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Alexus [3.1K]
3 years ago
7

Please could you guys help me. When an object falls from a height, the maximum energy transferred to its kinetic energy store is

Physics
1 answer:
lesya692 [45]3 years ago
3 0

Answer:

the energy transferred away from its gravitational potential energy store

Explanation:

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

K.E = \frac{1}{2}MV^{2}

Where, K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Potential energy can be defined as an energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

P.E = mgh

Where,

P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

When an object falls from a height, the maximum energy transferred to its kinetic energy store is equal to the energy transferred away from its gravitational potential energy store.

This ultimately implies that, while the object was at rest it possessed potential energy, which is then transformed into motion when it starts to fall.

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Forces normal to a particle's displacement do no work.
snow_tiger [21]

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Explanation:

The work done is defined as the product of force applied in the direction of displacement and the displacement.

W = F x d x Cosθ

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3 0
3 years ago
Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m
Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC

In words, the value of q₃ must be 5.3nC.

7 0
3 years ago
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