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kenny6666 [7]
3 years ago
8

a 2.35 water bucket is swung in a full cirlce of radius 0.824 m just fast enough so that the water doesn't fall out the top mean

ing n equals zero there. What is the speed of the water at the top
Physics
1 answer:
tiny-mole [99]3 years ago
4 0

Answer:

2.84 m/s

Explanation:

At the top position of the circular trajectory, the normal reaction is zero:

N = 0

So it means that the only force that is providing the centripetal force is the gravitational force (the weight of the bucket). Therefore we have:

mg = m \frac{v^2}{r}

where

m is the mass of the water bucket

g = 9.8 m/s^2 is the acceleration of gravity

v is the speed of the bucket

r = 0.824 m is the radius of the circle

Solving for v,

v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(0.824 m)}=2.84 m/s

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Hi there, it is your average pathetic Junior. Anyways, I need help on 8&9 ASAP, this assignment is beyond overdue but hey wh
zheka24 [161]

Answer:

a) -31.36 m/s

b) 50.176 m

Explanation:

<h2>a) Velocity of the bag</h2>

This is a problem of motion in one direction (specifically vertical motion), and the equation that best fulfills this approach is:

V_{f}=V_{o} +a.t  (1)

Where:

V_{f} is the final velocity of the supply bag

V_{o}=0 is the initial velocity of the supply bag (we know it is zero because we are told <u>it was "dropped", this means it goes to ground in free fall</u>)

a=g=-9.8m/s^{2} is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)

t=3.2s is the time

Knowing this, let's solve (1):

V_{f}=0+(-9.8m/s^{2})(3.2s)  (2)

Hence:

V_{f}=-31.36m/s  Note the negative sign is because the direction of the bag is downwards as well.

<h2>b) Final height of the bag</h2>

In this case we will use the following equation:

y=V_{o}t-\frac{1}{2}gt^{2} (3)

Where:

y is the distance the bag has fallen

V_{o}=0 remembering <u>the bag was dropped</u>

g=-9.8m/s^{2} is the acceleration due gravity (downwards)

t=3.2 s is the time

Then:

y=-\frac{1}{2}gt^{2} (3)

y=-\frac{1}{2}(-9.8m/s^{2})(3.2)^{2} (4)

Finally:

y=50.176 m

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