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julia-pushkina [17]
4 years ago
7

Can an object be at rest if forces are being applied to it

Physics
1 answer:
notka56 [123]4 years ago
6 0

Answer:No

Explanation: No, If forces are being applied to the object the object will not remain at rest.

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There are five basic health-related components that one must have in order to be physically fit.
kirza4 [7]

Answer: the answer is true :-)

Explanation:

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A ball of mass 2 kg is being pulled sideways by a force 3 N to the right, and a force of 8N to the left - these are the only for
Alla [95]

Answer:

Explanation:

1.) What is the net force in the horizontal (x) direction?

Fnet = 8 - 3 = 5 N left

2.) what is the acceleration in the horizontal (x) direction?​

a = Fnet/m = 5/2 = 2.5 m/s² left

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Calculate the total force on the Earth due to Venus, Jupiter, and Saturn, assuming all four planets are in a line. The masses ar
Furkat [3]
You need to consider the following:
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<span>
Formula:
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The fraction ratio is  1/36800
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3 0
3 years ago
Read 2 more answers
Tech A says that a relay is a one-way electrical check valve used in alternators to change AC into DC. Tech B says that a relay
Setler [38]

Answer:

The correct answer is from Tech B.

Explanation:

The relay is an electromagnetic device, which is stimulated by a very weak electrical current to open or close a circuit.  It functions as a switch controlled by an electrical circuit in which, by means of a coil and an electromagnet, a sequence of one or several contactors is activated that allow to open or close other electrical circuits independent of the one that controls it.

Have a nice day!

7 0
3 years ago
two point charges of magnitude 4.0 μc and -4.0 μc are situated along the x-axis at x1 = 2.0 m and x2 = -2.0 m, respectively. wha
user100 [1]

The electric potential at the origin of the xy coordinate system is negative infinity

<h3>What is the electric field due to the 4.0 μC charge?</h3>

The electric field due to the 4.0 μC charge is E = kq/r² where

  • k = electric constant = 9.0 × 10 Nm²/C²,
  • q = 4.0 μC = 4.0 × 10 C and
  • r = distance of charge from origin = x₁ - 0 = 2.0 m - 0 m = 2.0 m

<h3>What is the electric field due to the -4.0 μC charge?</h3>

The electric field due to the -4.0 μC charge is E = kq'/r² where

  • k = electric constant = 9.0 × 10 Nm²/C²,
  • q' = -4.0 μC = -4.0 × 10 C and
  • r = distance of charge from origin = 0 - x₂ = 0 - (-2.0 m) = 0 m + 2.0 m = 2.0 m

Since both electric fields are equal in magnitude and directed along the negative x-axis, the net electric field at the origin is

E" = E + E'

= -2E

= -2kq/r²

<h3>What is the electric potential at the origin?</h3>

So, the electric potential at the origin is V = -∫₂⁰E".dr

= -∫₂⁰-2kq/r².dr

Since E and dr = dx are parallel and r = x, we have

= -∫₂⁰-2kqdxcos0/x²

= 2kq∫₂⁰dx/x²

= 2kq[-1/x]₂⁰

= -2kq[1/x]₂⁰

= -2kq[1/0 - 1/2]

= -2kq[∞ - 1/2]

= -2kq[∞]

= -∞

So, the electric potential at the origin of the xy coordinate system is negative infinity

Learn more about electric potential here:

brainly.com/question/26978411

#SPJ11

3 0
2 years ago
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