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3 years ago

For every action there is an equal and opposite reaction. What does these words imply?

Physics

2 answers:

Finger [1]3 years ago

7 0

Answer:

Explanation:

"For every action, there is an equal but opposite reaction" is the 3rd Netwon's law of motion.

Examples of Newton's third law of motion are found in our everyday life. For example,

• Gun Firing: when someone fire gun the reaction force push the gun backward.

• When you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air.

Engineers find numerous applications in Newton's third law when designing rockets and other projectile devices.

MrRa [10]3 years ago

3 0

Answer: Newton's third law of motion

Explanation:

Newton's third law of motion states that, "for every action, there is an equal and opposite reaction"

This means that in every interaction, a pair of forces usually acts on the two interacting objects. Every action that happens or takes place, has an opposite reaction to it, which is also equal.

Example, take a look at the motion of a vehicle on your way to school. The vehicle has wheels that spin. As the wheels spin, they grip the road and push the road backwards. The road then pushes the wheels forward, since forces results from mutual interactions. The size of the force on the road is equal to the size of the force on the car; the direction of the force on the wheels (forwards) is opposite the direction of the force on the road (backwards)

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4 years ago

Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

Arlecino [84]

Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
$d({\rm C}) = \sqrt{3} \times (0.050\; \rm m)$ .

Let $k$ denote Coulomb's constant ( $k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}$ .)

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm A})}$ .

Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

Hence, the electric field at the midpoint of $\rm AB$ due to all these three charges would be:

$\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}$ .