Explain why it is impossible to whirl a bung around

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

Inessa [10]

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:

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

4 0

3 years ago

By quadrupling the mass of the body hanging on the thread what will become the period of oscillation?

nikklg [1K]

Well, the tension in the thread will probably quadruple, but the hanging body will continue to just hang there.

The question gives us no evidence that it is doing any oscillating, and there's no reason for it to start just because it suddenly got heavier.

6 0

3 years ago

shusha [124]

Answer: The unit of impulse is applied to an object produces an equivalent vector change in its linear momentum, also in the same direction.

Explanation:

4 0

2 years ago

Which of the following is an example of speed?

Arturiano [62]

The cat has a speed. The truck and the bicycle both have velocity. We can't be sure of what the plane has.

6 0

2 years ago

As the concentration of a solute in a solution increases, the freezing point of the solution ________ and the vapor pressure of

kykrilka [37]

Answer:

As the concentration of a solute in a solution increases, the freezing point of the solution decrease and the vapor pressure of the solution decrease .

Explanation:

Depression in freezing point :

$\Delta T_f=K_f\times m$

where,

$\Delta T_f$ =depression in freezing point =

$K_f$ = freezing point constant

m = molality ( moles per kg of solvent) of the solution

As we can see that from the formula that higher the molality of the solution is directly proportionate to the depression in freezing point which means that:

If molality of the solution in high the depression in freezing point of the solution will be more.
If molality of the solution in low the depression in freezing point of teh solution will be lower .

Relative lowering in vapor pressure of the solution is given by :

$\frac{p_o-p_s}{p_o}=\chi_{solute}$

$p_o$ = Vapor pressure of pure solvent

$p_s$ = Vapor pressure of solution

$\chi_{solute}$ = Mole fraction of solute

$p_s\propto \frac{1}{\chi_{solute}}$

Vapor pressure of the solution is inversely proportional to the mole fraction of solute.

Higher the concentration of solute more will the be solute's mole fraction and decrease in vapor pressure of the solution will be observed.
lower the concentration of solute more will the be solute's mole fraction and increase in vapor pressure of the solution will be observed.

8 0

2 years ago