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2 years ago

What measures the amount of displacement in a transverse wave

1 answer:

4 0

Unlike a longitudinal wave, a transverse wave moves about, perpendicular to the direction of propagation. The particles in a transverse wave do not travel along the direction of propagation, but only oscillate up and down on its equilibrium position. With this, the displacement can be determined by measuring (in the case of electronic waves, using an oscilloscope or spectrum analyzer) and setting the desired units to measure the wave in.

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A4) A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.3

Klio2033 [76]

Answer:

$\theta=30^{\circ}$

Explanation:

It is given that,

Length of the wire, L = 0.6 m

Current flowing inside the wire, I = 2 A

Uniform magnetic field, B = 0.3 T

Force experienced by the wire in the magnetic field, F = 0.18 N

To find,

The angle made by the wire with the magnetic field.

Solve,

We know that the magnetic force acting on the wire inside the magnetic field is given by :

$F=ILB\ sin\theta$

$sin\theta=\dfrac{F}{ILB}$

$sin\theta=\dfrac{0.18}{2\times 0.6\times 0.3}$

$\theta=30^{\circ}$

Therefore, the wire makes an angle of 30 degrees with respect to magnetic field.

5 0

2 years ago

Un carro de montaña rusa parte del reposo desde una primera cumbre, desciende una distancia vertical de 45 metros y luego sube u

garri49 [273]

A roller coaster car starts from the rest from a first summit, descends a vertical distance of 45 meters and then climbs a second summit, reaching the top with a speed of 15m / s. How high is the second summit? Do not consider friction

3 0

2 years ago

1. Bone has a Young’s modulus of about

Blababa [14]

As we know that

$Y = \frac{stress}{strain}$

now plug in all data into this

$1.8\times 10^{10} = \frac{1.68 \times 10^8}{strain}$

$strain = 9.33 \times 10^{-3}$

now from the formula of strain

$strain = \frac{\Delta L}{L}$

$9.33 \times 10^{-3} = \frac{\Delta L}{0.54}$

$\Delta L = 5.04 \times 10^{-3} m$

$\Delta L = 5.04 mm$

As we know that

pressure * area = Force

here we know that

$Area = 3.53 \times 11.6 = 40.95 m^2$

$P = 0.2 atm = 0.2 \times 1.01 \times 10^5 = 2.02\times 10^4 Pa$

now force is given as

$F = 40.95 \times (2.02\times 10^4) = 8.27 \times 10^5 N$

As we know that density of water will vary with the height as given below

$\rho = \frac{\rho_0}{1 - \frac{\Delta P}{B}}$

here we know that

$\Delta P = 2600 atm = 2600 \times 1.01 \times 10^5 = 2.63\times 10^8 Pa$

$B = 2.3 \times 10^9 N/m^2$

now density is given as

$\rho = \frac{1050}{1 - \frac{2.63\times 10^8}{2.3 \times 10^9}}$

$\rho = 1185.3 kg/m^3$

as we know that pressure changes with depth as per following equation

$P = P_o + \rho g h$

here we know that

$P = 3 P_0$

now we will have

$3P_0 = P_0 + \rho g h$

$2P_0 = \rho g h$

$2(1.01 \times 10^5) = 1025 (9.81)(h)$

here we will have

$h = 20.1 m$

so it is 20.1 m below the surface

Here net buoyancy force due to water and oil will balance the weight of the block

so here we will have

$mg = \rho_1V_1g + \rho_2V_2g$

$A(0.0476)979 = 922(A)(0.0476 - x) + 1000(A)(x)$

$46.6 = 43.89 - 922x + 1000x$

$x = 3.48 cm$

so it is 3.48 cm below the interface

5 0

3 years ago

What magnification is best for the observation of most tissue slides? Explain why this has proved to be true?

Komok [63]

Answer:

The 10X objective is use for the identification of actual size of histology tissues and 4X magnification is best for observation of most tissues slides

Explanation:

4X magnification is best for observation of most tissues slides because it has an objective lens that have lower power and have great high field overview which make it very easier to locate specimens on the slide. It is use to get the overview of histology slides. It is use to showcase more detailed observations about histology.

The 40X objective is use majorly to identify tissue , to observe the finer details and study tissue organization on the histology slide.

7 0

3 years ago

A gold sphere of radius R=100 μm and density 19g/cm^3 falls through water. Given the viscosity of water is about 10^-3 Pa s and

icang [17]

The terminal velocity of gold sphere is 39.2 cm/s

<h3>What is terminal velocity?</h3>

Terminal velocity is the maximum velocity attainable for an object as it falls through a fluid.

<h3>How to calculate the terminal velocity of the gold sphere?</h3>

The terminal velocity of the gold sphere is given by v = 2gr²(ρ - σ)/9η where

g = acceleration due to gravity = 9.8 m/s²,
r = radius of sphere = 100 μm = 100 × 10⁻⁶ m = 10⁻⁴ m = 10⁻² cm,
ρ = density of sphere = 19 g/cm³,
σ = density of water = 1.0 g/cm³ and
η = viscosity of water = 10⁻³ Pa-s

So, susbtituting the values of the variables into the equation, we have that

v = 2gr²(ρ - σ)/9η

v = 2 × 9.8m/s²× (10⁻² cm)²(19 g/cm³ - 1.0 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 9.8 m/s² × 10⁻⁴ cm² × (18 g/cm³)/(9 × 10⁻³ Pa-s)

v = 2 × 980 cm/s² × 10⁻⁴ cm² × 2 g/cm³/(1 × 10⁻³ Pa-s)

v = 3920 g/s² × 10⁻⁴/(1 × 10⁻³ Pa-s)

v = 392 cm/s × 10³ × 10⁻⁴

v = 392 × 10⁻¹ cm/s

v = 39.2 cm/s

So, the terminal velocity is 39.2 cm/s

Learn more about terminal velocity of sphere here:

brainly.com/question/21684177

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4 0

1 year ago