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Ira Lisetskai [31]
3 years ago
7

Can all waves be distorted, deflected, or changed? Explain your response.

Physics
1 answer:
kow [346]3 years ago
8 0
Yes, all waves can be distorted, deflected, or changed

<span>Waves are a means by which energy travels. Many different particles move in waves. </span>All waves can be changed through interference with waves of similar wavelengths.
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An inductor with an inductance of 2.30H and a resistance of 8.00 Ω isconnected to the terminals of a battery with an emf of 6.00
Basile [38]

Answer:

(a). The initial rate is 2.60 A/s.

(b). The rate of current increases is 0.8658 A/s.

(c). The current is 0.435 A.

(d). The final steady-state current is 0.75 A.

Explanation:

Given that,

Inductance = 2.30 H

Resistance = 8.00 Ω

Voltage = 6.00 V

(a). We need to calculate the initial rate of increase of current in the circuit.

Using formula of initial rate

V=initial\ rate\times inductance

initial\ rate=\dfrac{V}{L}

Put the value into the formula

initial \rate=\dfrac{6.00}{2.30}

initial\ rate=2.60\ A/s

The initial rate is 2.60 A/s.

(b). We need to calculate the rate of increase of current at the instant when the current is 0.500

Using formula of rate of increase of current

rate\ of \ current\ increase=initial\ rate\times e^{\dfrac{-t}{T}}....(I)

Where, T=\dfrac{L}{R}

T=\dfrac{2.30}{8.00}

T=0.2875

Using formula of current

i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})

e^{\dfrac{-t}{T}}=1-(i\times\dfrac{R}{V})

e^{\dfrac{-t}{T}}=1-(0.500\times\dfrac{8.00}{6.00})

e^{\dfrac{-t}{T}}=0.333

The rate of current increases is

Put the value in the equation (I)

rate\ of\ current\ increase=2.60\times0.333

rate\ of\ current\ increase=0.8658\ A/s

The rate of current increases is 0.8658 A/s.

(c). We need to calculate the current

Using formula of current

i=\dfrac{V}{R}(1-e^{\dfrac{-t}{T}})

Put the value into the formula

i=\dfrac{6.00}{8.00}\times(1-e^{\dfrac{-0.250}{0.2875}})

i=0.435\ A

The current is 0.435 A.

(d). We need to calculate the final steady-state current

Using formula of steady state

i=\dfrac{V}{R}

i=\dfrac{6.00}{8.00}

i=0.75\ A

The final steady-state current is 0.75 A.

Hence, This is the required solution.

3 0
3 years ago
Heavier elements like gold are created in the most massive stars and spread through the universe when these stars explode. This
myrzilka [38]

Answer:

Big bang

Explanation:

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7 0
3 years ago
2. Compare and contrast light microscopes, electron microscopes,<br> and atomic force microscopes.
IRINA_888 [86]

Answer:

Answer is the explanation.

Explanation:

A light microscope (LM) is an instrument which uses the visible light and magnifying glasses to examine small objects that are invisible to the naked eye or have finer details than the naked eye allows. However, magnification is not the main problem in microscopy.

Electron microscopy (EM) is a technique for obtaining high-resolution images of biological and non-biological samples. It is used in biomedical research to study the detailed structure of tissues, cells, organelles and macromolecular complexes.

An atomic force microscope is a type of high-resolution scanning probe microscope with a resolution that can be measured in fractions of a nanometer. It was launched in 1986 by Nobel Prize winners Gerd Binnig, Calvin Quate and Christoph Gerber.

8 0
3 years ago
A current of 1.41 A in a long, straight wire produces a magnetic field of 5.61 uT at a certain distance from the wire. Find
pshichka [43]

Answer:

0.050 m

Explanation:

The strength of the magnetic field produced by a current-carrying wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0=4\pi \cdot 10^{-7} H/m is the vacuum permeability

I is the current in the wire

r is the distance from the wire

And the magnetic field around the wire forms concentric circles, and it is tangential to the circles.

In this problem, we have:

I=1.41 A (current in the wire)

B=5.61\mu T=5.61\cdot 10^{-6} T (strength of magnetic field)

Solving  for r, we find the distance  from the wire:

r=\frac{\mu_0 I}{2\pi B}=\frac{(4\pi \cdot 10^{-7})(1.41)}{2\pi (5.61\cdot 10^{-6})}=0.050 m

4 0
3 years ago
On another planet, the isotopes of titanium have the given natural abundances. Isotope Abundance Mass (u) 46Ti 77.600% 45.95263
allsm [11]

Answer: Average atomic mass of titanium on that planet is 46.52

Explanation:

Mass of isotope Ti-46 = 45.95263

% abundance of isotope Ti-46  = 77.600 % = \frac{77.600}{100}=0.776

Mass of isotope Ti- 48= 47.94795

% abundance of isotope Ti-48 = 16.100%= \frac{16.100}{100}=0.161

Mass of isotope Ti- 50 = 49.94479

% abundance of isotope Ti-50  = 6.300%= \frac{6.300}{100}=0.063

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

Z=\sum[(45.95263 \times 0.776)+(47.94795 \times 0.161)+(49.94479\times 0.063)

Z=46.52

Average atomic mass of titanium on that planet is 46.52

8 0
3 years ago
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