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Gnoma [55]
3 years ago
6

Name the regions of a electromagnetic spectrum from the shortest wavelength to the longest wavelength

Physics
1 answer:
kaheart [24]3 years ago
5 0

Answer:

From shortest wavelength to longest wavelength:

Gamma Rays

X-Rays

Ultraviolet

Visible Light

Infrared waves

Microwaves

Radio Waves

Explanation:

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Choose the ending that best completes the following sentence: If an object is traveling with a constant velocity… A. There are n
Margarita [4]

I think the answer is B...

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3 years ago
Review 1: A plane is located x = 40 miles (horizontally) away from an airport at an altitude of h miles. Radar at the airport de
-Dominant- [34]

Explanation:

Let h is the height of the plane above ground. x is the horizontal distance between the ground and the airport. Let s(t) is the distance between the plane and the airport. So,

s(t)^2={h^2+x^2}...........(1)

Given, h = 4, x = 40 and s(t) = -20 mph

Differentiate equation (1) wrt t

2s(t)s'(t)=2x(t)x'(t)

x'(t)=\dfrac{s(t)s'(t)}{x(t)}

When x = 40, s(t)=\sqrt{40^2+4^2}=40.19\ m

x'(t)=\dfrac{-240s(t)}{x(t)}

x'(t)=\dfrac{-240\times 40.19}{40}

x'(t)=-241.14\ m/s

So, the speed of the airplane is 241.14  m/s. Hence, this is the required solution.

8 0
4 years ago
A rock layer that stores and transmits water​
lukranit [14]
The ability of the ground water to pass through the pore spaces in the rock is described as the rock's permeability. Permeable layers of rock that store and transport water are called aquifers.
5 0
4 years ago
The ball is moving at a constant speed of 0.5 m/s for 2.3 seconds how far does it go?
yarga [219]

Distance = (speed) x (time)

Distance = (0.5 m/s) x (2.3 s)

Distance = (0.5 x 2.3) m

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7 0
4 years ago
In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu
Ronch [10]

Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

3 0
3 years ago
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