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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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adoni [48]

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351 ohm

720 ohm

Explanation:

When c and d are open:

Terminals c and d are open.. If you  redraw the circuit as below, you can see that the two resistors in the first  column are in parallel as, they are connected together at both pairs of terminals  (due to the short).

Hence, we have a pair of parallel resistors:

Req1 = (R1*R2)/ (R1 + R2) = 360*540/(360+540) = 216 ohms

Req2 = (R3*R4)/ (R3 + R4) = 180*540/(180+540) = 135 ohms

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Answer: 351 ohms

When c and d are shorted:

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Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th
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a.\rm -1.49\ m/s^2.

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<u>Given:</u>

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<h2>(a):</h2>

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\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.

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