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3 years ago

Which of the following is an example of the conclusion phase of the scientific method?

Physics

1 answer:

Burka [1]3 years ago

5 0

Answer:

a scientist examines the results and answers the lab question- last choice

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Using a refracting telescope, you observe the planet Mars when it is 2.11 × 10 11 m 2.11×1011 m from Earth. The diameter of the

Norma-Jean [14]

Answer:

y = 150 Km

Explanation:

given,

Distance of the planet Mars from earth D = 2.11 × 10¹¹ m

The diameter of the telescope's objective lens d = 0.949 m

The wavelength of light λ = 553 nm

= 553 x 10⁻⁹ m

resolution of the telescope = ?

$y = \dfrac{1.22\lambda\ D}{d}$

$y = \dfrac{1.22\times 553\times 10^{-9}\times 2.11 \times 10^{11}}{0.949}$

y = 150003 m

y = 150 Km

the minimum feature size is equal to y = 150 Km

8 0

3 years ago

A diffraction grating has 300 lines per mm. If light of wavelength 630 nm is sent through this grating, what is the highest orde

True [87]

Answer:

The order of maximum is $n = 5$

Explanation:

From the question we are told that

The diffraction grating is k = 300 lines per mm = 300000 lines per m

The wavelength is $\lambda = 630 \ nm = 630 *10^{-9} \ m$

Generally the condition for constructive interference is mathematically represented as

$dsin \theta = n * \lambda$

Here n is the order maximum

d is the distance the grating which is mathematically represented as

$d = \frac{1}{k}$

=> $d = \frac{1}{300000}$

=> $d = 3.3*10^{-6}\ m$

$n = \frac{dsin \theta}{ \lambda}$

at maximum $sin\theta = 1$

$n = \frac{d}{\lambda}$

=> $n = \frac{3.3*10^{-6}}{630 *10^{-9}}$

=> $n = 5$

8 0

3 years ago

Melting Point of Substances

frez [133]

<span>(kg) Melting Point of Tin

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5 0

3 years ago

Read 2 more answers

A 10.0 cm3 sample of copper has a mass of 89.6

Romashka-Z-Leto [24]

Density is mass divides by volume, so
89.6g / 10cm^3 =8.96g /cm^3

*cm^3 is a standard unit of volume*

4 0

4 years ago

A small laser used as a pointer produces a beam of red light 5 mm in diameter, and has a power output of 5 milliwatts. What is t

Nata [24]

Answer:

434.0 V/m

Explanation:

The power output of the laser is:

$P=5 mW = 0.005 W$

while the radius of the beam is

$r=\frac{5 mm}{2}=2.5 mm = 0.0025 m$

so the cross-sectional area is

$A=\pi r^2 = \pi (0.0025 m)^2=2.0\cdot 10^{-5} m^2$

So the intensity of the laser beam is

$I=\frac{P}{A}=\frac{0.005 W}{2.0\cdot 10^{-5} m^2}=250 W/m^2$

The intensity of a laser beam is related to the magnitude of the electric field by

$I=\frac{1}{2}c\epsilon_0 E^2$

where

c is the speed of light

$\epsilon_0$ is the vacuum permittivity

Solving the formula for E, we find

$E=\sqrt{\frac{2I}{c\epsilon_0}}=\sqrt{\frac{2(250 W/m^2)}{(3\cdot 10^8 m/s)(8.85\cdot 10^{-12} F/m)}}=434.0 V/m$

4 0

3 years ago