The Hertzsprung-Russel diagram is used to show the relationship between which two characteristics of stars?

To solve a problem using the equation for keplerâs third law, enrico must convert the average distance of mars from the sun from

Natali5045456 [20]

1 astronomical unit = 149597870700m
Enrico should divide distance in meters with this number.

3 0

3 years ago

Read 2 more answers

A professor's office door is 0.91 m wide, 2.0 m high, and 4.0 cm thick; has a mass of 25 kg; and pivots on frictionless hinges.

olasank [31]

Answer:

F= 5.71 N

Explanation:

width of door= 0.91 m

door closer torque on door= 5.2 Nm

In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.

so wee need to exert 5.2 Nm torque on the door.

If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.

T= r x F

T= r F sin∅

F= T/ (r * sin∅)

F= 5.2/ (0.91 * 1)

F= 5.71 N

5 0

3 years ago

PLZ HELP ASAP WILL GIVE BRAINLIEST!!!

igomit [66]

the answer is the forth one treatment of cancer

3 0

3 years ago

Read 2 more answers

Properties can be observed only when the substance in a sample of matter are changing into different substance

Zolol [24]

The complete statement is "chemical properties can be observed only when the substance in a sample of matter are changing into different substance".

It states a key concept in chemistry.

A chemical property is the ability of a substance, element or compound, to <em>transform</em> into other substances either <em>by decomposing or by combining</em> with one or more substances.

This transformation is defined as chemical reaction.

During chemical reactions some chemical bonds are broken and others are formed leading to the formation of one or more different substances called products.

Some examples of chemical properties are: reactivity with oxygen, reactivity with water, acidity, basicity, oxidation, reduction. The only way to tell if a substance has certain chemical property is by letting it react and so observe the change of the original substance into one or more different substances.

8 0

3 years ago

A camera with a 50.0-mm focal length lens is being used to photograph a person standing 3.00 m away. (a) How far from the lens m

kirill [66]

a) 50.8 mm

b) The whole image (1:1)

c) It seems reasonable

Explanation:

a)

To project the image on the film, the distance of the film from the lens must be equal to the distance of the image from the lens. This can be found by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length of the lens

p is the distance of the object from the lens

q is the distance of the image from the lens

In this problem:

f = 50.0 mm = 0.050 m is the focal length (positive for a convex lens)

p = 3.00 m is the distance of the person from the lens

Therefore, we can find q:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.050}-\frac{1}{3.00}=19.667m^{-1}\\q=\frac{1}{19.667}=0.051 m=50.8 mm$

b)

Here we need to find the height of the image first.

This can be done by using the magnification equation:

$\frac{y'}{y}=-\frac{q}{p}$

where:

y' is the height of the image

y = 1.75 m is the height of the real person

q = 50.8 mm = 0.0508 m is the distance of the image from the lens

p = 3.00 m is the distance of the person from the lens

Solving for y', we find:

$y'=-\frac{qy}{p}=-\frac{(0.0508)(1.75)}{3.00}=-0.0296 m=-29.6mm$

(the negative sign means the image is inverted)

Therefore, the size of the image (29.6 mm) is smaller than the size of the film (36.0 mm), so the whole image can fit into the film.

c)

This seems reasonable: in fact, with a 50.0 mm focal length, if we try to take the picture of a person at a distance of 3.00 m, we are able to capture the whole image of the person in the photo.

3 0

3 years ago