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zhuklara [117]
2 years ago
9

A student on an amusement park ride moves in circular path with a radius of 3.5 m once every 4.5 s. What is the tangential veloc

ity of the student?
Physics
1 answer:
nata0808 [166]2 years ago
6 0

Answer:

the tangential velocity of the student is 4.89 m/s.

Explanation:

Given;

the radius of the circular path, r = 3.5 m

duration of the motion, t = 4.5 s

let the student's tangential velocity = v

The tangential velocity of the student is calculated as follows;

v = \frac{2\pi r}{t} \\\\v = \frac{2 \pi \times 3.5}{4.5} \\\\v = 4.89 \ m/s

Therefore, the tangential velocity of the student is 4.89 m/s.

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A 5.30kg block hangs from a spring with a spring constant 1700 N/m. The block is pulled down 4.50cm from the equilibrium positio
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To solve this problem it is necessary to apply the concepts related to the frequency in a spring, the conservation of energy and the total mechanical energy in the body (kinetic or potential as the case may be)

PART A) By definition the frequency in a spring is given by the equation

f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}

Where,

m = mass

k = spring constant

Our values are,

k=1700N/m

m=5.3 kg

Replacing,

f = \frac{1}{2\pi} \sqrt{\frac{1700}{5.3}}

f=2.85 Hz

PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2

Where,

k = Spring constant

m = mass

y = Vertical compression

v = Velocity

This expression is equivalent to,

kA^2 =mV^2 +ky^2

Our values are given as,

k=1700 N/m

V=1.70 m/s

y=0.045m

m=5.3 kg

Replacing we have,

1700*A^2=5.3*1.7^2 +1700*(0.045)^2

Solving for A,

A^2 = \frac{5.3*1.7^2 +1700*(0.045)^2}{1700}

A ^2 = 0.011035

A=0.105 m \approx 10.5 cm

PART C) Finally, the total mechanical energy is given by the equation

E = \frac{1}{2}kA^2

E=\frac{1}{2}1700*(0.105)^2

E= 9.3712 J

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