PLS HELP

Effectus [21]

A sedentary lifestyle is when you don't make any exercise. For example, if I spent a whole week just sitting down answering questions, that would be an example of a sedentary lifestyle, because I'm not running or jogging. Having that type of lifestyle can make you fat and get health problems.

Have a nice day! :)

7 0

3 years ago

Astar is 10 light years away from the earth. Suppose it brightens up suddenly today, after how long can we see this change?

Blababa [14]

The phrase "light year" is a <u><em>distance</em></u> ... it's the distance that light travels through vacuum in one year.

When you look at an object located 1 light year away from you, you see it as it was 1 year ago.

If a star located 10 light years away from us suddenly brightens, or dims, or explodes, we see the event <em>10 years later.</em>

7 0

3 years ago

A cylinder is fitted with a piston, beneath which is a spring, as in the drawing. The cylinder is open to the air at the top. Fr

astraxan [27]

Answer:

x = 0.0734 m = 7.34 cm

Explanation:

First we shall calculate the area of the piston:

$Area = \pi radius^2\\Area = \pi (0.028\ m)^2\\Area = 0.00246\ m^2$

Now, we will calculate the force on the piston due to atmospheric pressure:

$Atmospheric\ Pressure = \frac{Force}{Area}\\\\Force = (Atmospheric\ Pressure)(Area)\\Force = (101325\ N/m^2)(0.00246\ m^2) \\Force = F = 249.56\ N$

Now, for the compression of the spring we will use Hooke's Law as follows:

$F = kx\\$

where,

k = spring constant = 3400 N/m

x = compression = ?

Therefore,

8 0

3 years ago

in a solar system far, far away the sun's intensity is 200 w/m2 for an inner planet located a distance r away. what is the sun's

GarryVolchara [31]

The sun's intensity for an outer planet located at a distance 6r from the sun is 5.55 W/m². The result is obtained by using the inverse square law formula.

<h3>What is the Inverse Square Law formula?</h3>

The Inverse Square Law formula describes the intensity of light is inversely proportional to the square of the distance. It can be expressed as

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

Where

I₁ = Intensity at distance 1 (W/m²)
I₂ = Intensity at distance 2 (W/m²)
d₁ = distance 1 from a light source (m)
d₂ = distance 2 from a light source (m)

Given the case the sun's intensity is 200 W/m² for an inner planet at the distance r. If an outer planet is at a distance 6r, what is the sun's intensity?

By using the inverse square law formula, the sun's intensity for an outer planet is

$\frac{I_{1} }{I_{2} } = \frac{d_{2}^{2} }{d_{1}^{2}}$

$\frac{200 }{I_{2} } = \frac{(6r)^{2} }{r^{2}}$