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Alexeev081 [22]

3 years ago

6

A 1400 kg wrecking ball hangs from a 20-m-long cable. the ball is pulled back until the cable makes an angle of 30.0 ∘ with the

vertical.part aby how much has the gravitational potential energy of the ball changed?

2 answers:

anastassius [24]3 years ago

5 0

From the geometry of the problem, the 20 m-long cable creates the hypotenuse of a right triangle, with the extended of the other two sides of size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased by 20 m - 17.3 m = 2.7 m.

The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 * 1.6 m , or about 37044 joules.

Bas_tet [7]3 years ago

3 0

The gravitational potential energy of the ball changes by about

37000 J

$\texttt{ }$

<h3>Further explanation</h3>

Let's recall the formula of Gravitational Potential Energy as follows:

$\large {\boxed {E_p = mgh }$

<em>Ep = Gravitational Potential Energy ( Newton )</em>

<em>m = Object's Mass ( kg )</em>

<em>g = Gravitational Acceleration ( m )</em>

<em>h = Height ( m )</em>

Let us now tackle the problem !

$\texttt{ }$

<u>Given:</u>

mass of wrecking ball = m = 1400 kg

length of the cable = L = 20 m

angle with the vertical = θ = 30.0°

<u>Asked:</u>

gravitational potential energy = Ep = ?

<u>Solution:</u>

$Ep = m g h$

$Ep = m g (L - L\cos \theta)$

$Ep = 1400 ( 9.8 ) ( 20 - 20 \cos 30^o )$

$Ep = 13720 ( 20 - 10\sqrt{3} )$

$Ep \approx 37000 \texttt{ J}$

$\texttt{ }$

<h3>Conclusion:</h3>

The gravitational potential energy of the ball changes by about

37000 J

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Newton's Law of Motion: brainly.com/question/10431582
Example of Newton's Law: brainly.com/question/498822

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

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Explanation:

From the question we are told that

The open circuit voltage is $V = 40.0V$

The internal resistance is $R = 2 \Omega$

The emf of each battery is $e = 6.00 V$

The internal resistance of the battery is $r = 0.300V$

The charging current is $I = 4.00 \ A$

Let assume the the additional resistance to to added to the circuit is $R_z$

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The total resistance in the circuit is

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Substituting values

$R_T = 2.6 +R_z$

And the difference in potential in the circuit is

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=> $E = 40 - (2 * 6)$

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Now according to ohm's law

$I = \frac{E}{R_T}$

Substituting values

$4 = \frac{28}{R_z + 2.6}$

Making $R_z$ the subject of the formula

So $R_z = \frac{28 - 10.4}{4}$

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Substituting values

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The increase rate of internal energy at the batteries is mathematically represented as

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Substituting values

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Substituting values

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$Z_3 = 70.4 W$

Generally the increase rate of the chemically energy in the battery is mathematically represented as

$C = 2 * e * I$

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