The air pressure inside the balloon increases as the number of particles increases.
When a satellite is revolving into the orbit around a planet then we can say
net centripetal force on the satellite is due to gravitational attraction force of the planet, so we will have


now we can say that kinetic energy of satellite is given as


also we know that since satellite is in gravitational field of the planet so here it must have some gravitational potential energy in it
so we will have

so we can say that energy from the fuel is converted into kinetic energy and gravitational potential energy of the satellite
Highest to lowest number:
-less than 1 solar mass
-between 1 and 10 solar masses
-between 10 and 30 solar masses
-between 30 and 60 solar masses
<h3>What is Stellar masses ?</h3>
Stellar mass is a phrase that is used by astronomers to describe the mass of a star.
- It is usually enumerated in terms of the Sun's mass as a proportion of a solar mass ( M ☉). Hence, the bright star Sirius has around 2.02 M ☉.
- Stellar masses are not fixed, although they change for single stars only on long periods.
Learn more about Stellar masses here:
brainly.com/question/1128503
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Answer:
B. +m
Explanation:
The magnification of an image is defined as the ratio between the size of the image and of the object:

where we have
y' = size of the image
y = size of the object
There are two possible situations:
- When m is positive, y' has same sign as y: this means that the image image is upright
- When m is negative, y' has opposite sign to y: this means that the image is upside down
Therefore, the correct option representing an upright image is
B. +m
I'm actually going ahead in the book (DC Circuits) so this isn't really homework but I figured the tag was appropriate....the name of the chapter is Ohm's Law and Watt's Law.
<span>Problem: Calculate the power dissipated in the load resistor, R, for each of the circuits.Circuit (a): V = 10V; I = 100mA; R = ?; Since I know
V and
I use formula
P = IV: P = IV = (100mA)(10V) = 1 W.</span>
The next question is what I'm not sure about:
Question: What is the power in the circuit (a) above if the voltage is doubled? (Hint: Consider the effect on current).
What I did initially was: P = IV = (100mA)(2V) = 2 W
But then I looked at the answer and it said 4 W, then I looked at the Hint again. Then I remembered in the book early on it said "If the voltage increases across a resistor, current will increase."
So question is: When solving problems I have to increase (or decrease) current (I) every time voltage (V) is increased (decreased) in a problem, right? How about the other way around, when increasing current (I), you need to increase voltage (V). I'm pretty sure that's how they got 4 W, but want to make sure before I head to the next section of the book.
P = IV = (200mA)(2V) = 4 W