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2 years ago

Can EM waves pass through solid objects

Physics

2 answers:

a_sh-v [17]2 years ago

8 0

Answer:

no.

Explanation:

It’s true that sound travels fastest through solids, but solid objects actually block sound waves from reaching a given space. The reason behind this is very simple: you see, when sound originates from a point, travels through a medium, and then encounters a solid object, it loses some of its energy.

jarptica [38.1K]2 years ago

8 0

Answer:

These changing fields form electromagnetic waves. Electromagnetic waves differ from mechanical waves in that they do not require a medium to propagate. This means that electromagnetic waves can travel not only through air and solid materials, but also through the vacuum of space.

Explanation:

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he tune-up specifications of a car call for the spark plugs to be tightened to a torque of 47 N⋅m . You plan to tighten the plug

S_A_V [24]

Answer:

207.4 N

Explanation:

The torque $\tau$ on a body is

$\tau = r* F$ where r is the radius vector from the point of rotation to the point at which force F is applied.

The product of r and F is equal to the product of magnitude of r and F multiplied by the sine of angle between both vectors.

Therefore, torque is also given by

$\tau = rF\sin \theta$

Where $\theta$ is the angle between r and F.

Use the expression of torque.

Substitute L for r in the equation $\tau = rF\sin \theta$

$\tau = LF\sin \theta$

Where L is the length of the wrench.

Making F the subject

$F = \frac{\tau }{{L\sin \theta }}$

Force required to pull the wrench is given as,

$F = \frac{\tau }{{L\sin \theta }}$

Substitute $47{\rm{ N}} \cdot {\rm{m}}$ for $\tau$ , 25 cm for L, and 115o for $\theta$

$\begin{array}{c}\\F = \frac{{47{\rm{ N}} \cdot {\rm{m}}}}{{\left( {25{\rm{ cm}}} \right)\sin {{115}^{\rm{o}}}}}\left( {\frac{{1{\rm{ cm}}}}{{{{10}^{ - 2}}{\rm{ m}}}}} \right)\\\\ = 207.435{\rm{ N}}\\\\ \approx 207.4{\rm{ N}}\\\end{array}$

6 0

2 years ago

Read the scenario and solve these two problems.

Burka [1]

Answers:

a) 5400000 J

b) 45.92 m

Explanation:

a) The kinetic energy $K$ of an object is given by:

$K=\frac{1}{2}mV^{2}$

Where:

$m=12000 kg$ is the mass of the train

$V=30 m/s$ is the speed of the train

Solving the equation:

$K=\frac{1}{2}(12000 kg)(30 m/s)^{2}$

$K=5400000 J$ This is the train's kinetic energy at its top speed

b) Now, according to the Conservation of Energy Law, the total initial energy is equal to the total final energy:

$E_{i}=E_{f}$

$K_{i}+P_{i}=K_{f}+P_{f}$

Where:

$K_{i}=5400000 J$ is the train's initial kinetic energy

$P_{i}=0 J$ is the train's initial potential energy

$K_{f}=0 J$ is the train's final kinetic energy

$P_{f}=mgh$ is the train's final potential energy, where $g=9.8 m/s^{2}$ is the acceleration due gravity and $h$ is the height.

Rewriting the equation with the given values:

$5400000 J=(12000 kg)(9.8 m/s^{2})h$

Finding $h$ :

$h=45.918 m \approx 45.92 m$

7 0

3 years ago

Read 2 more answers

Consider two thin, coaxial, coplanar, uniformly charged rings with radii a and b푏 (a

Wittaler [7]

Answer:

electric potential, V = -q(a²- b²)/8π∈₀r³

Explanation:

Question (in proper order)

Consider two thin coaxial, coplanar, uniformly charged rings with radii a and b (b < a) and charges q and -q, respectively. Determine the potential at large distances from the rings

<em>consider the attached diagram below</em>

the electric potential at point p, distance r from the center of the outer charged ring with radius a is as given below

Va = q/4π∈₀ [1/(a² + b²)¹/²]

$Va = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} }$

Also

the electric potential at point p, distance r from the center of the inner charged ring with radius b is

$Vb = \frac{-q}{4\pi e0} * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

Sum of the potential at point p is

V = Va + Vb

that is

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } + \frac{-q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * \frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{q}{4\pi e0 } * \frac{1}{(b^{2} + r^{2} )^{1/2} }$

$V = \frac{q}{4\pi e0} * [\frac{1}{(a^{2} + r^{2} )^{1/2} } - \frac{1}{(b^{2} + r^{2} )^{1/2} }]$

the expression below can be written as the equivalent

$\frac{1}{(a^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + a^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} }$

likewise,

$\frac{1}{(b^{2} + r^{2} )^{1/2} } = \frac{1}{(r^{2} + b^{2} )^{1/2} } = \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }$

hence,

$V = \frac{q}{4\pi e0} * [\frac{1}{{r(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{r(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

1/r is common to both equation

hence, we have it out and joined to the 4π∈₀ denominator that is outside

$V = \frac{q}{4\pi e0 r} * [\frac{1}{{(1^{2} + \frac{a^{2} }{r^{2} } )}^{1/2} } - \frac{1}{{(1^{2} + \frac{b^{2} }{r^{2} } )}^{1/2} }]$

by reciprocal rule

1/a² = a⁻²

$V = \frac{q}{4\pi e0 r} * [{(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} - {(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2}]$

by binomial expansion of fractional powers

where $(1+a)^{n} =1+na+\frac{n(n-1)a^{2} }{2!}+ \frac{n(n-1)(n-2)a^{3}}{3!}+...$

if we expand the expression we have the equivalent as shown

${(1^{2} + \frac{a^{2} }{r^{2} } )}^{-1/2} = (1-\frac{a^{2} }{2r^{2} } )$

also,

${(1^{2} + \frac{b^{2} }{r^{2} } )}^{-1/2} = (1-\frac{b^{2} }{2r^{2} } )$

the above equation becomes

$V = \frac{q}{4\pi e0 r} * [((1-\frac{a^{2} }{2r^{2} } ) - (1-\frac{b^{2} }{2r^{2} } )]$

$V = \frac{q}{4\pi e0 r} * [1-\frac{a^{2} }{2r^{2} } - 1+\frac{b^{2} }{2r^{2} }]$

$V = \frac{q}{4\pi e0 r} * [-\frac{a^{2} }{2r^{2} } +\frac{b^{2} }{2r^{2} }]\\\\V = \frac{q}{4\pi e0 r} * [\frac{b^{2} }{2r^{2} } -\frac{a^{2} }{2r^{2} }]$