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Marysya12 [62]
3 years ago
14

3. An object with a mass of 10 kg is moving along a horizontal surface. At a certain point it has 40 J of kinetic energy. If the

coefficient of friction between the object and the surface is 0.40, how far will the object go beyond that point before coming to a stop? Use g = 10 m/s2
Physics
1 answer:
damaskus [11]3 years ago
7 0

Answer: 100cm

Explanation:

The force of friction on a surface normal to gravity where µ is the coefficient of friction is

F = µmg

Where

F = the friction force

µ = coefficient of friction

m = mass of the object

g = acceleration due to gravity

Also, the Kinetic Energy of the object, E = Fs, where

E = Kinetic Energy

s = stopping distance. So that,

E = µmgs

40 J = 0.4 * 10 kg * 10 m/s² * s

40 J = 40 kgm/s² * s

s = 40 J / 40 kgm/s²

s = 1 m or 100 cm

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Artist 52 [7]

Answer:

(a) 3.807 s

(b) 145.581 m

Explanation:

Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.

The distance traveled by car after Δt (seconds) at v_c = 23m/s speed is

s_c = \Delta t v_c = 23\Delta t

The distance traveled by the motorcycle after Δt (seconds) at m_m = 23 m/s speed and acceleration of a = 8 m/s2 is

s_m = \Delta t v_m + a\Delta t^2/2

s_m = 23\Delta t + 8\Delta t^2/2 = 23 \Delta t + 4 \Delta t^2

We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:

s_m = s_c + 58

23 \Delta t + 4 \Delta t^2 = 23\Delta t + 58

4 \Delta t^2 = 58

\Delta t^2 = 14.5

\Delta t = \sqrt{14.5} = 3.807s

(b)

s_m = 23 \Delta t + 4 \Delta t^2

s_m = 23*3.807 + 58 = 145.581 m

5 0
3 years ago
[7 pts] 1. Suppose that a car starts from rest, its engine providing an acceleration of 4 ft/s2 while air resistance provides 0.
kumpel [21]

Answer:

a) Initial Value Problem

dv/dt = 4 - 0.1v

v(0) = 0

b) solution to the IVP

v(t) = 40(1 - e^(-t/10))

c) Limiting velocity

Vo = 40 ft/s

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X = 14,390 ft

Explanations:

The complete explanations of each of the sections contained in the question are in the files attached to this solution.

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A train has a final velocity of 110 km/h. It accelerated for 36 s at 0.50 m/s2 . What was its initial velocity
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The initial velocity of the train  is 12.56 m/s.

<h3>Initial velocity</h3>

The initial velocity of the train can be determined by using the first kinematic equation as shown below;

v = u + at

u = v - at

where;

  • v is the final velocity = 110 km/h = 30.56 m/s
  • u is the initial velocity

u = 30.56 - (36 x 0.5)

u = 12.56 m/s

Thus, the initial velocity of the train  is 12.56 m/s.

Learn more about initial velocity here: brainly.com/question/19365526

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Romashka-Z-Leto [24]
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postnew [5]

Answer: option c: It orbits beyond the Earth's atmosphere to avoid scattering of light.

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