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chubhunter [2.5K]
3 years ago
5

Your friend, Tim, is playing with his sled. He ties a rope to his sled and attaches it to Lar's snowmobile. Tim has a mass of 71

kg, the tension in the rope is 220 N (horizontally), and a coefficient of kinetic friction is 0.1. Unfortunately, Tim can only hang on for 8 s.
(a) What is Tim's maximum speed?
Physics
1 answer:
kykrilka [37]3 years ago
7 0

The maximum speed of Tim is 16.95 m/s.

The given parameters:

  • Mass of the rope, m = 71 kg
  • Tension on the rope, T = 220 N
  • Coefficient of kinetic friction, = 0.1
  • Time of motion, t = 8 s

<h3>What is Newton's second law of motion?</h3>
  • Newton's second law of motion states that, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

The net force on Tim is calculated by applying Newton's second law of motion as follows;

T - \mu _k F_n = ma\\\\T - \mu _k F_n = m\frac{v}{t} \\\\T - \mu_k mg = m\frac{v}{t} \\\\t(\frac{T - \mu_k mg}{m} )= v\\\\8 (\frac{220 \ -\  0.1 \times 71 \times 9.8}{71} ) =v \\\\ 16.95 \ m/s = v

Thus, the maximum speed of Tim is 16.95 m/s.

Learn more about net horizontal force here: brainly.com/question/21684583

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7 0
4 years ago
A pendulum is made up of a small sphere of mass 0.500 kg attached to a string of length 0.950 m. The sphere is swinging back and
Semenov [28]

Answer:

W = 0.842 J

Explanation:

To solve this exercise we can use the relationship between work and kinetic energy

         W = ΔK

In this case the kinetic energy at point A is zero since the system is stopped

         W = K_f                (1)

now let's use conservation of energy

starting point. Highest point A

          Em₀ = U = m g h

Final point. Lowest point B

         Em_f = K = ½ m v²

energy is conserved

         Em₀ = Em_f

         mg h = K

to find the height let's use trigonometry

at point A

            cos 35 = x / L

            x = L cos 35

so at the height is

            h = L - L cos 35

            h = L (1-cos 35)

we substitute

           K = m g L (1 -cos 35)

we substitute in equation 1

           W = m g L (1 -cos 35)

let's calculate

           W = 0.500 9.8 0.950 (1 - cos 35)

           W = 0.842 J

7 0
3 years ago
Will give correct answer brainliest
Grace [21]

Answer:

all will help you!!! !!!!!!!

7 0
3 years ago
A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
jeka57 [31]

Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

a.

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A²  ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

b.

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

c.

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J  = 120 J decrease

d.

The missing energy has turned into interned energy in the completely inelastic collision

3 0
4 years ago
What do pitch and loudness have in common?
sladkih [1.3K]

Answer:

both are subject to a person's interpretation.

6 0
3 years ago
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