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3 years ago

What is the significance of the discovery of exoplanets?

Physics

2 answers:

polet [3.4K]3 years ago

5 0

For only through exploration of the unknown we have come as far as we are now.

Discoveries are what laid the foundation for the technology and all advancement of the society we have today. Similar to how we continue to voyage into the unknowns of this world to find resources to supply for our needs, studies through new materials, species we encounter will all benefit the most common problems we have such as energy resource, medical needs and cure for the incurable diseases. These explanetary exploration may be the next key which holds potential to solve these problems.

sveticcg [70]3 years ago

3 0

Sorry, i could not help you with the question, but I just wondering do you know what is Biomarker discovery.

You might be interested in

Conclusion on ohm's law experiment

Free_Kalibri [48]

Answer:

Explanation:

In our previous articles, we observed the theoretical formulas of Ohm’s law, its calculations in the lab report, and experiment. Today you’ll learn the verification of theory vs experimental results on a 1 kΩ resistor.

The theoretical results are obtained from the formula of Ohm’s law: V = IR. The experimental verification is provided for a metal film 1 kΩ (±0.05%). We have used a high-quality resistor with negligible tolerance value so as to reduce the tolerance error.

The little problem in our calculations arises due to improper handling of multimeter probes. You can learn the complete method to perform the Ohm’s experiment here and can calculate the current values by using the Ohm’s law calculator.

6 0

3 years ago

What did Thomson’s and Rutherford’s experiments have in common? They both used charged particles in their experiments. They both

Triss [41]

Answer:

Both Thomson and Rutherford used charged particles in their experiments.

Explanation:

6 0

3 years ago

Read 2 more answers

Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib

Alexus [3.1K]

Answer:

a) 318.2 W/m^2

b) 2.5 x 10^-4 J

c) 1.55 x 10^-8 v/m

Explanation:

Power of laser P = 1 mW = 1 x 10^-3 W

exposure time t = 250 ms = 250 x 10^-3 s

If beam diameter = 2 mm = 2 x 10^-3 m

then

cross-sectional area of beam A = $\pi d^{2} /4$ = (3.142 x $(2*10^{-3} )^{2}$ )/4

A = 3.142 x 10^-6 m^2

a) Intensity I = P/A

where P is the power of the laser

A is the cros-sectional area of the beam

I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2

b) Total energy delivered E = Pt

where P is the power of the beam

t is the exposure time

E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J

c) The peak electric field is given as

E = $\sqrt{2I/ce_{0} }$

where I is the intensity of the beam

E is the electric field

c is the speed of light = 3 x 10^8 m/s

$e_{0}$ = 8.85 x 10^9 m kg s^-2 A^-2

E = $\sqrt{2*318.2/3*10^8*8.85*10^9}$ = 1.55 x 10^-8 v/m

6 0

3 years ago

Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,

EastWind [94]

Answer:

F = 3.6 kN, direction is 9.6º to the North - East

Explanation:

The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.

Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.

Wind

X axis

F₁ = 2.50 kN

Tide

cos 30 = F₂ₓ / F₂

sin 30 = F_{2y} / F₂

F₂ₓ = F₂ cos 30

F_{2y} = F₂ sin 30

F₂ₓ = 1.20cos 30 = 1.039 kN

F_{2y} = 1.20 sin 30 = 0.600 kN

the resultant force is

X axis

Fₓ = F₁ₓ + F₂ₓ

Fₓ = 2.50 +1.039

Fₓ = 3,539 kN

F_y = F_{2y}

F_y = 0.600

to find the vector we use the Pythagorean theorem

F = $\sqrt{F_x^2 +F_y^2}$

F = $\sqrt{ 3.539^2 + 0.600^2 }$

F = 3,589 kN

the address is

tan θ = F_y / Fₓ

θ = tan⁻¹ $\frac{F_y}{F_x}$

θ = tan⁻¹ $\frac{0.6}{3.539}$ 0.6 / 3.539

θ = 9.6º

the resultant force to two significant figures is

F = 3.6 kN

the direction is 9.6º to the North - East

7 0

2 years ago

Everything is in the pic.

nikitadnepr [17]

The answer is A

Good luck!

3 0

3 years ago