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zavuch27 [327]
3 years ago
12

A sling-thrower puts a stone (0.250 kg) in the sling's pouch (0.010 kg) and then begins to make the stone and pouch move in a ve

rtical circle of radius 0.725 m. The cord between the pouch and the person's hand has negligible mass and will break when the tension in the cord is 39.5 N or more. Suppose the sling-thrower could gradually increase the speed of the stone. (a) Will the breaking occur at the lowest point of the circle or at the highest point? (b) At what speed of the stone will that breaking occur?
Physics
1 answer:
yan [13]3 years ago
7 0

Answer:

Lowest.

v=10.5m/s

Explanation:

The breaking will occur at the lowest point of the circle because in that position the weight acts against the tension, to the tension must be higher for one to get the same centripetal force as when it is at the highest point.

The equations of centripetal force is:

F_{cp}=ma_{cp}=m\frac{v^2}{r}

Where we also used the equatio for centripetal acceleration.

We calculate then the speed when the centripetal force is 39.5N, which is when the breaking will occur:

v=\sqrt{\frac{F_{cp}r}{m}}=\sqrt{\frac{(39.5N)(0.725)}{(0.25Kg+0.01Kg)}}=10.5m/s

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To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

Where,

\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

x = 14.397*(2\pi*\frac{0.25}{2})

x = 11.3m

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3 years ago
Garrett Hardin's "Tragedy of the Commons" describes the tendency of people to overuse publicly-owned natural resources until the
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Answer:

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3 years ago
10,500 J of GPE, weight 539 N, how tall is the hill you are sitting on?
maks197457 [2]

Answer:

19.48 m

Explanation:

Gravitational potential energy = mgh

Current weight = 539 N

Weight = mg = 539 N

Mass x Acceleration = 539 N

Mass x 9.81 = 539

Mass = 54.94 kg

Gravitational potential energy = mgh = 10500 J

                    54.94 x 9.81 x h = 10500

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6 0
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So A fold is a Bend? in a rock. Maybe.

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3 years ago
Read 2 more answers
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blagie [28]

Answer:

Option A is correct.

(The faster object encounters more resistance)

Explanation:

Option A is correct. (The faster object encounters more resistance)

Air resistance depends on various factors:

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Formula:

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