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marissa [1.9K]
1 year ago
12

A rock of mass 200 g is attached to a 0.75 m long string and swung in a vertical plane.

Physics
2 answers:
Ainat [17]1 year ago
5 0

Hello!

a) Assuming this is asking for the minimum speed for the rock to make the full circle, we must find the minimum speed necessary for the rock to continue moving in a circular path when it's at the top of the circle.

At the top of the circle, we have:

- Force of gravity (downward)

*Although the rock is still connected to the string, if the rock is swinging at the minimum speed required, there will be no tension in the string.

Therefore, only the force of gravity produces the net centripetal force:

\Sigma F = F_g\\\\F_c = F_g\\\\\frac{mv^2}{r} = mg

We can simplify and rearrange the equation to solve for 'v'.

\frac{v^2}{r} = g\\\\v^2 = gr\\\\v = \sqrt{gr}

Plugging in values:

v = \sqrt{9.8 * 0.75} = \boxed{2.711 m/s^}

b)
Let's do a summation of forces at the bottom of the swing. We have:
- Force due to gravity (downward, -)

- Tension force (upward, +)

The sum of these forces produces a centripetal force, upward (+).

\Sigma F = T - F_g\\\\F_c = T - F_g\\\\\frac{mv^2}{r} = T - mg

Rearranging for 'T":
T =   \frac{mv^2}{r} +  mg\\\\

Plugging in the appropriate values:
T =  \frac{(0.2)(2.711^2)}{(0.75)} + 0.2(9.8) = \boxed{3.92 N}

AVprozaik [17]1 year ago
3 0
Hello!!
The correct answer would be number B)!! I took the test hope it helps
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A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,
viktelen [127]
Part A)
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r_A = 11.0 cm = 0.110 m
r_B = 16.5 cm=0.165 m
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Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: Q=3.67 \cdot 10^{-9}C. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
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If you need anymore help feel free to ask me!

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