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3 years ago

How many degrees are in each quadrant A: 90° B: 30° C: 180° D: 360°

Physics

1 answer:

Airida [17]3 years ago

8 0

In one quadrant there are 90 degrees.

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What does the text mean when it says “naming an elements is like creating history

dolphi86 [110]

Answer: When you name an element, it's usually because the element is newly discovered. So by naming an element you are creating something that will last throughout history.

Explanation:

5 0

3 years ago

The sun's radiation has different wavelengths. the shorter is the radiation's wavelength, the weaker is the radiation's energy.

NeX [460]

It is false. Because the amount of energy carried in the wave is inversely related to the length of the waves wavelength. To correct the statement it should be that the shorter the radiation's wavelength the stronger is the radiation's energy.

7 0

3 years ago

On Earth, the number flux of solar neutrinos from the p-p chain is:

Nataly [62]

Answer:

Explanation:

The volume of the tank = 50 kton

50 kton = 5 × 10⁷ kg

Since 18 grams of water will contain: 10 electrons × 6.023 × 10²³

Then;

5× 10⁷ kg will contain $( \dfrac{5 \times 10^7 \times 10^3}{18}) \times 10 \times 6.023 \times 10^{23}$

= 1.67 × 10³⁴ electrons

(b)

Suppose:

$f_{neutrino} = \dfrac{2f_o}{26.2 MeV} = 6.7\times 10^{10} \ s^{-1} cm^{-2}$

Then;

10⁻⁶ of $f_{neutrino} = 6.7 \times 10^{10} \times 10^{-6} \ s^{-1} cm^{-2}$

$=6.7 \times 10^{4}\ s^{-1} cm^{-2}$

Thus, the number of high energy neutrinos which will interact with water is:

= $6.7 \times 10^4 \times \sigma$

= $6.7 \times 10^4 \times 10^{-43}$

= $6.7 \times 10^{-39} s^{-1}$

For 1.67 × 10³⁴ electrons, the detection rate is:

$6.7 \times 10^{-39} \times 1.67 \times 10^{34}$

$= 11.19 \times 10^{-5} \ s^{-1}$

= 9.668 per day

3 0

3 years ago

Find time when boy catches the girl or when they are at their closest separation..

soldier1979 [14.2K]

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\implies{Time=10\:sec\:and\:30\:sec}}} \\\\$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}} \\ \\$

$\green{\underline{\bold{Given :}}} \\ \\ \:\:\:\: \bullet\:\:\tt\red{ Velocity \: of \: boy = 50 \: m/s} \\ \\ \:\:\:\: \bullet\:\: \tt\orange{Velocity \: of \: girl = 30 \: m/s }\\ \\ \:\:\:\: \bullet\:\:\tt\green{ Acceleration \: of \: boy = {1 \: m/s}^{2}} \\ \\ \:\:\:\:\bullet\:\: \tt\blue{Acceleration \: of \:girl= {2\: m/s}^{2} }\\ \\ \:\:\:\: \bullet \:\:\tt\purple{Sepration \: between \: them = 150 \: m }\\ \\ \\ \red{\underline{\bold{To \: Find :}}} \\ \\ \:\:\:\: \bullet\:\: \tt\blue{Time \: taken \: to \: catch \: the \: girl }\\$

<u>According to given question</u> :

$\\ \green{\star} \: \text{Using \: relative \: motion \: method} \\ \\ \green{ \circ} \: \tt Net \: velocity = 50 - 30 = 20 \: m/s \\ \\ \green{ \circ} \: \tt Net \: acceleration = 1 - 2 = - 1\: m /{s}^{2} \\\\ \\ \star\:\bold\red{\underline{\:As \: we \: know \: that\:}} \\\\ \tt\purple{: \implies s = ut + \frac{1}{2} {at}^{2}} \\ \\ \tt\green{: \implies 150 = 20 \times t + \frac{1}{2} \times -1 \times {t}^{2}} \\ \\ \tt\purple{: \implies 300 = 40t - {t}^{2}} \\ \\ \tt\green{: \implies {t}^{2} - 40t + 300 = 0} \\ \\ \tt\purple{: \implies t = \frac{ - ( - 40) \pm\sqrt{ { (- 40)}^{2} - 4 \times 1 \times 300} }{2 \times 1} }\\ \\ \tt\green{: \implies t = \frac{40 \pm \sqrt{1600 - 1200} }{2} }\\ \\ \tt\purple{: \implies t = \frac{40 \pm 20}{2} }\\ \\ \green{\tt: \implies t = 10 \: sec \: and \: 30 \: sec}$

7 0

3 years ago

Which will change if you turn up a radios volume

lord [1]

The sound of the radio

3 0

3 years ago