 kg
c. 179 kg
d. 630. kg
Physics
1 answer:
satela [25.4K]3 years ago
8 0

Answer:

179 kg

Explanation:

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The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
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Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

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Explanation:

Given, the position of the particle along the x axis is

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The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

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Therefore, unit of \rm c= meter/ second^2.

(b):

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Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

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(e):

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  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
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Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

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\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

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