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What's the simplest approach to solve a physics dimensional question

Darina [25.2K]

Explanation:

<em><u>If Q is the unit of a derived quantity represented by Q = MaLbTc, then MaLbTc is called dimensional formula and the exponents a, b and, c are called the dimensions.</u></em>

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7 0

3 years ago

What does the word resolution mean in the sentence "By 1915, the British Roentgen Society had adopted a resolution to protect pe

iris [78.8K]

Resolution in this sentence refers to a solution to the problem

5 0

3 years ago

Hi i have homework can you help me

soldier1979 [14.2K]

Explanation:

The particle will be at rest when its velocity $v(t)$ is equal to zero. Recall that the velocity is simply the derivative of the position $x(t)$ with respect to time:

$v(t) = \dfrac{dx}{dt}$

Since $x(t) = t^2 - 2t + 2$

then

$v(t) = \dfrac{d}{dt}(t^2 - 2t + 2) = 2t - 2 = 0$

Solving for t, we find that the particle will be at rest at

$t = 1\:\text{s}$

6 0

2 years ago

An N-slit system has slit separation d and slit width a. Plane waves with intensity I and wavelength O are incident normally on

Leokris [45]

Answer:

E. d and O

Explanation:

"Light passing through a single slit forms a diffraction pattern somewhat different from those formed by double slits or diffraction gratings".

According to Huygens’s principle, "for each element of the wavefront in the slit emits wavelets. These are like rays that start out in phase and head in all directions. (Each ray is perpendicular to the wavefront of a wavelet.) Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel".

The destructive interference for a single slit is given by:

$d sin \theta = m\lambda , m=1,-1,2,-2,3,...$

Where

d is the slit width

$\lambda=O$ is the light's wavelength

$\theta$ is the angle relative to the original direction of the light

m is the order od the minimum

I represent the intensity

When the intensity and the wavelength are incident normally the angular as we can see on the expression above the angular separation just depends of the distance d and the wavelength O.

7 0

2 years ago

During a test, a NATO surveillance radar system, operating at 37 GHz at 182 kW of power, attempts to detect an incoming stealth

Marta_Voda [28]

(a) $2.68\cdot 10^{-6} W/m^2$

The intensity of an electromagnetic wave is given by

$I=\frac{P}{A}$

where

P is the power

A is the area of the surface considered

For the waves in the problem,

$P=182 kW = 1.82\cdot 10^5 W$ is the power

The area is a hemisphere of radius

$r=104 km=1.04\cdot 10^5 m$

so

$A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2$

So, the intensity is

$I=\frac{1.82\cdot 10^5 W}{6.8\cdot 10^{10}m^2}=2.68\cdot 10^{-6} W/m^2$

(b) $5.9\cdot 10^{-7} W$

In this case, the area of the reflection is

$A=0.22 m^2$

So, if we use the intensity of the wave that we found previously, we can calculate the power of the aircraft's reflection using the same formula:

$P=IA=(2.68\cdot 10^{-6} W/m^2)(0.22 m^2)=5.9\cdot 10^{-7} W$

(c) $8.7\cdot 10^{-18} W/m^2$

We said that the power of the waves reflected by the aircraft is

$P=5.9\cdot 10^{-7} W$

If we assume that the reflected waves also propagate over a hemisphere of radius

$r=104 km=1.04\cdot 10^5 m$

which has an area of

$A=2\pi r^2=2\pi (1.04\cdot 10^5 m)^2=6.8\cdot 10^{10} m^2$

Then the intensity of the reflected waves at the radar site will be

$I=\frac{P}{A}=\frac{5.9\cdot 10^{-7} W}{6.8\cdot 10^{10} m^2}=8.7\cdot 10^{-18} W/m^2$

(d) $8.1\cdot 10^{-8} V/m$

The intensity of a wave is related to the maximum value of the electric field by

$I=\frac{1}{2}c\epsilon_0 E_0^2$

where

c is the speed of light

$\epsilon_0$ is the vacuum permittivity

$E_0$ is the maximum value of the electric field vector

Solving the equation for $E_0$ ,

$E_0=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(8.7\cdot 10^{-18} W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=8.1\cdot 10^{-8} V/m$

(e) $1.9\cdot 10^{-16} T$

The maximum value of the magnetic field vector is given by

$B_0 = \frac{E_0}{c}$

Substituting the values,

$B_0 = \frac{(8.1\cdot 10^{-8} V/m)}{3\cdot 10^8 m/s}=2.7\cdot 10^{-16} T$

And the rms value of the magnetic field is given by

$B_{rms} = \frac{B_0}{\sqrt{2}}=\frac{2.7\cdot 10^{-16} T}{\sqrt{2}}=1.9\cdot 10^{-16} T$

7 0

3 years ago