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JulsSmile [24]
4 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?
Physics
2 answers:
Ber [7]4 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]4 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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FinnZ [79.3K]

Answer:

Explanation:

When beam is balanced and not rotating with Suki standing on it , let reaction force on the supports be R₁ and R₂. Then

R₁ +R₂ = 336 + 590

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Now the moment beam begins to tip , reaction on distant support R₁ = 0

only R₂ will exists on the support near to Suki.

Taking torque about this support of weight of beam acting from the middle point and weight of suki of 590N ,who is x distance from the support towards the other end.

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3 years ago
An object floats in water with 5 8 of its volume submerged. The ratio of the density of the object to that of water is
ki77a [65]

Answer:

\dfrac{5}{8}

Explanation:

m = Mass of object = \rho v

m' = Mass of water = \rho' v'

\rho = Density of object

\rho' = Density of water

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According to the question

v'=\dfrac{5}{8}v

In the case of floating objects

W=W'\\\Rightarrow mg=m'g\\\Rightarrow \rho vg=\rho'v'g\\\Rightarrow \rho v=\rho' \dfrac{5}{8}vg\\\Rightarrow \rho=\rho' \dfrac{5}{8}\\\Rightarrow \dfrac{\rho}{\rho'}=\dfrac{5}{8}

The ratio of the density of the object to that of water is \dfrac{5}{8}

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3 years ago
If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how high above the earth's surface
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Answer: h = 3R

Explanation:

Using the law of conservation of energy,

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Total energy at height h above the Earth where speed of the projectile is half the escape velocity:

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(1)=(2)

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Answer:

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