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gregori [183]
3 years ago
8

The value of gravitational acceleration of a body on earth is 9.8m/s^2. The gravitational potential energy for a 1.00 kilogram o

bject is found to be 12.5 joules. How high is this object above the ground, where PE=0?
Physics
1 answer:
emmasim [6.3K]3 years ago
4 0

Answer:

The object is 1.2755 meters above the ground

Explanation:

Recall the formula for gravitational potential energy for an object of mass "m" at a height "h" above the ground:

PE=m\,*\,g\,*\,h

In this case, since we are given the mass of the object and the object's potential energy, we can estimate the only unknown (height "h") from the formula shown above. Also since all units are given in the SI system, the result for the object's height will result in meters:

PE=m\,*\,g\,*\,h\\12.5\,J=1.0\,kg\,*9.8\,\frac{m}{s^2} \,*\,h\\\frac{12.5}{9.8} \,meters= h\\h=1.2755\ meters

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The diagram represents two charges, q1 and q2 separated by a distanced "d" which change would produce the greatest increase in t
ollegr [7]

Answer:

The diagram represents two charges, q1 and q2, separated by a distance d. Which change would produce the greatest increase in the electrical force between the two charges? *

Explanation:

doubling charge q1, only

4 0
2 years ago
Which statement is true about Car A?
alexira [117]

Answer:

Car A has a velocity of 40 km/hr  west.

Explanation:

The velocity of an object is a vector quantity. It has both magnitudes as well as direction.

Arrows are used to show direction.

Car A is moving towards west. Car B is moving towards east-south direction. Car C is moving towards South direction. Car E is moving towards North direction.

So, the correct option is (a) i.e. Car A has a velocity of 40 km/hr  west.

6 0
2 years ago
Two ice skaters, one with a mass of 45 kg and the other with a mass of 60 kg, push off against one another starting from a stati
Lina20 [59]

Answer:

0.25714 m/s

Explanation:

m_1 = Mass of 45 kg skater

m_2 = Mass of 60 kg skater

v_1 = Speed of 45 kg skater

v_2 = Speed of 60 kg skater

As the momentum is conserved

m_1v_1+m_2v_2=0\\\Rightarrow 45v_1+60v_2=0\\\Rightarrow v_1=-\frac{60v_2}{45}\\\Rightarrow v_1=-\frac{4}{3}v_2

v_r = Relative speed = 0.6 m/s

v_r=v_2-v_1\\\Rightarrow v_r=v_2-\left(-\frac{4}{3}v_2\right)\\\Rightarrow 0.6=\frac{7}{3}v_2\\\Rightarrow v_2=\frac{0.6\times 3}{7}\\\Rightarrow v_2=0.25714

The speed of the 60kg skater is 0.25714 m/s

4 0
3 years ago
You are playing a violin, where the fundamental frequency of one of the strings is 440 Hz, as you are standing in front of the o
Natalka [10]

Answer:

a)   L = 440 cm

Explanation:

In the open tube on one side and cowbell on the other, we have a maximum in the open part and a node in the closed part, therefore the resonance frequencies are

             λ₁ = 4L             fundamental

             λ₃ = 4L / 3       third harmonic

             λ₅ = 4L / 5       five harmonic

             

The violin string is a fixed cure in its two extracts, so both are nodes, their length from resonance wave are

              λ₁ = 2L                    fundamental

             λ₂ = 2L / 2              second harmonic

             λ₃ = 2L / 3              third harmonic

             λ₄= 2L / 4               fourth harmonic

They indicate that resonance occurs in the fourth harmonic, let's look for the frequency

              v =λ f

for the fundamental

              v = λ₀ f₀

              V = 2L f₀

for the fourth harmonica

              v = λ₄  f ’

              v = L / 2  f'

             2L f₀ = L / 2 f ’

             f ’= 4 f₀

             f ’= 4 440

             f ’= 1760 Hz

for this frequency it has the resonance with the tube

           f ’= 4L

           L = f ’/ 4

           L = 1760/4

           L = 440 cm

b) let's find the frequency of the next harmonic in the tube

             λ₃ = 4L / 3

             λ₃ = 4 400/3

             λ₃ = 586.6 cm

            v = λf

            f = v / λlam₃

            f₃3 = 340 / 586.6

            f3 = 0.579

as the minimum frequency on the violin is 440 Beam there is no way to reach this value, therefore there are no higher resonances

6 0
3 years ago
During a total lunar eclipse, the moon remains visible because ________.
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During a total lunar eclipse, the Moon is still visible to the naked eye due to the fact that the atmosphere of our planet, which is Earth bends, which results to the scattering of some sunlight, which actually reaches the Moon during a Lunar Eclipse, which is why the Moon remains visible despite the occurrence of a full lunar eclipse.

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3 years ago
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