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bonufazy [111]
3 years ago
14

Using Figure 25-2, determine how Giant stars differ from main sequence stars.

Physics
2 answers:
erastova [34]3 years ago
8 0
Main sequence stars lay on a diagonal going from upper left angle to lower right angle. Part of main sequence is our Sun.

When we observe this diagram we need to compare giant stars and main sequence stars. GIant stars are positioned to upper right when compared to main sequence. From labels on coordinate axis we can see that giant stars generaly have lower temperature than main sequence stars. They also have higher brightness and lower magnitude (meaning that they are more bright in night sky).
lora16 [44]3 years ago
6 0

Answer:

Giant stars differ from main sequence stars in having greater absolute magnitudes for the same temperatures.

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Please, I need help with this.
solmaris [256]

this is the answer :)

6 0
3 years ago
Read 2 more answers
A loaded ore car has a mass of 950 kg. and rolls on rails ofnegligible friction. It starts from rest ans is pulled up a mineshaf
stiks02 [169]

(a) 10241 W

In this situation, the car is moving at constant speed: this means that its acceleration along the direction parallel to the slope is zero, and so the net force along this direction is also zero.

The equation of the forces along the parallel direction is:

F - mg sin \theta = 0

where

F is the force applied to pull the car

m = 950 kg is the mass of the car

g=9.8 m/s^2 is the acceleration of gravity

\theta=30.0^{\circ} is the angle of the incline

Solving for F,

F=mg sin \theta = (950)(9.8)(sin 30.0^{\circ})=4655 N

Now we know that the car is moving at constant velocity of

v = 2.20 m/s

So we can find the power done by the motor during the constant speed phase as

P=Fv = (4655)(2.20)=10241 W

(b) 10624 W

The maximum power is provided during the phase of acceleration, because during this phase the force applied is maximum. The acceleration of the car can be found with the equation

v=u+at

where

v = 2.20 m/s is the final velocity

a is the acceleration

u = 0 is the initial velocity

t = 12.0 s is the time

Solving for a,

a=\frac{v-u}{t}=\frac{2.20-0}{12.0}=0.183 m/s^2

So now the equation of the forces along the direction parallel to the incline is

F - mg sin \theta = ma

And solving for F, we find the maximum force applied by the motor:

F=ma+mgsin \theta =(950)(0.183)+(950)(9.8)(sin 30^{\circ})=4829 N

The maximum power will be applied when the velocity is maximum, v = 2.20 m/s, and so it is:

P=Fv=(4829)(2.20)=10624 W

(c) 5.82\cdot 10^6 J

Due to the law of conservation of energy, the total energy transferred out of the motor by work must be equal to the gravitational potential energy gained by the car.

The change in potential energy of the car is:

\Delta U = mg \Delta h

where

m = 950 kg is the mass

g=9.8 m/s^2 is the acceleration of gravity

\Delta h is the change in height, which is

\Delta h = L sin 30^{\circ}

where L = 1250 m is the total distance covered.

Substituting, we find the energy transferred:

\Delta U = mg L sin \theta = (950)(9.8)(1250)(sin 30^{\circ})=5.82\cdot 10^6 J

8 0
3 years ago
PLEASE HELP!!!
Simora [160]
I only know what number 1. is and its Mechanical Energy.
8 0
3 years ago
TRUE OR FALSE
katen-ka-za [31]

Answer:

True

Explanation:

When an object is held higher, it has more potential energy because more energy is stored from its higher position to swing further than it would have, had it been held lower.

7 0
3 years ago
Read 2 more answers
Which statement is true regarding a chemical reaction?
liberstina [14]

Answer:

The total number of atoms does not change, so mass is conserved in the reaction.

Explanation:

If you like my answer than please mark me brainliest thanks

7 0
2 years ago
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