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Alenkinab [10]
3 years ago
6

Question 3 You see the coordinates 5 E longitude, 10 N latitude. You do not need to look at a map in order to deduce that this l

ocation is a. near both the equator and the prime meridian Selected: b. near the International Date Line as well as the North PoleThis answer is incorrect. c. near the International Date Line and the prime meridian d. near the equator but quite far from the prime meridian e. near both the equator and the International Date Line
Physics
2 answers:
olchik [2.2K]3 years ago
8 0

Answer:

a. near both the equator and the prime meridian

Explanation:

Lets first understand what is equator and prime meridian.

We have drawn an imaginary grid on the Earth for navigational and other purposes. This grid has two coordinates : Latitude and Longitude. Equator is the largest latitude circle that divides the Earth in two equal hemispheres. This is taken as 0° for latitude. Latitudes increase as we move towards North/South pole which lies at 90°N and 90° S respectively. Longitudes are the lines running from north pole to south pole. The longitude that crosses from Greenwich is taken as 0°. This longitude is also known as prime meridian. The value of longitudes vary from 0° to 180° East and 180° West.

Now, the given coordinates are: 5° E and 10° N. It is quite evident that this location will be close to equator and prime meridian.

san4es73 [151]3 years ago
5 0

Answer:

a. Near both the equator and the prime meridian.

Explanation:

The  equator is at 0 degrees latitude and the prime meridian is 0 degrees longitude.

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Fed [463]
You have to use the specific heat equation. 

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So we can substitute our variables into the equation.

30000J = (390g)(3.9J*g/C)ΔT

Solving for ΔT, we get:

30000J/[(390g)*(3.9J*g/C) = ΔT

ΔT = 19.72386588C

I'm assuming the temperature is C, since it was not specified.

Hope this helps!
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3 years ago
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By calculating its wavelength (in nm), show that the second line in the Lyman series is UV radiation.
Rashid [163]

Answer:

 λ = 102.78  nm

This radiation is in the UV range,

Explanation:

Bohr's atomic model for the hydrogen atom states that the energy is

           E = - 13.606 / n²

where 13.606 eV   is the ground state energy and n is an integer

an atom transition is the jump of an electron from an initial state to a final state of lesser emergy

            ΔE = 13.606 (1 / n_{f}^{2} - 1 / n_{i}^{2})

the so-called Lyman series occurs when the final state nf = 1, so the second line occurs when ni = 3, let's calculate the energy of the emitted photon

            DE = 13.606 (1/1 - 1/3²)

            DE = 12.094 eV

let's reduce the energy to the SI system

            DE = 12.094 eV (1.6 10⁻¹⁹ J / 1 ev) = 10.35 10⁻¹⁹ J

let's find the wavelength is this energy, let's use Planck's equation to find the frequency

            E = h f

             f = E / h

            f = 19.35 10⁻¹⁹ / 6.63 10⁻³⁴

            f = 2.9186 10¹⁵ Hz

now we can look up the wavelength

           c = λ f

           λ = c / f

           λ = 3 10⁸ / 2.9186 10¹⁵

           λ = 1.0278  10⁻⁷ m

let's reduce to nm

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This radiation is in the UV range, which occurs for wavelengths less than 400 nm.

5 0
3 years ago
Is Smoke Abiotic or Biotic?
nlexa [21]

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6 0
2 years ago
block of mass 5kgriding on a horizontal frictionlessxy-plane surface is subjected tothree applied forces:→F1= 12√2N[ 45◦]→F2= (8
dsp73

Answer:

(i) See attached image for the drawing

(ii) net force given in component form: (20, 20)N with magnitude: \sqrt{800} \,\,\,N

Explanation:

First try to write all forces in  vector component form:

The force F1 acting at 45 degrees would have multiplication factors of \frac{\sqrt{2} }{2} on both axes, to take care of the sine and cosine projections. Therefore, the:

x-component of F1 is    F1_x=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

y-component of F1 is    F1_y=12\,\sqrt{2} \frac{\sqrt{2} }{2} =12\,\,N

As far as force F2, it is given already in x and y components, then:

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x-component of R = 12 + 8 = 20 N

y-component of R = 12 + 14 - 6 = 20 N

Therefore, the acceleration that the mass receives due to this force is given in component form as:

x-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

y-component of acceleration: 20 N / 5 kg = 4\,\,\,m/s^2

Now we can calculate the components of the velocity of this mass after 2 seconds of being accelerated by this force, using the formula of acceleration times time:

x-component of the velocity is:     v_x=4\,*\,2=8\,m/s

y-component of the velocity is:     v_y=4\,*\,2=8\,m/s

7 0
3 years ago
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