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3 years ago

Why is it important for scientist to do research?

Physics

2 answers:

Brums [2.3K]3 years ago

5 0

Answer:

becuz without research they do not get observations then they can prove nothing

sweet-ann [11.9K]3 years ago

3 0

Answer:

Because if they dont research first they will be unprepared

Explanation:

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Consider the following three statements: For any electromagnetic radiation, the product of the wavelength and the frequency is a

boyakko [2]

Answer:

The correct answer is B. Statements (i) and (ii) are true.

Explanation:

Fisrt statement:

Electromagnetic radiation such as wave, wavelength λ and oscillation frequency ν are related by a constant, the speed of light. The equation is given by:

$c= \lambda \nu$

So the first statement is true.

Second statement:

$c=\lambda \nu\\c= 3.0 A \times 1.0\times 10^{18} Hz\\c= 3.0\times 10^{-10} m \times 1.0\times 10^{18} s^{-1}\\c=3\times 10^8 \frac{m}{s}$

The value of c in the vacuum is 3×10⁸ m/s. Hence, the second statement is true.

Third statement:

The speed of any electromagnetic radiation is constant regardless the type of radiation.

Hence, the third statement is false.

8 0

3 years ago

What is the wavelength of light (nm) that has a frequency of 6.44 x 1013 s-1?

lesantik [10]

4660

Hope this helped!
STSN

4 0

3 years ago

Read 2 more answers

How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n

grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

$dSin\theta = m\lambda$

Where

m= Any integer which represent the number of repetition of spectrum

$\lambda$ = Wavelength

d = Distance between the slits.

$\theta$ = Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

$d = \frac{m\lambda}{sin\theta }$

$d = \frac{2*536*10^{-9}}{sin(24)}$

$d = 2.63*10^{-6}m$

Therefore the number of lines per millimeter would be given as

$\frac{1}{d} = \frac{1}{2.63*10^{-6} }$

$\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})$

$\frac{1}{d} = 379.4 lines/mm$

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0

3 years ago

3. Which of the following magnetic scenarios will repel each

Nastasia [14]

Answer:

option C (1 and 4)

Explanation:

Like poles repel each other, unlike poles attract each other

7 0

4 years ago

Read 2 more answers

An uncrewed mission to the nearest star, Proxima Centauri, is launched from the Earth's surface as a projectile with an initial

Anna [14]

Answer:

42.96 km/s

Explanation:

From the conservation of Energy

$(PE+KE)_i=(PE+KE)_f\\\Rightarrow -\frac{GmM}{R}+\frac{1}{2}mv_i^2=0+\frac{1}{2}mv_f^2$

Mass gets cancelled

$-\frac{GM}{R}+\frac{1}{2}v_i^2=0+\frac{1}{2}v_f^2\\\Rightarrow -2\frac{GM}{R}+v_i^2=v_f^2\\\Rightarrow -v_e^2+v_i^2=v_f^2\\\Rightarrow v_f=\sqrt{v_i^2-v_e^2}$

$v_e=\sqrt{\frac{2Gm}{R}}$ = Escape velocity of Earth = 11.2 km/s

$v_i$ = Velocity of projectile = 44.4 km/s

$v_f=\sqrt{44.4^2-11.2^2}\\\Rightarrow v_f=42.96\ km/s$

The velocity of the spacecraft when it is more than halfway to the star is 42.96 km/s

6 0

3 years ago