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ra1l [238]
2 years ago
12

a 3-newton force and a 4-newton force are acting concurrently on a point. which force could not produce equilibrium with these t

wo forces?
Physics
1 answer:
Paladinen [302]2 years ago
7 0

A force that cannot produce equilibrium with these two forces must be less than 7 N or greater 7 N.

The given parameters;

  • <em>3 Newton force </em>
  • <em>4 Newton force</em>

A body is said to be in equilibrium state when the net force on the body is zero.

A force that produce equilibrium among two forces acting concurrently is known as equilibrant force. This force is usually equal to sum of the two concurrent forces.

F_1 + F_2 = F_3

where;

  • F₁ and F₂ are the concurrent forces
  • F₃ is the equilibrant force

From the given parameters;

F_1 + F_2 = 3 \ N + 4\ N  = 7\ N

Thus, a force that cannot produce equilibrium with these two forces must be less than 7 N or greater 7 N.

Learn more here:brainly.com/question/12582625

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Three point charges are arranged along the x-axis. Charge q1 = +3.00 uC is at the origin, and charge q2= -5.00 uC is at x= 0.200
artcher [175]
We have all the charges for q1, q2, and q3. 
Since k = 8.988x10^2, and N=m^2/c^2

F(1) = F (2on1) + F (3on1)

F(2on1) = k |q1 q2| / r(the distance between the two)^2
k^ | 3x10^-6 x -5 x 10^-6 |   / (.2m)^2
F(2on1) = 3.37 N

Since F1 is 7N,

F(1) = F (2on1) + F (3on1)
7N = 3.37 N + F (3on1)

Since it wil be going in the negative direction,
-7N = 3.37 N + F (3on1)
F(3on1) = -10.37N

F(3on1) = k |q1 q3| / r(the distance between the two)^2 
r^2 x F(3on1) = k |q1 q3| 
r = sqrt of k |q1 q3| / F(3on1) 
= .144 m (distance between q1 and q3)
0 - .144m 

So it's located in -.144m

Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help. 
6 0
3 years ago
Read 2 more answers
Its mass is 20 grams, and its density is 7.87 g/cm3. What’s the larger cube’s volume?
Alik [6]
Volume = mass / density
Volume = 20 / 7.87
Volume = 2.54 (2 s.f)
3 0
3 years ago
Is an object speeding up or slowing down of the V final is greater than the V initial?
sertanlavr [38]

Answer:

V is greater

Explanation:

because v intial at that time V final is the that speed which it is going at that time

7 0
3 years ago
A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b
Softa [21]

Answer:

  The net force on the block  F(net)  = mgsinθ).

   Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw  is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

                    F(net)  = mgsinθ

The net force on the block  F(net)  = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

                           n=mgcosθ

The horizontal component of normal force on the block is equal to force

                           Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

                           Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

6 0
3 years ago
Please help me with this​
masha68 [24]

Answer:

>400N is needed to balance that lever

7 0
2 years ago
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