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maks197457 [2]
3 years ago
9

A steel cable has a cross-sectional area 4.49 × 10^-3 m^2 and is kept under a tension of 2.96 × 10^4 N. The density of steel is

7860 kg/m^3. Note that this value is not the linear density of the cable. At what speed does a transverse wave move along the cable?
Physics
1 answer:
Lemur [1.5K]3 years ago
7 0

Answer:

The transverse wave will travel with a speed of 25.5 m/s along the cable.

Explanation:

let T = 2.96×10^4 N be the tension in in the steel cable, ρ  = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.

then, if V is the volume of the cable:

ρ = m/V

m = ρ×V

but V = A×L , where L is the length of the cable.

m = ρ×(A×L)

m/L = ρ×A

then the speed of the wave in the cable is given by:

v = √(T×L/m)

  = √(T/A×ρ)

  = √[2.96×10^4/(4.49×10^-3×7860)]

  = 25.5 m/s

Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.

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Answer:

A

Explanation:

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0.88 x was used to convert water to steam.

heat carried by steam = 40% × 0.88 x = 0.352 x

efficiency of the heat -to- work conversion = work output / work input = 0.352 x / x = 0.352 × 100 = 35.2 % which is less than 40 %

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3 years ago
what term describes the situation where the frequency of the wind match the frequency of the building? a. velocity b. resonance
saul85 [17]
The answer is B) resonance
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What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?
kvv77 [185]

Answer:

W = 78.48N\\W =0.0784kN\\W = 8kgf\\

W= 3.6lbf\\W= 115.2 lbm*ft/s2

Explanation:

if

m=8kg=3.6lb\\

and g=9.81 m/s2=32.16 ft/s2

and

W=m*g

we can just replace de mass and gravity and we have

W = 8kg * 9.81 \frac{m}{s^{2} } =78.48N\\W = \frac{78.48N}{1000} =0.0784kN\\W = 8kgf\\

W= 3.6lbf\\W= 3.6lbm *32.16 ft/s2 =115.2 lbm*ft/s2

5 0
3 years ago
Two blocks of masses 20 kg and 8.0 kg are connected togetherby
OleMash [197]

Answer:

14 N

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So tension in the second string = total mass x acceleration

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F = Magnetic Force
B = Magnetic Field
V = Velocity

*The vectors from the photo you get doing the left-hand rule.

The magnetic force is always perpendicular to the magnetic field.

And as told in the statement, the electron is moving perpendicular to a magnetic field, that is, the Velocity forms an 90 degree angle / Right angle with the magnetic field.

The formula to find the Magnetic Force is:

f = |q| \times v \times b \times sin \: \theta

Where "q" is the Charge and the sin theta is the angle formed by the Velocity and Magnetic Field, in this case it's 90°. Sin 90° = 1.

f = |- 1.6 \times {10}^{ - 19} | \times 2.5 \times {10}^{5} \times 1.5 \times 1 \\ f = 6 \times {10}^{ - 19 + 5} \\ f = 6 \times {10}^{ - 14} \: newtons
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4 0
3 years ago
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