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Travka [436]
3 years ago
11

A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static and kinetic friction

between the hockey puck and the metal ramp are μs = 0.40 and μk = 0.30, respectively. The puck's initial speed is 26 m/s. What vertical height does the puck reach above its starting point?

Physics
1 answer:
nikitadnepr [17]3 years ago
6 0

Answer:

71.76 m

Explanation:

We will solve this question using the work energy theorem.

The theorem explains that, the change in kinetic energy of a particle between two points is equal to the workdone in moving the particle from the one point to the other.

ΔK.E = W

In the attached free body diagram for the question, the forces acting on the puck are given.

ΔK.E = (final kinetic energy) - (initial kinetic energy)

Final kinetic energy = 0 J (since the puck comes to a stop)

Initial kinetic energy = (1/2)(m)(v²) = (1/2)(0.2)(26²) = 67.6 J

ΔK.E = 0 - 67.6 = - 67.6 J

W = Workdone between the starting and stopping points = (work done by the force of gravity) + (work done by frictional force)

Work done by the force of gravity = - mgh = - (0.2)(9.8)(h) = - 1.96 h

Workdone by the frictional force = F × d

F = μ N

μ = coefficient of kinetic friction = 0.30 (kinetic frictional force is the only frictional force that moves a distance of d, the static frictional force doesn't move any distance, so it does no work)

N = normal reaction of the plane surface on the puck = mg cos 30° = (0.2)(9.8)(0.866) = 1.697 N

F = μ N = 0.3 × 1.697 = 0.509 N

where d = distance along the incline that the puck travels.

d = h/sin 30° = 2h (from trigonometric relations)

Workdone by the frictional force = F × d = 0.509 × 2h = 1.02 h

ΔK.E = W = (work done by the force of gravity) + (work done by frictional force)

- 67.6 = - 1.96h + 1.02h

-0.942h = - 67.6

h = 71.76 m

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Answer:

scalar,magnitude

Explanation:

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8 0
2 years ago
A 0.54 kg particle has a speed of 5.0 m/s at point A and kinetic energy of 7.5 J at point B. What is
Wewaii [24]

<u>Answers</u>

(a)  6.75 Joules.

(b)  5.27 m/s

(c) 0.75 Joules


<u>Explanation</u>

Kinetic energy is the energy possessed by a body in motion.

(a) its kinetic energy at A?

K.E = 1/2 mv²

       = 1/2 ×  0.54 × 5²

       = 6.75 Joules.

(b) its speed at point B?

K.E = 1/2 mv²

7.5 = 1/2 × 0.54 × V²

V² = 7.5 ÷ 0.27

     = 27.77778

V = √27.77778

   = 5.27 m/s

(c) the total work done on the particle as it moves from A to B?

Work done = 7.5 - 6.75

                 = 0.75 Joules

4 0
3 years ago
Review. As an astronaut, you observe a small planet to be spherical. After landing on the planet, you set off, walking always st
anyanavicka [17]

To find the mass of the planet we will apply the relationship of the given circumference of the planet with the given data and thus find the radius of the planet. From the kinematic equations of motion we will find the gravitational acceleration of the planet, and under the description of this value by Newton's laws the mass of the planet, that is,

The circumference of the planet is,

\phi = 25.1m

Under the mathematical value the radius would be

\phi = 2\pi r

r = \frac{25}{2\pi}

r = 3.9788km

Using second equation of motion

x = \frac{1}{2} at^2

Replacing the values given,

1.4 = \frac{1}{2} a (29.2)^2

Rearranging and solving for 'a' we have,

a = 0.003283m/s^2

Using the value of acceleration due to gravity from Newton's law we have that

a = \frac{GM}{r^2}

Here,

r = Radius of the planet

G = Gravitational Universal constant

M = Mass of the Planet

\frac{(6.67*10^{-11})*M}{(3.9788*10^3)^2} = 0.003283

M = 7.79201*10^{14}kg

Therefore the mass of this planet is 7.79201*10^{14}kg

5 0
3 years ago
A mass of (200 g) of hot water at (75.0°C) is mixed with cold water of mass M at (5.0°C). The final temperature of the mixture i
SIZIF [17.4K]

The mass of the cold water, given the data from the question is 500 g

<h3>Data obtained from the question</h3>
  • Mass of warm water (Mᵥᵥ) = 200 g
  • Temperature warm water (Tᵥᵥ) = 75 °C
  • Temperature of cold water (T꜀) = 5 °C
  • Equilibrium temperature (Tₑ) = 25 °C
  • Specific heat capacity of the water = 4.184 J/gºC
  • Mass of cold water (M꜀) =?

<h3>How to determine the mass of the cold water </h3>

Heat loss = Heat gain

MᵥᵥC(Tᵥᵥ – Tₑ) = M꜀C(Tₑ – T꜀)

200 × 4.184 (75 – 25) = M꜀ × 4.184(25 – 5)

41840 = M꜀ × 83.68

Divide both side 83.68

M꜀ = 41840 / 83.68

M꜀ = 500 g

Learn more about heat transfer:

brainly.com/question/6363778

#SPJ1

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2 years ago
Which of the following is an example of a measurement of velocity?
tatyana61 [14]
 velocity is magnitude and direction. its unit is meter per second
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