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gayaneshka [121]
4 years ago
15

At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci

ng the axis (middle), with their backs against the outer wall. At one instant the outer wall moves at a speed of 3.2 m/s, and an 83kg person feels a 560N force pressing against his back. What is the radius of a chamber
Physics
1 answer:
lana [24]4 years ago
5 0

To solve this problem we will apply the concepts related to centripetal acceleration and Newton's second law. In the case of acceleration, we will define how it is shaped, and we will equate that acceleration to that stipulated by Newton in his second law. We will clear the variable of the 'radius'. Centripetal acceleration is described as,

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

Here,

v =Velocity

r = Radius

Now Newton's second law is,

F = ma_R

Here,

m = mass

a_R = Centripetal acceleration

Rearranging in function of the acceleration,

a_R = \frac{F}{m}

Rearranging the first equation in function of the radius

a_R = \frac{v^2}{r} \rightarrow r=\frac{v^2}{a_R}

Equating,

r = \frac{mv^2}{F}

Replacing,

r = \frac{(83kg)(3.2m/s^2)}{560N}

r = 1.518m

Therefore the radius of the  chamber is 1.518m

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What is the frequency of a wave that has a period of vibration of 2 seconds?
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Answer:

The answer is 0.5 Hz

Explanation:

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You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,
lbvjy [14]

Answer:

The value is T_c  =  12 .1 ^oC

Explanation:

From the question we are told that

The mass of the ice cube is m_i  =  7.50 *10^{-3} \  kg

The temperature of the ice cube is T_i = 0^o C

The mass of the copper cube is m_c  =  0.540 \  kg

The final temperature of both substance is T_f  =  0^oC

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

Q =  m_c  *  c_c *  [T_c  -  T_f ]

The specific heat of copper is c_c  = 385J/kg \cdot  ^oC

Generally the heat gained by the ice cube is mathematically represented as

Q_1 =  m_i * L

Here L is the latent heat of fusion of the ice with value L  =  3.34 * 10^{5} J/kg

So

Q_1 =  7.50 *10^{-3} * 3.34 * 10^{5}

=> Q_1 =  2505 \ J

So

2505  =  0.540  *  385 *  [T_c  - 0 ]

=>    T_c  =  12 .1 ^oC

4 0
3 years ago
S A block of mass M is connected to a spring of mass m and oscillates in simple harmonic motion on a frictionless, horizontal tr
Elina [12.6K]

The period of oscillation is T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

<h3>What is period of oscillation?</h3>

This is the time in seconds it takes to complete one oscillation. where an oscillation is a repetitive to and fro motion. period if the inverse of frequency and both are basic when calculation motion in simple harmonic motion.

The period of oscillation is given as T

T = 2 * pi * sqrt ( m / k )

where

m = mass on this case mass of the spring will be inclusive to the mass of the block such that we have:

m1 = mass of the block

m2 = mass pf the spring

k = force constant of the spring

including the two masses to the period gives

T = 2 * pi * sqrt ( ( m2/3 + m1) / k )

Read more on period of oscillation here: brainly.com/question/22499336

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7 0
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