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Licemer1 [7]
3 years ago
6

The other name for 'net force' is 'unbalanced force'. What is the name of the force that could be applied to an object that woul

d balance a net force and create a state of equilibrium?
Physics
1 answer:
jeka943 years ago
6 0

Answer:

Friction

Explanation:

This is because friction tends to oppose the motion of an object. Since the unbalance force or net force causes a motion of the object, the frictional force would tend to oppose the object until it is large enough to balance the net force. When this is done, equilibrium is achieved.

At this instance, there is no net force acting on the object.

So, the forces on the object become balanced and in a state of equilibrium.

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A box has a 20 N force applied to it to move it 5 m. What is the work done on the box? 4 J 4 N 25 J 100 J
Sunny_sXe [5.5K]
100 J

Explanation:
multiply the force by the distance
20 N x 5 meters = 100 J
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7 0
3 years ago
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Two point charges are fixed on the y axis: a negative point charge q1 = -25 μC at y1 = +0.18 m and a positive point charge q2 at
dedylja [7]

Answer:

50.91 \mu C

Explanation:

The magnitude of the net force exerted on q is known, we have the values and positions for q_{1} and q. So, making use of coulomb's law, we can calculate the magnitude of the force exerted byq_{1} on q. Then we can know the magnitude of the force exerted by q_{2} about q, finally this will allow us to know the magnitude of q_{2}

q_{1} exerts a force on q in +y direction, and q_{2} exerts a force on q in -y direction.

F_{1}=\frac{kq_{1} q }{d^2}\\F_{1}=\frac{(8.99*10^9)(25*10^{-6}C)(8.4*10^{-6}C)}{(0.18m)^2}=58.26 N\\

The net force on q is:

F_{T}=F_{1} - F_{2}\\25N=58.26N-F_{2}\\F_{2}=58.26N-25N=33.26N\\\mid F_{2} \mid=\frac{kq_{2}q}{d^2}

Rewriting for q_{2}:

q_{2}=\frac{F_{2}d^2}{kq}\\q_{2}=\frac{33.26N(0.34m)^2}{8.99*10^9\frac{Nm^2}{C^2}(8.4*10^{-6}C)}=50.91*10^{-6}C=50.91 \mu C

8 0
3 years ago
A man on the moon throws a ball vertically upwards and it is noticed that the ball travels 3.0m less in the fifth second of its
sdas [7]
<h2>Acceleration due to gravity in moon is 1.5 m/s²</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Here the ball travels 3 m less distance in fifth second compared to third second.

That is

           s₃ = s₅ + 3

Now we have

Distance traveled in third second, s₃ = u x 3 - 0.5 x g x 3² -  u x 2 - 0.5 x g x 2²

           s₃ = u - 2.5 g

Also

Distance traveled in fifth second, s₅ = u x 5 - 0.5 x g x 5² -  u x 4 - 0.5 x g x 4²

           s₅ = u - 4.5 g    

That is

           u - 2.5 g = u - 4.5 g + 3

             2 g = 3

                g = 1.5 m/s²

Acceleration due to gravity in moon = 1.5 m/s²

8 0
3 years ago
Which statement is true?
Fynjy0 [20]
If it's Kepler's law of equal areas you're talking about,
then the first of the four statements is true.
4 0
3 years ago
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Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti
liq [111]

Answer:

The speed of water must be expelled at 6.06 m/s

Explanation:

Neglecting any drag effects of the surrounding water we can assume the linear momentum in this case is conserves, that is, the total initial momentum of the octopus and the water kept in it cavity should be equal to the total final linear momentum. That's known as conservation of momentum, mathematically expressed as:

p_f=p_i

with Pi the total initial momentum and Pf the final total momentum. The total momentum is the sum of the momentums of the individual objects, in our case the octopus and the mass of water that will be expelled:

p_{of}+p_{wf}=p_{oi}+p_{wi}

with Po the momentum of the octopus and Pw the momentum of expelled water. Linear momentum is defined as mass times velocity:

m_o*v_{of}+m_w*v_{wf}=m_o*v_{oi}+m_w*v_{wi}

Note that initially the octopus has the water in its cavity and both are at rest before it sees the predator so v_{oi}=v_{wi} = 0\frac{m}{s}:

m_o*v_{of}+m_w*v_{wf}=0

We should find the final velocity of water if the final velocity of the octopus is 2.70 m/s, solving for v_{wf}:

v_{wf}=-\frac{m_o*v_{of}}{m_w}=-\frac{(6.00-1.85)*(2.70)}{1.85}

v_{wf}=-6.06\frac{m}{s}

The minus sign indicates the velocity of the water is opposite the velocity of the octopus.

3 0
3 years ago
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