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2 years ago

if you push your chair across the floor at a constant velocity how does the force of friction compare with the force you exert?

2 answers:

7 0

The force of friction is equal but opposite to the force that you exert. This is due to newtons 3rd law 'For every action there is an equal but opposite reaction'

Alik [6]2 years ago

5 0

-- If velocity is constant, then there is no net force
on the chair.

-- If there is no net force on the chair, then friction
must exactly balance out your push.

-- The force of friction is exactly equal in magnitude
to your push, and in exactly the opposite direction.

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When red-headed woodpeckers (melanerpes erythrocephalus) strike the trunk of a tree, they can experience an acceleration ten tim

kifflom [539]

98.1 m/s² × (1 in)/(0.0254 m) ≈ 3862 in/s²

8 0

3 years ago

Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if

klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:

"The electrostatic force $F_{E}$ between two point charges $q_{1}$ and $q_{2}$ is proportional to the product of the charges and inversely proportional to the square of the distance $d$ that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

$F_{E}= K\frac{q_{1}.q_{2}}{d^{2}}$ (1)

Where:

$F_{E}=60 N$ is the electrostatic force

$K=8.99(10)^{9} Nm^{2}/C^{2}$ is the Coulomb's constant

$q_{1}$ and $q_{2}$ are the electric charges

$d=3 m$ is the separation distance between the charges

Then:

$60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}}$ (2)

Isolating $q_{1}$ and $q_{2}$ :

$q_{1}q_{2}=6(10)^{-8} C^{2}$ (3)

Now, if we keep the same charges but we decrease the distance to $d_{1}=1 m$ , (1) is rewritten as:

$F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}}$ (4)

Then, the new electrostatic force will be:

$F_{E}= 539.4 N$ (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0

2 years ago

The electric field between two parallel plates connected to a 45-volt battery is 500. volts per meter. The distance between the

kotykmax [81]

Electric field = potential difference
-----------------------------
distance between plates

Distance between plates = 45
----------
500

= 0.09 meters.

8 0

2 years ago

If a 5 N force pushes a 20 kg mass on a frictionless surface, how fast is the mass going in 5 seconds.

stira [4]

Explanation:

Let me know if you have questions

3 0

3 years ago

how many times does the kinectic energy of a car increase when traveling 60 mph as opposed to traveling 30 mph

Bumek [7]

The car at 60 kph has 9 times more kinetic energy than the car traveling at 20 kph. This assumes that both cars have the same mass. Kinetic energy depends on the square of thee speed so if one car is going 3 times faster, its kinetic energy will be 3^2 ( = 9 ) greater. The car going at 60 kph will have 4 times the KE of the car going at 30 kph ( again assuming that the cars have the same mass.)

3 0

3 years ago