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2 years ago

Compare the events in the life of the Sun with those of a star that starts with less

Physics

1 answer:

GarryVolchara [31]2 years ago

6 0

A star with greater mass will die out faster than the Sun.

<h3>What factors star is dependent on?</h3>

A star's future relies upon its mass. For the most part, the more huge the star, the quicker it consumes its fuel supply, and the more limited its life. The most huge stars can wear out and detonate in a cosmic explosion after two or three million years of combination.

Our Sun is a typical estimated star: there are more modest stars and bigger stars, even up to multiple times bigger. Numerous other planetary groups have different suns, while our own simply has one. The Sun is made for the most part out of hydrogen and helium gas.

In this manner, one correlation in the occasions in the existence of the Sun with those of a star that beginnings with a mass multiple times more prominent than the Sun's is a star that has a more noteworthy mass will vanish quicker.

Learn more about Star.

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Describe a good level of body fat

m_a_m_a [10]

Answer: a good level of body fat can be found using your weight and height as a reference.

Explanation:

6 0

4 years ago

A student pushed a 100 N bicycle over a distance of 15 m in 5 s. calculate the power generated.

Umnica [9.8K]

The catch in this one is: We don't know how much <u>force</u> the student used to push the bike.

It wasn't necessarily the 100N. That's just the weight of the bike. But you know that you can push a car, a wagon, or a bicycle hard, you can push it not so hard, you can give it a little push, you can give it a big push, you can push it strong, you can push it weak, you can push it medium. The harder you push, the more it'll accelerate, but it's completely up to you how hard you want to push. That's what's so great about wheels ! That's why they were such a great invention ! This is where I made my biggest mistake. This guy came into my store one day and said he's got this great invention, it's definitely going to take off, it'll be a winner for sure, he called it a "wheel". I looked at it, I turned it over and I looked on all sides. I thought it was too simple. I didn't know then it was elegant. I threw him out. I was so dumb. I could have invested money in that guy, today I would have probably more than a hundred dollars.

Anyway, can we figure out how much force the student used to push with ? Stay tuned:

-- The bike covered 15 meters in 5 seconds. Its average speed during the whole push was (15m/5s) = 3 meters/sec.

-- If the bike started out with no speed, and its average speed was 3 m/s, then it must have been moving at 6 m/s at the end of the push.

-- If its speed increased from zero to 6 m/s in 5 seconds, then its acceleration was (6m/s / 5 sec) = 1.2 m/s²

-- The bike's weight is 100N.

(mass) x (gravity) = 100N

Bikemass = (100N) / (9.8 m/s²)

Bikemass = 10.2 kilograms

-- F = m A

Force = (mass) x (acceleration)

Force = (10.2 kg) x (1.2 m/s²)

Force = 12.24 N

-- Work = (force) x (distance)

Work = (12.24 N) x (15 m)

Work = 183.67 Joules

-- Power = (work done) / (time to do the work)

Power = (183.67 joules) / (5 seconds)

<em>Power = 36.73 watts</em>

7 0

3 years ago

Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s

Zielflug [23.3K]

Answer:

there will be collision

Explanation:

$v_{s}$ = speed of sue = 34 m/s

$v_{v}$ = speed of van = 5.20 m/s

$v_{sv}$ = speed of sue relative to van = $v_{s} - v_{v}$ = 34 - 5.20 = 28.8 m/s

$d_{s}$ = stopping distance after brakes are applied

$D$ = distance between sue and van = 160 m

$v_{f}$ = final speed of sue = 0 m/s

$a$ = acceleration = - 1.80 m/s²

Using the kinematics equation

$v_{f}^{2} = v_{o}^{2} + 2 a d_{s}$

$0^{2} = 28.8^{2} + 2 (1.80) d_{s}$

$d_{s} = 230.4$ m

Since $d_{s} < D$

hence there will be collision

7 0

4 years ago

Before hanging new William Morris wallpaper in her bedroom, Brenda sanded the walls lightly to smooth out some irregularities on

Allushta [10]

To solve the exercise it is necessary to apply the equations necessary to apply Newton's second law and the concept related to frictional force.

An angle of 30 degrees is formed on the vertical at an applied force of 2.3N

In this way the frictional force, opposite to the movement will be given by

$f_k = \mu_k N$

where,

$\mu_k =$ Kinetic friction constant

N = Normal Force (Mass*gravity)

The friction force is completely vertical and opposes the rising force of 2.3 N. The Normal force acts perpendicular to the surface (vertical) therefore corresponds to the horizontal component of the applied force.

The ascending force would be given by

$F_v = 2.3N*Cos30 = 1.99N$