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Vera_Pavlovna [14]
3 years ago
15

While water skiing behind her father’s boat, Letty is pulled at constant speed by a force of 164 N from the tow rope that makes

an angle of 10.0° (from the horizontal). If Letty’s mass is 65.0 kg, what is the coefficient of friction between her skis and the water?
Physics
1 answer:
kap26 [50]3 years ago
7 0

Answer:

0.265

Explanation:

Draw a free body diagram.  There are four forces:

Normal force Fn pushing up.

Weight force mg pulling down.

Tension force T at an angle θ.

Friction force Fn μ pushing left.

Sum the forces in the y direction:

∑F = ma

Fn + T sin θ − mg = 0

Fn = mg − T sin θ

Sum the forces in the x direction:

∑F = ma

T cos θ − Fn μ = 0

Fn μ = T cos θ

μ = T cos θ / Fn

μ = T cos θ / (mg − T sin θ)

Given T = 164 N, θ = 10.0°, m = 65.0 kg, and g = 9.8 m/s²:

μ = (164 N cos 10.0°) / (65.0 kg × 9.8 m/s² − 164 N sin 10.0°)

μ = 0.265

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What is the magnitude of the net force ∑ F on a 1.9 kg bathroom scale when a 74 kg person stands on it?
dexar [7]

Answer:

725.2‬ N

Explanation:

Since it is not stated the scale, the person or both accelerated or experience weightlessness, the net force acting on the bathroom scale is the weight of the person acting downward as the person stands on the scale .

                       Weight = mass of a body × acceleration due to gravity

                                    = 74 kg × 9.8 m/s²

                                    = 725.2‬ N    

6 0
3 years ago
Plz answer this question
Lady_Fox [76]

The answer will be pressure.

Pressure is force per unit area.

8 0
3 years ago
A 10kg object is 15 meters up a hill. Find its potential energy
Tema [17]

Answer:

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PE = mgh = 10(9.8)(15) = 1470 J

5 0
2 years ago
Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. Take T = 1.3 kN·m. De
My name is Ann [436]

Answer:

τ (bc. max) =25.37 MPa

Explanation:

From the question, T = 1.3Kn.m;

(G = 77.2 GPa) and from the image of this solid shaft system i attached;

d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m

So ΣT = 0 → Ta + Tc = 1.3Kn.m

So the system is statically indeterminate.

Let's check at the equation that makes it compatible ;

ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0

[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =

ΣT = 0 and T(bc) = T(c)

ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m

Now,

[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0

So, T(c) = 273374 Nmm = 273.374Nm

T(a) = 1300Nm - 273.374Nm = 1026. 63Nm

From the beginning we saw that;

T(bc) = T(c) = 273.374Nm

Now let's find the maximum shear stress in shaft BC;

τ (bc. max) = {τ (bc) x r(bc)} / J(bc)

τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707

= 25.37 MPa

3 0
3 years ago
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