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3 years ago

15

A girl sits on a tire swing and is pushed in a circular motion by her father. Her tangential speed is 2.8 m/s and the girl trave

ls in a circle with a diameter of 4.0 m. What is the girls centripetal acceleration?

2 answers:

Ber [7]3 years ago

4 0

Answer:

The girl's centripetal acceleration is 3.92 m/s².

Explanation:

Given that,

Tangential speed = 2.8 m/s

Diameter = 4.0 m

We need to calculate the centripetal acceleration

Using formula of centripetal acceleration

$\alpha=\dfrac{v^2}{r}$

Where, v = tangential speed

r = radius

Put the value into the formula

$\alpha=\dfrac{2.8^2}{2.0}$

$\alpha=3.92\ m/s^2$

Hence, The girl's centripetal acceleration is 3.92 m/s².

Jet001 [13]3 years ago

3 0

Formula: Ca=Vt^2/r

centripetal acceleration(Ca)= ?
tangential speed(Vt)= 2.8m/s
Radius(r)=2m

Substitute: Ca=2.8^2/2
Ca=3.92m/s^2

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xxMikexx [17]

Answer:

$T_f=24.71$

Explanation:

From the question we are told that:

Mass of block $m=50$

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Generally the equation for equilibrium stage is mathematically given by

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2 years ago

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the s

yawa3891 [41]

Hello!

A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?

Data:

$E_{pe}\:(elastic\:potential\:energy) = 5184\:J$

$K\:(constant) = 16200\:N/m$

$x\:(displacement) =\:?$

For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

$E_{pe} = \dfrac{k*x^2}{2}$

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.

Solving:

$E_{pe} = \dfrac{k*x^2}{2}$

$5184 = \dfrac{16200*x^2}{2}$

$5184*2 = 16200*x^2$

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$x^{2} = \dfrac{10368}{16200}$

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$x = \sqrt{0.64}$

$\boxed{\boxed{x = 0.8\:m}}\end{array}}\qquad\checkmark$

Answer:

The displacement of the spring = 0.8 m

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

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