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Reil [10]
3 years ago
15

One horse is pulling a 755 kg sled straight ahead applying a force of 1988 N. If the acceleration of the sled is 1.36 m/s2, what

is the coefficient of kinetic friction between the sled and the ground?
Physics
1 answer:
Inessa [10]3 years ago
4 0

Answer:

The coefficient of kinetic friction is 0.13

Explanation:

Newton's second law states that the acceleration of an object is proportional to the net force on it, the factor of proportionality is the mass. So, we can express that law mathematically as:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

With F the net force, m the mass and a the acceleration of the object. In our case we're interested on what's happening to the sled, then we have to analyze the forces on it, those forces are the weight and the normal force on the vertical direction and the pulling force and frictional force in the horizontal direction. So, because (1) is a vector equation we can express that in their vertical (y) and horizontal (x) components:

F_y=ma_y (2)

F_x=ma_x (3)

On y we have that the acceleration is zero because the sled is not moving upward or downward, remember that the net force on y is the weight (W) pointing downward and the normal force pointing upward:

F_y=W+n=0

Following the convention that positive is upward and negative downward, W=mg=(755)(-9.81):

F_y=(755)(-9.81)+n=0

n=7406.55 N (4)

Now on the x direction we have the sum of the forces is the pulling force (T) and friction force (f)

F_x=F+f=ma_x

Choosing the direction where the horse is pulling F=1988N and the acceleration should be positive too, then:

1988+f=m(1.36)

f=(755)(1.36)-1988=-961.2 N

The negative sign means it's in the opposite direction the horse is pulling

The frictional force is related with the coefficient of kinetic friction in the next way:

|f|=\mu_k n

with μk the coefficient of kinetic friction, and n the normal force that we already found on (4), so we simply solve the last equation for μk:

\mu_k=\frac{|f|}{n}=\frac{961.2}{7406.55}=0.13

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7 0
3 years ago
The diagram does not represent a real electric field because the field lines, can someone help explain this for me
Papessa [141]

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It is the graphical way to represent the electric field variation

If we draw the tangent to electric field line then it will give the direction of net electric field at that point

So whenever we draw the electric field lines of a charge distribution then it will always follow this basic properties

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now as we can see that here two positive charges are placed nearby so the electric field must be like it can not intersect at any point because at intersection of two lines the direction of electric field not defined

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4 0
3 years ago
A proton moves from a location where V = 87 V to a spot where V = -40 V. (a) What is the change in the proton's kinetic energy?
Art [367]

Answer: a) 127 eV; b) there is no change of kinetic energy.

Explanation: In order to explain this problem we have to use the change of potentail energy ( conservative field) is equal to changes in kinetic energy. So for the proton ther move to lower potential then they gain kinetic energy from the electric field.  This means the electric force do work in this trayectory and then the protons increased changes its speed.

If we replace the proton by a electron we have a very different situaction, the electrons are located in a lower potental then  they can not move to higher potential  if any  external force does work on the system.

In resumem, the electrons do not move from a point with V=87 to other point with V=-40 V. The electric force point to high potential so the electrons  can not move to lower potential region (V=-40V).

6 0
3 years ago
A process with a negative change in enthalpy and a negative change in entropy will generally be:________
valentina_108 [34]

A process with a negative change in enthalpy and a negative change in entropy will generally be: <u>spontaneous</u>.

<h3>Gibbs free energy:</h3>

Since the Gibbs free energy is a parameter that tells us whether a chemical reaction is spontaneous (Gibbs free energy less than 0) or nonspontaneous (Gibbs free energy greater than 0) in this situation, we can describe it mathematically as:

                                      ΔG = ΔH - TΔS

Therefore, any process with a negative change in enthalpy and a positive change in entropy will be spontaneous. If the enthalpy and the entropy are both negative, the subtraction becomes always negative, for which the Gibbs free energy is also negative.

One of the most crucial thermodynamic functions for the characterization of a system is the Gibbs free energy. It influences results like the voltage of an electrochemical cell and the equilibrium constant for a reversible reaction, among others.

Learn more about spontaneous here:

brainly.com/question/16975806

#SPJ4

3 0
1 year ago
12
Harrizon [31]

Answer:

B. 6

Explanation:

i think... im in 7th grade and haven't really leaned this but im like 60% sure but i migjt be wrong

7 0
2 years ago
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