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Nesterboy [21]
3 years ago
5

2. A 0.8 kg tetherball hangs on the end of a cord. It is hit by a child and rises 2.1 m above the ground. a. What is the maximum

gravitational potential energy of the ball?
Physics
1 answer:
skad [1K]3 years ago
3 0

Answer:

E = 16.464 J

Explanation:

Given that,

Mass of tetherball, m = 0.8 kg

It is hit by a child and rises 2.1 m above the ground, h = 21. m

We need to find the maximum gravitational potential energy of the ball. The formula for the gravitational potential energy is given by :

E = mgh

g is acceleration due to gravity

E = 0.8 kg × 9.8 m/s² × 2.1 m

= 16.464 J

So, the maximum potential energy of the ball is 16.464 J.

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Answer:

amitude doesnt change

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The movement of electrons through a wire is called.....
Elenna [48]

Answer:

Electric Current

Explanation:

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2 years ago
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Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8
umka2103 [35]

Answer:

a) F=2.048\times 10^{-7}\ N

b) a=0.1138\ m.s^{-2}

Explanation:

Given:

  • mass of raindrops, m=1.8\times 10^{-6}\ kg
  • charge on the raindrops, q=+21\times 10^{-12}\ C
  • horizontal distance between the raindrops, r=0.0044\ m

A)

<u>From the Coulomb's Law the force between the charges is given as:</u>

F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}

we have:

\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}

<em>Now force:</em>

F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}

F=2.048\times 10^{-7}\ N

B)

<u>Now the acceleration on the raindrops due to the electrostatic force:</u>

a=\frac{F}{m}

a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}

a=0.1138\ m.s^{-2}

7 0
3 years ago
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A 2.0-kg object moving at 5.0 m/s encounters a 30-Newton restive force
sveta [45]

The impulse experienced by the object is 3 N s.

<u>Explanation:</u>

Impulse is also termed as change in the momentum of the object. So, it is directly proportional to the force acting on the object and the time for which the force is acting on that object.

Thus, impulse experienced by an object is the product of force acting on the object for a given time period. So, it is the sudden influence of force on the given volume.

As the force is given as 30 N and the duration or the time is given as 0.1 seconds. Then, the impulse will be product of force with duration.

Impulse = Force × ΔTime = Force × Duration

Impulse = 30 × 0.1 = 3 N s.

Thus, the impulse experienced by the object is 3 N s.

6 0
3 years ago
The center of gravity of a loaded truck depends on how the truck is packed. If it is 4.0 m high and 2.4 m wide, and its CG is 2.
Vitek1552 [10]

The slope of the road can be given as the ratio of the change in vertical

distance per unit change in horizontal distance.

  • The maximum steepness of the slope where the truck can be parked without tipping over is approximately <u>54.55 %</u>.

Reasons:

Width of the truck = 2.4 meters

Height of the truck = 4.0 meters

Height of the center of gravity = 2.2 meters

Required:

The allowable steepness of the slope the truck can be parked without tipping over.

Solution:

Let, <em>C</em> represent the Center of Gravity, CG

At the tipping point, the angle of elevation of the slope = θ

Where;

tan\left(\theta \right) = \dfrac{\overline{AM}}{\overline{CM}}

The steepness of the slope is therefore;

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{\overline{AM}}{\overline{CM}} \times 100

Where;

\overline{AM} = Half the width of the truck = \dfrac{2.4 \, m}{2} = 1.2 m

\overline{CM} = The elevation of the center of gravity above the ground = 2.2 m

\mathrm{The \ steepness \  of  \ the  \ slope}= \dfrac{1.2}{2.2} \times 100 \approx 54.55\%

tan\left(\theta \right) = \mathbf{\dfrac{2.2}{1.2}} = \dfrac{11}{6}

Elevation \ of \ the \ road \ \theta = arctan\left( \dfrac{6}{11} \right)  \approx 28.6 ^{\circ}

The maximum steepness of the slope where the truck can be parked is <u>54.55 %</u>.

Learn more here:

brainly.com/question/20793607

3 0
2 years ago
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