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3 years ago

10

What is amplitude?

1 answer:

nydimaria [60]3 years ago

8 0

1. Frequency is the number of complete waves that pass a point in a second. 2.Wavelength is the distance between two crests or two troughs. 3.Time period <span> is the time it takes for one complete wave to pass a given point. 4. Amplitude is the height of the wave. Hence option 4 is correct. </span>

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A country would place a tariff on imported steel to

Whitepunk [10]

. . . 'protect' its domestic steel industry, by
increasing the price of imported steel.

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3 years ago

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Which two terms indicate the same area of the body? *

Kruka [31]

Answer:

For example, the toes are anterior to the heel, and the popliteus is posterior to the patella. Superior and inferior, which describe a position above (superior) or below (inferior) another part of the body. For example, the orbits are superior to the oris, and the pelvis is inferior to the abdomen.

Explanation:

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2 years ago

A long, thin solenoid has 450 turns per meter and a radius of 1.06 . The current in the solenoid is increasing at a uniform rate

kirill115 [55]

Answer:9.34 A/s

Explanation:

Given

radius of solenoid $R=1.06 m$

Emf induced $E=8.50\times 10^{-6} V/m$

no of turns per meter n=450

we know Induced EMF is given by

$\int Edl=-\frac{\mathrm{d} \phi}{\mathrm{d} t}=-\frac{\mathrm{d} B}{\mathrm{d} t}A$

Magnetic Field is given by

$B=\mu _0ni$

thus $\frac{\mathrm{d} B}{\mathrm{d} t}=-\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}$

Area of cross-section

$A=\pi R^2$ where

solving integration we get

$E.\cdot 2\pi r=\mu _0n\frac{\mathrm{d} i}{\mathrm{d} t}\pi R^2$

where r=distance from axis

R=radius of Solenoid

$\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{Er}{\mu _0nR^2}$

$\frac{\mathrm{d} i}{\mathrm{d} t}=\frac{8.50\times 10^{-6}\times 3.49\times 10^{-2}}{4\pi \times 10^{-7}\times 450\times 1.06^2}$

$\frac{\mathrm{d} i}{\mathrm{d} t}=9.34 A/s$

4 0

3 years ago

At what point on a roller coaster is the kinetic energy increasing? As it is going down the hill. At the top of the hill. As it

blagie [28]

Answer:

As it is going down the hill

Explanation:

8 0

3 years ago

The maximum stress in a section of a circular tube subject to a torque is τmax = 27 MPa . If the inner diameter is Di = 3.75 cm

DIA [1.3K]

Answer:

$T_{max} = 4.735\,kN\cdot m$

Explanation:

The shear stress due to torque can be calculed by using the following model:

$\tau_{max} = \frac{T_{max}\cdot r_{ext}}{J_{tube}}$

The maximum torque on the section is:

$T_{max} = \frac{\tau_{max}\cdot J_{tube}}{r_{ext}}$

The Torsion Constant for the circular tube is:

$J_{tube} = \frac{\pi}{32}\cdot (D_{ext}^{4}-D_{int}^{4})$

$J_{tube} = \frac{\pi}{4}\cdot [(0.053\,m)^{4}-(0.038\,m)^{4}]$

$J_{tube} = 4.560\times 10^{-6}\,m^{4}$

Now, the require output is computed:

$T_{max} = \frac{(27\times 10^{3}\,kPa)\cdot (4.560\times 10^{-6}\,m^{4})}{0.026\,m}$

$T_{max} = 4.735\,kN\cdot m$

7 0

3 years ago

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