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VLD [36.1K]
3 years ago
6

5. Sandor fills a bucket with water and whirls it in a vertical circle to demonstrate that the

Physics
1 answer:
just olya [345]3 years ago
3 0

Given that,

radius = 1.24 m

According to question,

The rope cannot push outwards. It must always have some slight tension or the bucket will fall.

We need to calculate the tension in the rope

At the top the force of gravity is

F=mg

The force needed to move the bucket in a circle is centripetal force.

So, if mg is ever greater than centripetal force then the bucket and the contents will start to fall.

The rope have a tension of less than zero.

We need to calculate the velocity of swing bucket

Using centripetal force

F=\dfrac{mv^2}{r}

mg=\dfrac{mv^2}{r}

g=\dfrac{v^2}{r}

v^2=gr

v=\sqrt{gr}

Put the value into the formula

v=\sqrt{9.8\times1.24}

v=3.49\ m/s

Hence, The minimum tension in the rope is less than zero .

The bucket swings with the velocity of 3.49 m/s.

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a force is applied to a block causing it to accelerate along a horizontal, frictionless surfaces. the energy gained by the block
GaryK [48]
Energy given to the block
3 0
3 years ago
A small sphere has a harge of 9uC and other small sphere has a charge of 4uC.
Helga [31]

Answer:

Electrical force, F = 90 N

Explanation:

It is given that,

Charge on sphere 1, q_1=9\ \mu C=9\times 10^{-6}\ C

Charge on sphere 2, q_1=4\ \mu C=4\times 10^{-6}\ C

Distance between two spheres, d = 6 cm = 0.06 m

Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,

F=k\dfrac{q_1q_2}{d^2}

F=9\times 10^9\times \dfrac{9\times 10^{-6}\times 4\times 10^{-6}}{(0.06)^2}

F = 90 N

So, the electrical force between them is 90 N. Hence, this is the required solution.

7 0
3 years ago
A research team developed a robot named Ellie. Ellie ran 1,000 meters for 200 seconds from the research building, rested for 100
Verizon [17]

Answer:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)

Explanation:

Attached you will find the plot of position vs time of Ellie´s movement.

The velocity is the displacement of the object over time relative to the system of reference. The speed, in change, is the traveled distance over time in disregard of the system of reference.

So, the velocity is calculated as follows:

v = Δx / Δt

where

Δx = final position - initial position

Δt = elapsed time

1) The average velocity of Ellie while running is:

v = 1000 m - 0 m / 200 s = 5 m/s

While resting:

v = 0 m - 0 m / 100 s = 0 m/s

And while walking back:

v = 0 m - 1000 m / 1000 s = - 1 m/s

Note that in this last case, the initial position is 1000 m because Ellie is 1000 m from the origin of the system of reference when she walks back. The final position will be the origin of the system of reference, 0 m.

Comparing with the graphic, the velocity is the slope of the function position(t).

Then:

1. Running velocity (5 m/s)

2. Resting velocity (0 m/s)

3. Walking velocity (-1 m/s)

2) The speed is the distance traveled over time:

Running speed = 1000 m / 200 s = 5m /s

Resting speed = 0 m / 100 s = 0 m/s

Walking speed = 1000 m/ 1000 s = 1 m/s

Then:

1. Running speed (5 m/s)

2. Walking speed (1 m/s)

3. Resting speed (0 m/s)  

4 0
3 years ago
Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.
padilas [110]

Explanation:

We have,

Semimajor axis is 4\times 10^{12}\ m

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

T^2=\dfrac{4\pi ^2}{GM}a^3

G is universal gravitational constant

M is solar mass

Plugging all the values,

T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s

Since,

1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}

So, the orbital period of a dwarf planet is 138.52 years.

3 0
3 years ago
A worker does 25 J of work lifting a bucket, then sets the bucket back down in the same place. What is the total net work done o
Vera_Pavlovna [14]

Answer:

0 J

Explanation:

As work is force times displacement, if no displacement occurs, no work occurs.

5 0
2 years ago
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