Physics, 26.06.2020 15:01 varonadestiny1109

A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on the ice and slides 115 m before coming to rest, determine the coefficient of kinetic friction between the puck and ice.

Answers: 2

Show answers

Answers

Answer from: SoccerHalo

The coefficient of kinetic friction between the hockey puck and ice is equal to 0.178

Given the following data:

Initial speed = 20 m/sFinal velocity = 0 m/s (since it came to rest)Distance = 115 m

Scientific data:

Acceleration due to gravity = 9.8 m/s^2

To determine the coefficient of kinetic friction between the hockey puck and ice:

First of all, we would calculate the acceleration of the hockey puck by using the third equation of motion.

$V^2 = U^2 + 2aS\\\\0^2 =20^2 + 2a(115)\\\\-400=230a\\\\a=\frac{-400}{230}$

Acceleration, a = -1.74 m/s^2

Note: The negative signs indicates that the hockey puck is slowing down or decelerating.

From Newton's Second Law of Motion, we have:

$\sum F_x = F_k + F_n =0\\\\F_k =- F_n\\\\\mu mg =-ma\\\\\mu = \frac{-a}{g}\\\\\mu = \frac{-(-1.74)}{9.8}\\\\\mu = \frac{1.74}{9.8}$

Coefficient of kinetic friction = 0.178

Read more: link

Answer from: lesliezavalaa16

μ_k = 0.1773

Explanation:

We are given;

Initial velocity;u = 20 m/s

Final velocity;v = 0 m/s (since it comes to rest)

Distance before coming to rest;s = 115 m

Let's find the acceleration using Newton's second law of motion;

v² = u² + 2as

Making a the subject, we have;

a = (v² - u²)/2s

Plugging relevant values;

a = (0² - 20²)/(2 × 115)

a = -400/230

a = -1.739 m/s²

From the question, the only force acting on the puck in the x direction is the force of friction. Since friction always opposes motion, we see that:

F_k = −ma - - - (1)

We also know that F_k is defined by;

F_k = μ_k•N

Where;

μ_k is coefficient of kinetic friction

N is normal force which is (mg)

Since gravity acts in the negative direction, the normal force will be positive.

Thus;

F_k = μ_k•mg - - - (2)

where g is acceleration due to gravity.

Thus,equating equation 1 and 2,we have;

−ma = μ_k•mg

m will cancel out to give;

-a = μ_k•g

μ_k = -a/g

g has a constant value of 9.81 m/s², so;

μ_k = - (-1.739/9.81)

μ_k = 0.1773

Answer from: Quest

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

$p(t)=0.3t^2+6t+80$

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

according to question, we have said that if the population has twice its initial population, then it becomes

$p(t)=2\times 80=160$

so, time taken to reach this above mentioned population is given by

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.

Answer from: Quest

The objects do not reach terminal velocity there is not enough time from the time the object falls from when it hits the ground to achieve terminal velocity. an object must be placed higher to achieve terminal velocity in order to be in free fall.

Another question on Physics

Physics, 22.06.2019 07:20

If 2 moles of co_2 at 2l and 500k are expanded reversibly to 20l, more work can be obtained from an adiabatic process than from an isothermal process. is the above statement true or false?

Answers: 3

Answer

Physics, 22.06.2019 17:40

The weights of bags filled by a machine are normally distributed with a standard deviation of 0.05 kilograms and a mean that can be set by the operator. at what level should the mean weight be set if it required that only 1% of the bags weigh less than 9.5 kilograms? round the answer to 2 decimal places.

Answers: 1

Answer

Physics, 23.06.2019 01:00

Learning goal: to understand newton's 1st law. newton's principia states this first law of motion: an object subject to no net force maintains its state of motion, either at rest or at constant speed in a right line. this law may be restated as follows: if the sum of all forces acting on an object is zero, then the acceleration of that object is zero. mathematically this is just a special case of the 2nd law of motion, f⃗ =ma⃗ , when f⃗ =0. when studying newtonian mechanics, it is best to remember the 1st law in two ways: if the net force (i.e., sum of all forces) acting on an object is zero, the object will keep moving with constant velocity (which may be zero). if an object is moving with constant velocity, that is, with zero acceleration, then the net force acting on that object must be zero. complete the following sentences to see if you can apply these ideas. part a if a car is moving to the left with constant velocity, one can conclude that

Answers: 1

Answer

Physics, 23.06.2019 02:30

If jason ii needs 15 kw of power at 300 volts to operate, what amperage of current must flow through the cable?

Answers: 2

Answer

You know the right answer?

A hockey puck on a frozen pond is given an initial speed of 20.0 m/s. If the puck always remains on...

Questions