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STatiana [176]
3 years ago
15

A baby stroller is a rest on top of a hill which is 10 m high. The stroller and baby have a mass of 20 kg. What is the potential

energy of the baby and stroller?
Formula: PE = m g h (Gravity near earth is 9.8 m/s²)
Physics
1 answer:
Anettt [7]3 years ago
6 0

The potential energy is defined as the energy is contained in the body due to the height over the surface of the earth and it is calculated from the equation

PE=mgh

<em>where:</em>

  • PE: the potential energy in Joules.
  • m: the mass of the body in kg.
  • g: the acceleration due to the gravity in m/s^2
  • h: the height of the body over the earth in meters.

<em>in our problem:</em>

PE=20*9.8*10=1960 J

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Answer:

q₁ = + 1.25 nC

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Theory of electrical forces

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Known data

q₃=5 nC

q₂=- 3 nC

d₁₃=  2 cm

d₂₃ = 4 cm

Graphic attached

The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.

For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So,  the charge q₁ must be positive(q₁+).

The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).

The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs.  F₂₃ is directed to the right (+x)

Calculation of q1

F₁₃ = F₂₃

\frac{k*q_{1}*q_3 }{(d_{13})^{2}  } = \frac{k*q_{2}*q_3 }{(d_{23})^{2}  }

We divide by (k * q3) on both sides of the equation

\frac{q_{1} }{(d_{13})^{2} } = \frac{q_{2} }{(d_{23})^{2} }

q_{1} = \frac{q_{2}*(d_{13})^{2}   }{(d_{23} )^{2}  }

q_{1} = \frac{5*(2)^{2} }{(4 )^{2}  }

q₁ = + 1.25 nC

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The height of the image formed is 18.89cm.

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