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3 years ago

Magnetic force... Is it a Conservative or Non-conservative field?

1 answer:

4 0

The correct answer is non-conservative

This is bewildering to people, however, since it shows that it has many aspects of a conservative fields, but is technically not one.

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State the name of electromagnetic wave other than visible light ?

Natali5045456 [20]

Answer:

-Radio waves

-Microwaves

-Infrared light (IR)

-Visible light

-Ultraviolet (UV)

-X-ray

-Gamma rays

(this is in order from the longest wavelength to shortest wavelength)

7 0

3 years ago

Two identical ambulances with loud sirens are driving directly towards you at a speed of 40 mph. One ambulance is 2 blocks away

Rasek [7]

Answer:

c) The pitch of the two sirens sounds the same to you

Explanation:

The pitch does not depend on the distance of the object from the observer.

As per the given data

pitch = frequency

Frequency = $f_{0}$ $\frac{V +- V_{0}}{V +- V_{s}}$

$f^{'}$ = $f_{0}$ $\frac{V }{V - V_{s}}$

Hence, the pitch of the two sirens remains the same for the observer.

5 0

3 years ago

Read 2 more answers

A scuba diver, submerged under water looks up and seessunlight

maw [93]

Answer:

$41.7975\ ^{\circ}$

Explanation:

$n_2$ = Refraction index of water = 1.333

$n_1$ = Refraction index of air = 1

$\theta_2$ = Angle of sunlight = 30°

From Snell's law we have

$\dfrac{n_1}{n_2}=\dfrac{sin\theta_2}{sin\theta_1}\\\Rightarrow \theta_1=sin^{-1}\dfrac{n_2sin\theta_2}{n_1}\\\Rightarrow \theta_1=sin^{-1}\dfrac{1.333sin30}{1}\\\Rightarrow \theta_1=41.7975\ ^{\circ}$

The angle measured from the vertical does the sunlight strike thesurface of the water is $41.7975\ ^{\circ}$

5 0

3 years ago

Suppose that a parallel-plate capacitor has circular plates with radius R = 25 mm and a plate separation of 4.7 mm. Suppose also

ozzi

Answer:

a) $B_{max} = 1.784*10^{-12}$

Explanation:

Given paraeters are:

R = 25 cm

d = 4.7 mm

f = 60 Hz

$V_m$ = 160 V

a) $V = V_msin(2\pi ft)$

Where $f = 60$ Hz and $V_m = 160$ V

$E =V/d= \frac{V_msin(2\pi ft)}{d}$

For $r = R$

$A = \pi R^2$

Since $\Phi_E = EA$

$\Phi_E=\frac{\pi R^2V_msin(2\pi ft) }{d}$

From Ampere's Law:

$\int B.ds = \mu_0\epsilon_0\frac{d\Phi_E}{dt} + \mu_0I_{encl}$ where $I_{encl}=0$

So at $r = R$ ,

$B.2\pi R = \mu_0\epsilon_0\frac{d\Phi_E}{dt}\\B.2\pi R = \mu_0\epsilon_0\frac{2\pi^2fR^2V_mcos(2\pi ft)}{d}\\B = \frac{\mu_0\epsilon_0\pi fRV_mcos(2\pi ft)}{d}$