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2 years ago

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Energy Calculation 4. What is the gravitational potential energy of an object that has a mass of 100kg it is 10m off of the grou

nd? The gravity on Earth is measured at 9.8N/kg

1 answer:

konstantin123 [22]2 years ago

3 0

100kg * 9.8N/kg * 10m = 9800

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The distance between the centers of the wheels of a motorcycle is 146 cm. The center of mass ofthe motorcycle, including the rid

Vitek1552 [10]

Answer:

9.12267515924 m/s²

Explanation:

Here the moment created by the wheels and the moment created by the center of gravity will balance each other.

h = Height of the center of mass = 78.5 cm

d = Distance from back wheel to the center of mass = $\dfrac{146\times 10^{-2}}{2}\ m$

g = Acceleration due to gravity = 9.81 m/s²

a = Horizontal acceleration

The equation is of the form

$mgd=Fh\\\Rightarrow mgd=mah\\\Rightarrow a=\dfrac{gd}{h}\\\Rightarrow a=\dfrac{9.81\times \dfrac{146\times 10^{-2}}{2}}{78.5\times 10^{-2}}\\\Rightarrow a=9.12267515924\ m/s^2$

The horizontal acceleration of the motorcycle that will make the front wheel rise off the ground is 9.12267515924 m/s²

8 0

3 years ago

Someone please help with this

SSSSS [86.1K]

Answer:

<em>The new force is 2/3 of the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

$\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}$

Factoring out 2/3:

$\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}$

Substituting the original force:

$F'=\frac{2}{3}F$

The new force is 2/3 of the original force

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