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3 years ago

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With what tension must a rope with length 3.00 mm and mass 0.105 kgkg be stretched for transverse waves of frequency 40.0 HzHz t

o have a wavelength of 0.790 mm

1 answer:

VladimirAG [237]3 years ago

4 0

Answer:

the tension of the rope is 34.95 N

Explanation:

Given;

length of the rope, L = 3 m

mass of the rope, m = 0.105 kg

frequency of the wave, f = 40 Hz

wavelength of the wave, λ = 0.79 m

Let the tension of the rope = T

The speed of the wave is given as;

$v = f\lambda = \sqrt{\frac{T}{\mu} } \\\\where;\\\\\mu \ is \ mass \ per \ unit \ length\\\\\mu = \frac{0.105}{3} = 0.035 \ kg/m\\\\v = f\lambda = 40 \times 0.79 = 31.6 \ m/s\\\\v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu\\\\T = (31.6^2)(0.035)\\\\T = 34.95 \ N$

Therefore, the tension of the rope is 34.95 N

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At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon

ArbitrLikvidat [17]

Answer:

The pressure at point 2 is $P_2 = 254.01 kPa$

Explanation:

From the question we are told that

The speed at point 1 is $v_1 = 3.57 \ m/s$

The gauge pressure at point 1 is $P_1 = 68.7kPa = 68.7*10^{3}\ Pa$

The density of water is $\rho = 1000 \ kg/m^3$

Let the height at point 1 be $h_1$ then the height at point two will be

$h_2 = h_1 - 18.5$

Let the diameter at point 1 be $d_1$ then the diameter at point two will be

$d_2 = 2 * d_1$

Now the continuity equation is mathematically represented as

$A_1 v_1 = A_2 v_2$

Here $A_1 , A_2$ are the area at point 1 and 2

Now given that the are is directly proportional to the square of the diameter [i.e $A= \frac{\pi d^2}{4}$ ]

which can represent as

$A \ \ \alpha \ \ d^2$

=> $A = c d^2$

where c is a constant

so $\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2}$

=> $\frac{A_1}{d_1^2} = \frac{A_2}{4d_1^2}$

=> $A_2 = 4 A_1$

Now from the continuity equation

$A_1 v_1 = 4 A_1 v_2$

=> $v_2 = \frac{v_1}{4}$

=> $v_2 = \frac{3.57}{4}$

$v_2 = 0.893 \ m/s$

Generally the Bernoulli equation is mathematically represented as

$P_1 + \frac{1}{2} \rho v_1^2 + \rho * g * h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho * g * h_2$

So

$P_2 = \rho * g (h_1 -h_2 )+P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

=> $P_2 = \rho * g (h_1 -(h_1 -18.3) + P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

substituting values

$P_2 = 1000 * 9.8 (18.3) )+ 68.7*10^{3} + \frac{1}{2} * 1000 ((3.57)^2 -0.893 ^2 )$

$P_2 = 254.01 kPa$

8 0

3 years ago

brilliants [131]

Answer: current I = 1.875A

Explanation:

If the resistors are connected in series,

Then the equivalent resistance will be

R = 6 + 18 + 15 + 9

R = 48 ohms

Using ohms law

V = IR

Make current I the subject of formula

I = V/R

I = 90/48

I = 1.875A

And if the resistors are connected in parallel, the equivalent resistance will be

1/R = 1/6 + 1/18 + 1/15 + 1/9

1/R = 0.166 + 0.055 + 0.066 + 0.111

R = 1/0.3999

R = 2.5 ohms

Using ohms law

V = IR

I = 90/2.5

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3 years ago

Two insulated wires, each 2.50 m long, are taped together to form a two-wire unit that is 2.50 m long. One wire carries a curren

liubo4ka [24]

Answer:

3.46 A

Explanation:

The force (F) exerted on a wire of a particular length (L) carrying current (I) through a magnetic field (B) at an angle (θ) to the magnetic field is given as

F = (B)(I)(L) sin θ

F = 3.13 N

B = 0.360 T

I = ?

L = 2.50 m

θ = 79°

3.13 = (0.360 × I × 2.5 × sin 79°)

0.8835 I = 3.13

I = 3.54 A

But this is the resultant current in this magnetic field.

Since the two wires are conducting current in opposite directions,

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3 years ago

When a flat slab of transparent material is placed under water, the critical angle for light traveling from the slab into water

Novosadov [1.4K]

Answer:

(a) 40.6 degree

Explanation:

When refraction takes place from slab to water, the critical angle is 60 degree.

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$\mu _{w}^{s}=\frac{Sin60}{Sin90}$

$\frac{\mu _{w}}{\mu _{s}}=0.866$

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Now for slab air interface, the critical angle is C.

$\mu _{a}^{s}=\frac{SinC}{Sin90}$

$\frac{\mu _{a}}{\mu _{s}}=\frac{SinC}{Sin90}$

1 / 1.536 = Sin C

C = 40.6 degree

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3 years ago

A solid block, with a mass of 0.15kg, on a frictionless surface is pushed directly onto a horizontal spring, with a spring const

iren [92.7K]

Answer:

16.1 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, the spring is compressed by x = 35 cm = 0.35 m, and it stores an elastic potential energy given by

$U=\frac{1}{2}kx^2$

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3 years ago