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Pie
3 years ago
6

A string is attached to a vibrating machine that has a frequency of 120 Hz. The other end of the string is passed over a pulley

of negligible mass and friction and is attached to a weight hanger which holds a mass m=0.5kg. What is the tension in newtons?
Physics
1 answer:
BabaBlast [244]3 years ago
4 0

Answer:

Explanation:

Given that,

The frequency of vibration is 120Hz

The mass of object attached is 0.5kg.

We want to find the tension In the string

The tension in the string is the weight of the object, it is how much gravity is pulling the object to the centre of the earth

Using newton second law.

F_net = m•a_y

The body is not accelerating the y-direction, then, a_y = 0

F_net = 0

Force acting on the string is the weight of the object and the tension in the string

T - W = 0

T = W

Where weight is mass × gravity

W = mg

Then,

T = W = mg

T = mg = 0.5 × 9.81

T = 4.905 N

The tension in the string is 4.905 N

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Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up
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Answer:

V = 90.51 m/s

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After the parachute opens:

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How fast is the racer can be determined by using the relation:

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6 0
3 years ago
A child\'s top is held in place, upright on a frictionless surface. The axle has a radius of r = 2.96 mm. Two strings are wrappe
sladkih [1.3K]

Answer:

Explanation:

Given

radius r=2.96 mm

Tension T=2.4 N

time taken=0.74 s

Let \alphabe the angular acceleration

2 T\times r=I\times \alpha

2\times 2.4\times 2.96\times 10^{-3}=0.5\times m\times (2.96\times 10^{-3})^2\times \alpha

\alpha =\frac{4\times 2.4}{m\times 2.96\times 10^{-3}}

\alpha =\frac{3.24\times 10^3}{m} rad/s^2

\omega =\omega _0+\alpha \cdot t

\omega =0+\frac{3.24\times 10^3}{m}\times 0.74

\omega =\frac{2.4\times 10^3}{m} rad/s

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L=0.5\times m\times (2.96\times 10^{-3})^2\times \frac{2.4\times 10^3}{m}

L=0.01051 kg-m^2/s

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3 years ago
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