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3 years ago

8

Which are examples of transverse waves? Check all that apply. earthquake P-waves earthquake S-waves radio waves sound visible li

ght X-rays

2 answers:

Mariulka [41]3 years ago

8 0

Earthquake S - Waves are examples of transverse waves. The correct option among all the options that are given in the question is the second option. Other good examples of transverse waves are an oscillating string and light waves. A wave is a kind of disturbance that or an oscillation that travels through space.

Artist 52 [7]3 years ago

5 0

Earthquake s waves

radio waves

visible light

x-rays

this is the correct answer on edge (i got it right)

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Helppp<br> In Static Electricity, The charges do not ____

Anna11 [10]

Answer:

basically I will tell you the definition

Explanation:

so when charges are unbalanced statistic energy is formed positive attract negative and negative attracts positive like repell while unlike attract .

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3 years ago

An aluminum bar 600mm long, with diameter 40mm, has a hole drilled in the center of the bar. The hole is 40mm in diameter and 10

s2008m [1.1K]

Answer:

<em>1.228 x </em> $10^{-6}$ <em> mm </em>

<em></em>

Explanation:

diameter of aluminium bar D = 40 mm

diameter of hole d = 30 mm

compressive Load F = 180 kN = 180 x $10^{3}$ N

modulus of elasticity E = 85 GN/m^2 = 85 x $10^{9}$ Pa

length of bar L = 600 mm

length of hole = 100 mm

true length of bar = 600 - 100 = 500 mm

area of the bar A = $\frac{\pi D^{2} }{4}$ = $\frac{3.142* 40^{2} }{4}$ = 1256.8 mm^2

area of hole a = $\frac{\pi(D^{2} - d^{2}) }{4}$ = $\frac{3.142*(40^{2} - 30^{2})}{4}$ = 549.85 mm^2

Total contraction of the bar = $\frac{F*L}{AE} + \frac{Fl}{aE}$

total contraction = $\frac{F}{E} * (\frac{L}{A} +\frac{l}{a})$

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3 years ago

A rectangular loop of area A is placed in a region where the magnetic field is perpendicular to the plane of the loop. The magni

mina [271]

Answer:

Induced emf, $\epsilon=-A\dfrac{B_{max}e^{-t/\tau}}{\tau}$

Explanation:

The varying magnetic field with time t is given by according to equation as :

$B=B_{max}e^{-t/\tau}$

Where

$B_{max}\ and\ t$ are constant

Let $\epsilon$ is the emf induced in the loop as a function of time. We know that the rate of change of magnetic flux is equal to the induced emf as:

$\epsilon=-\dfrac{d\phi}{dt}$

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$\epsilon=-A\dfrac{d(B)}{dt}$

$\epsilon=-A\dfrac{d(B_{max}e^{-t/\tau})}{dt}$

$\epsilon=A\dfrac{B_{max}e^{-t/\tau}}{\tau}$

So, the induced emf in the loop as a function of time is $A\dfrac{B_{max}e^{-t/\tau}}{\tau}$ . Hence, this is the required solution.

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2 years ago

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Sladkaya [172]

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3 years ago

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pishuonlain [190]

Answer:

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3 years ago