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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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3 years ago
Can someone help me with this question
Mademuasel [1]

Answer:

Net force: 20 N to the right

mass of the bag: 20.489 kg

acceleration:  0.976  m/s^2

Explanation:

Since the normal force and the weight are equal in magnitude but opposite in direction, they add up to zero in the vertical direction. In the horizontal direction, the 195 N tension to the right minus the 175 force of friction to the left render a net force towards the right of magnitude:

195 N - 175 N = 20 N

So net force on the bag is 20 N to the right.

The mass of the bag can be found using the value of the weight force: 201 N:

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and the acceleration of the bag can be found as the net force divided by the mass we just found:

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3 years ago
What is conductivity?​
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3 years ago
Explain why chlorine forms a -1 ion, but sulfur forms a -2 ion.
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3 years ago
Point PP is a distance d1d1 = 4.0 mm above a large sheet of metal that carries a current of 35 AA in the positive xx direction a
zhenek [66]

Answer:

The width of the sheet is   w =0.8046m

Explanation:

  From the question we are told that

             The distance of point P above a  large sheet of metal is D = 4.0mm =\frac{4}{1000}  = 0.004m

             The current on the large metal sheet  is  I =34A

              The  distance of the the point P below a long wire d = 3.0mm = \frac{3}{1000} = 0.003m

               The current on the long wire is  I_w = 0.41A

                The magnetic field at  P is  B = 0T

Generally magnetic field  of P long wire is mathematically represented as

                B_w = \frac{\mu_o I_w}{2\pi r}

Generally magnetic field  of P large sheet of meta is mathematically represented as

              B_m = \frac{\mu_o K}{2}

Where K is the current per unit width

 The total magnetic field at P is

                  \frac{\mu_o I_w}{2 \pi r}  = \frac{\mu_o K}{2}

Making K the subject of formula

                 K = \frac{2 I_w }{2 \pi r }

 Substituting values

                   K = \frac{2 *  0.41 }{2 * 3.142 * (0.0030) }

                        K = 43.4967 A/m

Generally K is mathematically represented as

                K = \frac{I}{w}

Where w is the width of the large sheet

Therefore the width  of the metal sheet   w = \frac{I}{K}

                                                                         = \frac{35}{43.4967}

                                                                         w =0.8046m

             

                                                               

                                                                 

                   

7 0
3 years ago
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