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pshichka [43]
2 years ago
14

A 1.00-kilogram ball is dropped from the top of a building. just before striking the ground, the ball's speed is 12.0 meters per

second. what was the ball's gravitational potential energy, relative to the ground, at the instant it was dropped? [neglect friction.]
Physics
1 answer:
Anarel [89]2 years ago
6 0
During the fall, the potential energy stored in the ball is converted into kinetic energy.
Thus,
PE = KE before hitting the ground
= 1/2 • mv^2
= 1/2 • 1 • 12^2
= 72J
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Answer:

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Explanation:

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Replacing (2) in (1) and replacing the given values:

a_c=\frac{(\frac{2\pi r}{T})^2}{r}\\a_c=\frac{4\pi^2 r}{T^2}\\a_c=\frac{4\pi^2 (1.85m)}{(2.75s)^2}\\a_c=9.66\frac{m}{s^2}

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kodGreya [7K]

Find the given attachment for solution

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1 year ago
Calculate the calories lost when 95 g of water cools from 45 ∘C to 29 ∘C. Express your answer to two significant figures and inc
lubasha [3.4K]

Answer:

1,520.00 calories

Explanation:

Water molecules are linked by hydrogen bonds that require a lot of heat (energy) to break, which is released when the temperature drops. That energy is called specific heat or thermal capacity (ĉ) when it is enough to change the temperature of 1g of the substance (in this case water) by 1°C. Water ĉ equals 1 cal/(g.°C).

Given that ĉ = Q / (m.ΔT),

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m= mass of the substance (unity: grams or g), and

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8 0
2 years ago
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Tasya [4]

Answer:

17 m/s south

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m_1 = Mass of dog = 10 kg

m_2 = Mass of skateboard = 2 kg

v = Combined velocity = 2 m/s

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u_2 = Velocity of skateboard

In this system the linear momentum is conserved

(m_1+m_2)v+m_1u_1+m_2u_2=0\\\Rightarrow u_2=-\dfrac{(m_1+m_2)v+m_1u_1}{m_2}\\\Rightarrow u_2=-\dfrac{(10+2)2+10\times 1}{2}\\\Rightarrow u_2=-17\ m/s

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The gap between electrodes in a spark plug is 0.060 cm. Producing an electric spark in a gasoline-air mixture requires an electr
VladimirAG [237]

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Given data,

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ΔV=-1800 V

The minimum potential difference must be supplied by the ignition circuit to start a car is -1800 V

6 0
2 years ago
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