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a 20 g bead is attached to a light 120 cm long string. This bead moves in a horizontal circle with a constant speed of 1.5 m/s. what is the tension in the string if the angle is 25 degrees

Respuesta :

opudodennis

Answer:

0.22 N

Explanation:

The tension in the spring is expressed as

[tex]Tcos\theta=mg[/tex] and making T the subject then

[tex]T=\frac {mg}{cos\theta}[/tex] where T is the tension in the string, m is the mass of the bead, g is acceleration due to gravity and [tex]\theta[/tex] is the angle which here is given as 25 degrees. Taking the value of g as 9.81 then

[tex]T=\frac {0.02\times 9.81}{cos 25}=0.216482748 N\approx 0.22 N[/tex]

hamzaahmeds

This question can be solved by using the formula of the centripetal force.

The tension in the string is "0.0375 N".

The tension in the string must be equal to the centrifpetal force.

Tension = Centripetal Force = [tex]\frac{mv^2}{r}[/tex]

where,

m = mass = 20 g = 0.02 kg

v = speed = 1.5 m/s

r = radius of circle = length of string = 120 cm = 1.2 m

Therefore,

[tex]Tension = \frac{(0.02\ kg)(1.5\ m/s)^2}{1.2\ m}[/tex]

Tension = 0.0375 N

Learn more about centripetal force here:

https://brainly.com/question/14021112?referrer=searchResults

The attached picture shows centripetal force.

Ver imagen hamzaahmeds

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