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ladessa [460]
3 years ago
10

Part III: If a mouse and an elephant both run with the same kinetic energy, can you say which is running faster? Use the space b

elow to thoroughly explain your answer (hint: use the book, slides, and other resources to pull information to support your answer - remember, KE = ½ m*v²).
Physics
1 answer:
Verdich [7]3 years ago
3 0

Answer:

The mouse runs faster to have the same kinetic energy as the elephant.

Explanation:

Note from the equation given, mass (m) is directly proportional to KE. This means an elephant with more mass will have more KE, therefore, for the mouse to compensate, it has to run faster because its KE is smaller because of its small mass. If both run at the same speed, the elephant would have thousands of times more kinetic energy than the mouse. So the mouse has to run faster so that its speed compansates for its smaller weight.

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As an object falls from rest, its gravitational energy is converted to kinetic energy

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4 years ago
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Two resistors, R1 and R2, are
Aleonysh [2.5K]

Answer:

The value of R₂ is equal to 24.75 ohms.

Explanation:

Given that,

Two resistors, R₁ and R₂, are  connected in parallel.

The equivalent resistance is 14.5 ohms

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When two resistors are connected in parallel. The equivalent resistance is given by :

\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\\\\\dfrac{1}{14.5}=\dfrac{1}{35}+\dfrac{1}{R_2}\\\\\dfrac{1}{R_2}=\dfrac{1}{14.5}-\dfrac{1}{35}\\\\\dfrac{1}{R_2}=0.04039\\\\R_2=\dfrac{1}{0.04039}\\\\R_2=24.75\ \Omega

So, the value of R₂ is equal to 24.75 ohms.

8 0
3 years ago
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of −25.0 nc? g?
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To calculate the force between two negative charges, we use the formula which is given by the Coulomb`s Law as

F=\frac{kq_{1}q_{2}  }{r^{2} }

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Given  q_{1} =q_{2} =-25 nC=-25\times 10^{-9}C and  r=8 cm=8\times10^{-2} m.

Substituting these values in above formula we get,

F= 8.99\times 10^9 N m^2/C^2\frac{(-25\times 10^{-9}C)(-25\times 10^{-9}C)}{(8\times10^{-2} m)^2} \\\\\ F= 877929.7\times 10^{-9}  N\\\\F=8.8\times10^{-4}N

Thus, the repulsive force between two pith balls is 8.8\times10^{-4}N.

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Answer:

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