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GarryVolchara [31]
1 year ago
9

How is force useful as well as harmful.Comment with the statement.​

Physics
1 answer:
vladimir1956 [14]1 year ago
7 0

No, we can't say force is useful or harmful. All else being equal, force is either not harmful or useful depending on the body that is applying it and where it is applying it.

Given how it affects motion, force is a crucial idea. It is a relationship that, in the absence of an opposing force, modifies an object's motion. However, a push or a pull that any object feels is the simplest definition of force. Due to the fact that force is a vector quantity, it possesses both a magnitude and a direction.

How force is useful

A body at rest can move with enough force.A body in motion may be slowed down or stopped by it.It has the power to quicken the pace of an object in motion.Along with its shape and size, it can also alter the direction of a moving body.

How force is harmful

  • Force has the power to alter an object's state of motion.
  • Moving objects can shift direction due to force.
  • Moving things' speeds can be increased by force.
  • Moving items can become slower due to force.
  • Force has the power to alter an object's shape.

Learn more about force here

brainly.com/question/14362949

#SPJ9

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How is electrolysis used in order to prevent materials from corrosion or rusting?​
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Two small charged spheres are located on the y-axis. One is at y = 1.00 m, the other is at y = −1.00 m, and they both have a cha
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Answer:

(a) 23.946 kV

(b) -0.077 J

Explanation:

(a) The electric potential is given by the following formula:

V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}   (1)

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

q1 = q2 = 1.60*10^{-6}C

r1 and r2 are the distance from the charges to the point in which electric potential is evaluated.

Firs, you calculate the distance r1 and r2 by taking into account the position of the charges

r_1=\sqrt{(1.00)^2+(0.670)^2}m=1.20m\\\\r_2=\sqrt{(-1.00)^2+(0.670)^2}m=1.20m

Next, you replace the values of the parameters to calculate V:

V=(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}+(8.98*10^9Nm^2/C^2)\frac{1.6*10^{-6}C}{1.20m}\\\\V=23946.66\ V=23.946\ kV

(b) The potential electric energy is given by:

U_T=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_T=k\frac{q_1q_2}{r_{1,2}}+k\frac{q_1q_3}{r_{1,3}}+k\frac{q_2q_3}{r_{2,3}}\\\\r_{1,2}=2.00m\\\\r_{1,3}=1.20m\\\\r_{2,3}=1.20m\\\\U_T=(8.98*10^9)[\frac{(1.6*10^{-6})^2}{2.00m}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}+\frac{(1.6*10^{-6})(-3.70*10^{-6})}{1.20}]J\\\\U_T=-0.077J

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3 years ago
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 To find the force, we multiply the area by the change in pressure: 2527 (231) = 583737N
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