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prohojiy [21]
3 years ago
10

At an amusement park, the wheelie carries passengers in a circular path of radius r = 11.2 m. If the angular speed of the wheeli

e is 0.550 revolutions/s, (a) What is the tangential velocity of the passengers due to the circular motion? (b) What is the acceleration of the passengers?
Physics
1 answer:
Dafna1 [17]3 years ago
8 0

Answer:

(a) Tangential velocity will be 38.648 m/sec

(b) Acceleration will be 133.617m/sec^2

Explanation:

We have given radius r = 11.2 m

Angular speed \omega =0.550rev/sec=0.550\times 2\pi =3.454rad/sec

(a) We have to find the tangential velocity

We know that tangential velocity is given by  

v_t=\omega r=3.454\times 11.2=38.684m/sec

(b) We know that acceleration is given by

a=\frac{v^2}{r}=\frac{38.684^2}{11.2}=133.617m/sec^2

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According to one set of measurements, the tensile strength of hair is 196 MPa , which produces a maximum strain of 0.380 in the
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Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

A=\dfrac{\pi}{4}\times d^2

Put the value into the formula

A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2

A=1.96\times10^{-9}\ m^2

We need to calculate the magnitude of the force

Using formula of force

F=\sigma A

Put the value into the formula

F=196\times10^{6}\times1.96\times10^{-9}

F=0.38416\ N

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

strain=\dfrac{\Delta l}{l_{0}}

\Delta l=strain\times l_{0}

Put the value into the formula

\Delta l=0.380\times l_{0}

Length after expansion is 12 cm

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Using formula of length

l=l_{0}+\Delta l

Put the value into the formula

I=l_{0}+0.380\times l_{0}

l=1.38l_{0}

l_{0}=\dfrac{l}{1.38}

l_{0}=\dfrac{12\times10^{-2}}{1.38}

l_{0}=0.0869\ m

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

4 0
3 years ago
A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
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Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
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W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

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