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2 years ago

Find the coefficient of kinetic friction μk. express your answer in terms of some or all of the variables d1, d2, and θ.

2 answers:

6 0

<span>internet tension = mass * acceleration internet tension = 23 – Friction tension = 14 * acceleration Friction tension = µ * 14 * 9.8 = µ * 137.2 23 – µ * 137.2 = 14 * acceleration Distance = undemanding speed * time undemanding speed = ½ * (preliminary speed + very final speed) Distance = ½ * (preliminary speed + very final speed) * time Distance = 8.a million m, preliminary speed = 0 m/s, very final speed = a million.8 m/s 8.a million = ½ * (0 + a million.8) * t Time = 8.a million ÷ 0.9 = 9 seconds Acceleration = (very final speed – preliminary speed) ÷ time Acceleration = (a million.8 – 0) ÷ 9 = 0.2 m/s^2 23 – µ * 137.2 = 14 * 0.2 resolve for µ</span>

topjm [15]2 years ago

3 0

Answer:

μ_k = (d_1)*sin(Q) / [(cos(Q))*(d_1) + (d_2)]

Explanation:

On the slope:

- The force of gravity has magnitude "mg" and points points straight down the slope and is given by the expression:

F_g,down = m*g*sin(Q)

- The work done by gravity W_g is the dot product of F_g and the distance traveled d_1, shown:

W_g = F_g . d_1

W_g = m*g*sin(Q). d_1

- The normal Force F_n acts perpendicular to the slope and the gravity components also acts in the opposite to its direction, apply Equilibrium condition in the axis perpendicular to the slope since, their is no acceleration in that direction:

F_perp = F_n - F_g,up = 0

F_n = F_g,up = m*g*cos(Q)

- Since there is no work done in the direction perpendicular to slope its zero.

- Frictional force F_f acts parallel to the slope opposite in direction of motion, hence, directed up the slope is dependent on coefficient of friction u_k and normal contact force as follows:

F_f = u_k . F_n

F_f = u_k (m*g*cos(Q))

- The work done by friction W_f is the dot product of F_f and the distance traveled d_1, shown:

W_f = F_f . d_1

W_f = u_k*m*g*cos(Q). d_1

- So in summary, the total work done on the sled during the downhill phase is:

W_downhill = (m*g)*(d_1)*sin(Q) – (m*g)*(cos(Q))*(μ_k)*(d_1)

The horizontal terrain:

- Since, the Force of gravity (F_g) and normal force (F_n) both act perpendicular to the direction of motion, they do zero work.

- However, friction force (F_f) acts parallel to the direction of motion but opposite in direction, the work done by friction is:

W_f = F_f*d_2

W_f = u_k*F_n * d_2

W_f = - u_k * m*g*d_2

- So the total work done during the horizontal phase is:

W_hor = W_f = - u_k * m*g*d_2

Total Work done:

- So that means the total work done on the sled overall is:

W = W_downhill + W_hor

= (m*g)*(d_1)*sin(θ) – (m*g)*(cos(Q))*(μ_k)*(d_1) – (m*g)*(μ_k)*(d_2)

- The total work done is zero because the Kinetic energy at the start and end of the journey is zero.

(m*g)*(d_1)*sin(θ) – (m*g)*(cos(Q))*(μ_k)*(d_1) – (m*g)*(μ_k)*(d_2) = 0

- Simplify:

(d_1)*sin(Q) = (cos(Q))*(μ_k)*(d_1) + (μ_k)*(d_2)

(d_1)*sin(Q) = [(cos(Q))*(d_1) + (d_2)]*(μ_k)

- And then:

μ_k = (d_1)*sin(Q) / [(cos(Q))*(d_1) + (d_2)]

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oksian1 [2.3K]

<span>First, water evaporates from the ocean to the atmosphere, then... put each step in the correct order.
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2 years ago

If you walk 30 meters forwards, and then turn around and walk 25 meters backwards, what is the distance that you walked? What di

xeze [42]

Given :

Walk in forward direction is 30 m .

Walk in backward direction is 25 m .

To Find :

The distance and displacement .

Solution :

We know , distance is total distance covered and displacement is distance between final and initial position .

So , distance travelled is :

D = 30 + 25 m = 55 m .

Now , we first move 30 m in forward direction and then 25 m in backward direction .

So , displacement is :

D = 30 - 25 m = 5 m .

Therefore , distance and displacement covered is 55 m and 5 m respectively .

Hence , this is the required solution .

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2 years ago

The volume flow rate in an artery supplying the brain is 4.00 10-6 m3/s. (a) If the radius of the artery is 4.50 10-3 m, determi

mr_godi [17]

Answer:

(a) 0.063 m/s

(b) 1.01 m/s

Explanation:

rate of volume flow, V = 4 x 10^-6 m^3/s

(a) radius, r = 4.5 x 10^-3 m

Let the speed of blood is v.

So, V = A x v

where A be the area of crossection of artery

4 x 10^-6 = 3.14 x 4.5 x 10^-3 x 4.5 x 10^-3 x v

v = 0.063 m/s

Thus, the speed of flow of blood is 0.063 m/s .

(b) Now r' = r / 4 = 4.5 /4 x 10^-3 m = 1.125 x 10^-3 m

Let the speed is v'.

So, V = A' x v'

4 x 10^-6 = 3.14 x 1.125 x 10^-3 x 1.125 x 10^-3 x v'

v' = 1.01 m/s

Thus, the speed of flow of blood is 1.01 m/s .

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2 years ago

Two equal point charges QQQ are separated by a distance ddd. One of the charges is released and moves away from the other due on

lys-0071 [83]

Answer:

The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

Explanation:

The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.

Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.

From the law of conservation of energy, U₁ + K₁ = U₂ + K₂

So, kQ²/d + 0 = kQ²/3d + K

K₂ = kQ²/d - kQ²/3d = 2kQ²/3d

So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd

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3 years ago

Middle school physics

Alecsey [184]

Answer:

8 V

Explanation:

There is no resistance between the left legs of voltmeters 2 and 3 and there is no resistance between the right legs of voltmeters 2 and 3. They are measuring the same voltage.

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2 years ago