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snow_lady [41]
3 years ago
11

A 16.0 kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 25.0 N. Starting from rest, the

sled attains a speed of 1.00 m/s in 8.00 m. Find the coefficient of kinetic friction between the runners of the sled and the snow. You Answered
Physics
1 answer:
Fudgin [204]3 years ago
4 0

Answer:

\mu_k = 0.15

Explanation:

according to the kinematic equation

v^{2} - u^{2} = 2aS

Where

u is initial velocity  = 0 m/s

a = acceleration

S is distance = 8.00 m

final velocity = 1.0 m/s

a = \frac {v^{2}}{2S}

a = \frac {1{2}}{2*8.6}

a = 0.058 m/s^2

from newton second law

Net force = ma

f_{net} = ma

F - f = ma

25 - \mu_kN = ma

25 - \mu_kmg = ma

\frac {25 - ma}{mg} =\mu_k

\frac {25 - 16*0.058}{16*9.81} = 0.15

\mu_k = 0.15

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three point charges are arranged in a line. charge q3=+5.00 nC and is located at the origin. charge q2=-3.00 nC and is located a
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Answer:

q_1=+0.375\ {10}^{-9}

Explanation:

Electrostatic Forces

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\displaystyle F=\frac{k\ q_1\ q_2}{d^2}

The forces are attractive if the charges have different signs and repulsive if they have equal signs.

The problem described in the question locates three point charges in a straight line. The charges have the values shown below

\displaystyle q_3=+5\ 10^{-9}\ c

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The distance between q_3 and q_2 is

\displaystyle d_2=4cm=0.04\ m

The distance between q_3 and q_1 is

\displaystyle d_1=2cm=0.02\ m

We must find the value of q_1 such that

\displaystyle |F_3|=0

Applying Coulomb's formula for q_1 is

\displaystyle F_{13}=\frac{k\ q_1\ q_3}{d_1^2}

Now for q_2

\displaystyle F_{23}=\frac{k\ q_2\ q_3}{d_2^2}

If the total force on q_3 is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:

\displaystyle F_{13}=F_{23}

\displaystyle \frac{k\ q_1\ q_3}{d_1^2}=\frac{k\ q_2\ q_3}{d_2^2}

Simplfying and solving for q_1

\displaystyle q_1=\frac{q_2\ d_1^2}{d_2^2}

\displaystyle q_1=\frac{3.10^{-9}\ 0.02^2}{0.04^2}

\boxed{\displaystyle q_1=+0.375\ {10}^{-9}}

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A =\dfrac{7.2\times 10^3 \times 9.8}{2\times 9977.5}

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