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Alex17521 [72]
3 years ago
5

On an essentially frictionless, horizontal ice rink, a skater moving at 3.0 m/s encounters a rough patch that reduces her speed

by 45% due to a friction force that is 25% of her weight. Use the work‑energy theorem to find the length of this rough patch.
Physics
1 answer:
JulsSmile [24]3 years ago
7 0

Answer: 1.25 m

Explanation:

Given

initial velocity(v_i )=3 m/s

Final velocity(v_f)=0.55\times 3=1.65 m/s

Change in kinetic Energy =work done by Friction

change in Kinetic Energy=\frac{m}{2}\left ( v_i^2-v_f^2\right )

work done by friction=\mu mgL

\frac{m}{2}\left ( 3^2-1.65^2\right )=0.25\cdot mg\times L

3.135=0.25\times 9.8\times L

L=1.25

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A person is sitting at the very back of a canoe of length L, when the front just bumps into the dock. show answer No Attempt 50%
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The distance of the canoeist from the dock is equal to length of the canoe, L.

<h3>Conservation of linear momentum</h3>

The principle of conservation of linear momentum states that the total momentum of an isolated system is always conserved.

v(m₁ + m₂) = m₁v₁ + m₂v₂

where;

v is the velocity of the canoeist and the canoe when they are together

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  • u₂ velocity of the canoeist
  • m₁ mass of the canoe
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<h3>Distance traveled by the canoeist</h3>

The distance traveled by the canoeist from the back of the canoe to the front of the canoe is equal to the length of the canoe.

Thus, the distance of the canoeist from the dock is equal to length of the canoe, L.

Learn more about conservation of linear momentum here: brainly.com/question/7538238

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A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim
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Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

F_D = -\frac{1}{2}\rho V^2 C_d A

Where

\rho is the density of the flow

V = Velocity

C_d= Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

F=ma=m\frac{dV}{dt}

Equating both equations we have:

m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A

m(dV)=-\frac{1}{2}\rho C_d A (dt)

\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)

Integrating

\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)

-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t

Here,

V_f = 60mph = 26.82m/s

V_i = 120.7m/s

m= 1600lbf = 725.747Kg

\rho = 1.21 kg/m^3

C_d = 0.3

\Delta t=7s

Replacing:

\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)

-0.029 = -5.4997r^2

r = 2.2963m

d= r*2 = 4.592m \approx 15.065ft

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