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Rudiy27
3 years ago
5

One way to search for planets around other stars is the doppler technique. another way uses transits of planets around other sta

rs. how does the transit technique work?
Physics
1 answer:
katovenus [111]3 years ago
3 0
The transit method requires watching the light output of a star over long periods of time. A transit occurs when the planet crosses in front of its star from earths point of view. Since there is a small object (the planet) now blocking some of the star, it appears to dim a little bit for a while until the planet passes. If we are in a position where that occurs regularly (most paths of planets do not happen to be on the line of sight between earth and their star) we can deduce the period of orbit. From the amount of dimming and the period you can estimate the mass
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3 years ago
PlzHELP
Naddika [18.5K]
 Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion.Displacement<span> is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

</span>To calculate displacement<span>, simply draw a vector from your starting point to your final position and solve for the length of this line. If your starting and ending position are the same, like your circular 5K route, then your </span>displacement<span> is 0. In physics, </span>displacement<span> is represented by Δs.
 
 For me to solve this I would need to know the time, but I can give you a handy displacement calculator I used that helped me. 
 
https://www.easycalculation.com/physics/classical-physics/constant-acc-displacement.php

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8 0
3 years ago
The distance that a spring will stretch varies directly as the force applied to the spring. A force of 8080 pounds is needed to
xxTIMURxx [149]

Answer:

F₂= 210 pounds

Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

F= K*x   Formula (1)

Where;

F  is the magnitude of the force applied to the spring in Newtons (Pounds)

K is the elastic spring constant, which relates force and elongation. The higher its value, the more work it will cost to stretch the spring. (Pounds/inch)

x the elongation of the spring (inch)

Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

F₁ =80 pounds

x₁= 8 inches

x₂= 21  inches

Problem development

We replace data in formula 1 to calculate  K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

We apply The formula 1 to calculate  F₂

F₂= K*x₂

F₂= (10)*(21)

F₂= 210 pounds

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