All categories

3 years ago

Does parallax affect the precision of a measurement that you make

2 answers:

8 0

Yes, parallax affects the precision of a measurement that you make. It introduces an error in the order of the parallax. It will cause the measurement to be different from the real answer. Hope this answers the question. Have a nice day.

sergij07 [2.7K]3 years ago

7 0

Answer:

Yes

Explanation:

Parallax is the effect caused when there is an error in observing a reading. This is caused due to the variation of line of sight of the observer.

Precision refers to the variability of the readings being taken. If the readings are close to one another then the readings are precise and if the readings are not close to one another then the readings are not precise.

Parallax may cause the readings not to close therefore affecting the precision.

You might be interested in

Which of these is not true about the proton?

bulgar [2K]

Answer:

Option C is the untrue statement.

5 0

3 years ago

Read 2 more answers

A sled with no initial velocity accelerates at a rate of 3.2 m/s2 down a hill. How long does it take the sled to go 15 m to the

grin007 [14]

S = ut + 1/2 at^2
a = 3.2 m/s^2
s = 15m
Find t
15 = 1/2(3.2)t^2
15 = 3.2t^2/2
30 = 3.2t^2
30/ 3.2 = 9.38
Square root of 9.38 = 3.06
It takes 3.06 seconds

3 0

3 years ago

1. Two-point charges, QA = +8 μC and QB = -5 μC, are separated by a distance r = 10 cm. What is the magnitude and direction of t

densk [106]

Answer:

F = 36 N

Explanation:

Given that,

Charge, q₁ = +8 μC

Charge, q₂ = -5 μC

The distance between the charges, r = 10 cm = 0.1 m

We need to find the magnitude of the electrostatic force. The formula for the electrostatic force is given by :

$F=\dfrac{kq_1q_2}{r^2}\\\\F=\dfrac{9\times 10^9\times8\times 10^{-6}\times 5\times 10^{-6}}{(0.1)^2}\\F=36\ N$

So, the magnitude of the electrostatic force is 36 N.

3 0

2 years ago

The energy levels of a particular quantum object are -11.7 eV, -4.2 eV, and -3.3 eV. If a collection of these objects is bombard

gogolik [260]

To solve this problem it is necessary to apply an energy balance equation in each of the states to assess what their respective relationship is.

By definition the energy balance is simply given by the change between the two states:

$|\Delta E_{ij}| = |E_i-E_j|$

Our states are given by

$E_1 = -11.7eV$

$E_2 = -4.2eV$

$E_3 = -3.3eV$

In this way the energy balance for the states would be given by,

$|\Delta E_{12}| = |E_1-E_2|\\|\Delta E_{12}| = |-11.7-(-4.2)|\\|\Delta E_{12}| = 7.5eV\\$

$|\Delta E_{13}| = |E_1-E_3|\\|\Delta E_{13}| = |-11.7-(-3.3)|\\|\Delta E_{13}| = 8.4eV$

$|\Delta E_{23}| = |E_2-E_3|\\|\Delta E_{23}| = |-4.2-(-3.3)|\\|\Delta E_{23}| = 0.9eV$

Therefore the states of energy would be

Lowest : 0.9eV

Middle :7.5eV

Highest: 8.4eV

8 0

3 years ago

The Sun’s surface temperature is about 5800 K and its spectrum peaks at 5000 Å. An O-type star’s surface temperature may be 40,0

nirvana33 [79]

(a) $7.25\cdot 10^{-8}m$

Wien's displacement law is summarized by the equation

$\lambda = \frac{b}{T}$

where

$\lambda$ is the peak wavelength

$b=2.898\cdot 10^{-3} m \cdot K$ is Wien's displacement constant

T is the absolute temperature at the surface of the star

For an O-type star, we have

T = 40,000 K

Therefore, its peak wavelength is

$\lambda = \frac{2.898\cdot 10^{-3}}{40000}=7.25\cdot 10^{-8}m$

(b) Ultraviolet

We can answer this part by looking at the wavelength range of the different parts of the electromagnetic spectrum:

gamma rays

X-rays 1 nm - 1 pm

ultraviolet 380 nm - 1 nm

visible light 750 nm - 380 nm

infrared $25 \mu m - 750 nm$

microwaves 1 mm - $25 \mu m$

radio waves > 1 mm

The peak wavelength of this star is

$\lambda=7.25\cdot 10^{-8}m=72.5 nm$

Therefore, it falls in the ultraviolet region.

(c) No

The Keck telescopes is actually a system of 2 telescopes in the Keck Observatory, located in Mauna kea, Hawai.

The two telescopes, thanks to several instruments, are able to detect much of the electromagnetic radiation in the visible ligth and infrared parts of the spectrum. However, they are not able to detect light in the ultraviolet region: therefore, they cannot observe the star mentioned in the previous part of the problem.

7 0

3 years ago