A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement?

g The electric field in a sinusoidal wave changes as E =125 N>C2cos 311.2 * 1011 rad>s2t +14.2 * 102 rad>m2x] (a) In wh

Elan Coil [88]

Answer:

a) the propagation direction is x, b) E₀ = 125 N / C² , c) B = 41.67 10⁻⁸ T ,

d) f = 4.95 10¹¹ Hz, e) λ = 4.42 10⁻³ m, f) the speed of light ,

Explanation:

The equation they give for the sine wave is

E = 125 cos (14.2 10² x - 311.2 10¹¹ t)

This expression must have the general shape of a traveling wave

E = Eo cos (kx - wt + Ф)

we can equal each term between the two equations

a) the propagation direction is x, since it is the term that accompanies the vector k

b) the amplitude is the coefficient before the cosine function

E₀ = 125 N / C²

c) to find the amplitude of the magnetic field we use that the two fields are in phase

C = E / B

B = E / c

B = 125/3 10⁸

B = 41.67 10⁻⁸ T

d) the angular velocity is

w = 311.2 10¹¹ rad / s

angular velocity and frequencies are related

w = 2π f

f = w / 2π

f = 311.2 10¹¹ / 2π

f = 4.95 10¹¹ Hz

e) the wavelength is obtained from the wave number

k = 2π /λ

k = 14.2 10² rad / m

λ = 2π / k

λ = 2π / 14.2 10²

λ = 4.42 10⁻³ m

f) the speed of an electromagnetic wave is the speed of light

g) what is a transverse wave

4 0

3 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

4 years ago

What is the volume of a marble with a diameter of 3.0

RideAnS [48]

Explanation:

You can solve for volume using radius or diameter.

Sphere Volume = 4/3 • π • r³ = ( π •d³)/6

We're given the diameter so let's use that.

Volume = PI * d^3 / 6

Volume = 3.14159 * 3.0^3 / 6

Volume = 3.14159 * 9 / 2

Volume = 14.137 cubic centimeters

7 0

3 years ago

ListenA person on a ledge throws a ball vertically downward, striking the ground below the ledge with 200 joules of kinetic ener

polet [3.4K]

Answer:

A. 200 J

Explanation:

The initial kinetic energy depends on the initial speed, while the gravitational potential energy depends on the height, both balls are thrown with the same initial speed and from the same height. Therefore, due to the law of conservation of energy, the balls must have the same mechanical energy (the sum of both energies) when both impact the ground. Since the potential energy is zero at this point, its final kinetic energy must also be the same.

8 0

3 years ago

How many molecules are in 3 moles of CO2

mote1985 [20]

Answer:

There are 1.806 X 1024 molecules in 3 moles of CO2. This concept is similar to counting 144 eggs and dividing by 12 to get the number of cartons.

8 0

3 years ago

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