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Lady_Fox [76]
2 years ago
13

A stuntman is being pulled along a rough road at a constant velocity by a cable attached to a moving truck. The cable is paralle

l to the ground. The mass of the stuntman is 107 kg, and the coefficient of kinetic friction between the road and him is 0.682. Find the tension in the cable.
Physics
1 answer:
Alex73 [517]2 years ago
6 0

Answer:

715 N

Explanation:

Since the system is moving at a constant velocity, the net force must be 0. The tension on the road is equal and opposite direction with the kinetic friction force created by the road and the stuntman.

Let g = 9.8 m/s2

Gravity and equalized normal force is:

N = P = mg = 107*9.8 = 1048.6 N

Kinetic friction force and equalized tension force on the rope is

T = F_{\mu} = N\mu = 1048.6 * 0.682 = 715.1452 N

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А______<br> is a unit of measurement for energy. (7 Letters)
a_sh-v [17]

Answer:

Joule or kilowatt/hour

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2 years ago
At the _______________ point, the particles in an object have not kinetic energy
Brums [2.3K]
The freezing point ..... :)
4 0
3 years ago
A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v
atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

y(x, t) = Acos(kx -wt)

The relationship between velocity and propagation constant is

v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\

v = 15.87m/s

Time taken, t = \frac{\lambda}{v}

= \frac{1.3}{15.87}\\\\=0.0819 sec

t = 0.0819s

2)

The velocity of transverse wave is given by

v = \sqrt{\frac{T}{\mu}}

v = \sqrt{\frac{W}{\frac{m}{\lambda}}}

mass of string is calculated thus

mg = 0.0125N

m = \frac{0.0125N}{9.8N/s}

m = 0.00128kg

\omega = \frac{v^2m}{\lambda}

\omega = \frac{(15.87^2)(0.00128)}{1.30}

\omega = 0.25N

3)

The propagation constant k is

k=\frac{2\pi}{\lambda}

hence

\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}

\lambda = 0.036 m

No of wavelengths, n is

n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\

n = 36

4)

The equation of wave travelling down the string is

y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]

without, unit\\\\y(x , t)= Acos[172x + 2730t]

7 0
2 years ago
1. You have a cat who has a mass of 10 kg and is
disa [49]

Answer:

1) F = 100N

2) a = 2 m/s²

3) m = 25 kg

Explanation:

1) F = ma ( F = ?, m = 10 kg, a = 10 m/s² )

  F = 10×10

  F = 100 N

2) F = ma ( F = 20N, m = 10 kg, a = ? )

   20 = 10×a

   10a = 20

   a = 20/10

   a = 2 m/s²

3)F = ma ( F = 100N, m = ?, a = 4 m/s² )

  100 = m×4

  4m = 100

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  m = 25 kg

Hope that helps! Good luck!

 

7 0
3 years ago
Problem 13.175 A 1-kg block B is moving with a velocity v0 of magnitude as it hits the 0.5-kg sphere A, which is at rest and han
Svetradugi [14.3K]

Explanation:

Mass of the Block, Mb = 1.0 kg Initial Velocity of Block, Vb = 2.0 m/s

Mass of the Ball, Ma = 0.5 kg Coefficient of Kinetic Friction, μ = 0.6

Coefficient of Restitution, e = 0.8 Gravity = 9.81 m/s^2

Sum of the forces of the X-axis components and Y-axis components are:

∑ Fx = Ff = Mb × a

Equation for frictional force,

Ff = μ × N

∑ Fy = N - mb × g = 0

Note:

N = mb × g

Therefore, to solve for the acceleration, we have:

μ × Mb × g = Mb × a

a = μ × g

= 0.6 × 9.81

= 5.88 m/s^2

Therefore, ratio of velocities using the coefficient of restitution, e:

e = (Vb2 – Va1)/ (Va – Vb)

Note: Va = zero (initially at rest)

Va2 – Vb2 = 0.8 × ( 0 - 2 m/s)

Vb2 – Va2 = -1.6 m/s

Vb2 = a2 – 1.6

Using Conservation of Momentum Equation:

Ma × Va + Mb × Vb = Ma × Va2 + Mb × Vb2

0 + 1kg × 2 m/s = 0.5 kg × Va2 + 1 kg × Vb2

Vb2 = 2 - 0.5 × Va2

Substitute in Vb2,

1.5 × Va2 = 3.6

Va2 = 2.4 m/s

Vbs = 0.8 m/s

.

mgh = 0.5 × Mv2

h = v2/ 2g

h = 0.294 m

B.

Using equation of motion,

Vf^2 = Vo^2 + 2a × S

Given:

Vf = 0 m/s

a = 5.88 m/s^2

0 = 0.82 + 2(5.88) × ΔS

Δx = 0.0544 m

4 0
3 years ago
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