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frez [133]
4 years ago
7

You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. What

is the direction of the force on the charge you are holding?
1. to the east
2. to the southwest
3. to the south
4. to the northeast
Physics
1 answer:
balandron [24]4 years ago
6 0

Answer:

southwest

Explanation:

Here, all the particles have same charge this means that only repulsive force is acting on the particles according to Coulomb's laws.

One particle is 1 m to my north and another particle is to my east.

The particle to my north will push the positive charge I am holding to the south and the particle to my east will push the positive charge I am holding to the west.

Hence, the resultant direction of the force will be southwest.

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Vera_Pavlovna [14]

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Hope this helps!! :))

5 0
3 years ago
Three point charges, two positive and one negative, each having a magnitude of 20 C are placed at the vertices of an equilateral
Daniel [21]

The resultant force on the positive charge  is mathematically given as

X=40N

<h3>What is the magnitude of the electrostatic force on the negative charge?</h3>

Question Parameters:

Three-point charges, two positive and one negative, each having a magnitude of 20

Generally, the -ve charge   is mathematically given as

Q+=\sqrt{x^2+x^2+2x.xcos120}\\\\Q+=\sqrt{2x^2+2x*(1/2)}

Q+=X

Therefore

x=\frac{Kq1q2}{r2}\\\\x=\frac{9*10^9*20*10^{-6}*20*10^{-6}}{(30*10^-2)^2}

X=40N

For more information on Force

brainly.com/question/26115859

5 0
2 years ago
A proud new Jaguar owner drives her car at a speed of 25 m/s into a corner. The coefficients of friction between the road and th
ehidna [41]

Answer:

ac = 3.92 m/s²

Explanation:

In this case the frictional force must balance the centripetal force for the car not to skid. Therefore,

Frictional Force = Centripetal Force

where,

Frictional Force = μ(Normal Force) = μ(weight) = μmg

Centripetal Force = (m)(ac)

Therefore,

μmg = (m)(ac)

ac = μg

where,

ac = magnitude of centripetal acceleration of car = ?

μ = coefficient of friction of tires (kinetic) = 0.4

g = 9.8 m/s²

Therefore,

ac = (0.4)(9.8 m/s²)

<u>ac = 3.92 m/s²</u>

5 0
3 years ago
A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The
Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

\sum {F = ma}

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg

Here, {\mu _{\rm{s}}} is the coefficient of static friction, mm is mass of the object, and g is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass {m_1}

Substitute  for in the expression of maximum static friction {F_{\rm{f}}} = {\mu _{\rm{s}}}mg

{F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute  for , [/tex]{m_1}[/tex] for in the equation .

{F_{\rm{f}}} = {m_1}a

Substitute {\mu _{\rm{s}}}{m_1}g for {F_{\rm{f}}} in the equation {F_{\rm{f}}} = {m_1}a

{\mu _{\rm{s}}}{m_1}g = {m_1}a

Rearrange for a.

a = {\mu _{\rm{s}}}g

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right)a

Substitute for in the above equation .

{F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0
3 years ago
An object that is moving must have a change in A) its speed. B) its position. C) its acceleration. D) its applied force.
denis-greek [22]

An object that's moving doesn't necessarily change its speed or acceleration. Also, the force applied to it doesn't need to change ... in fact, a moving object doesn't need ANY force applied to it in order to keep moving.

But any moving object WILL have a change in its position ... THAT's how you know it's moving, and that's WHY you say "It's moving !". (choice-B)

8 0
3 years ago
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