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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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Read the scenario below and answer the question that follows. Jane is a psychologist. She submits a study on sleep to a prestigi

Setler79 [48]

A:the other options are examples of unethical behavior

3 0

3 years ago

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At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci

ozzi

Answer:

$r=2.4m$

Explanation:

We have to use the centripetal force equation

$Fc=\frac{mv^{2} }{r}$

we need the radious so we have to isolate "r" and we get

$r=\frac{mv^{2} }{Fc}$

replacing m=65 kg, v= 4.1 m/s and Fc=455N we get

$r=\frac{65*4.1^{2} }{455}$

$r=2.4m$

The radius of the amusement park chamber is 2.4m

4 0

3 years ago

A cannon fires a 0.2 kg shell with initial velocity vi = 9.2 m/s in the direction θ = 46 ◦ above the horizontal. The shell’s tra

Sedbober [7]

Answer:

∆h = 0.071 m

Explanation:

I rename angle (θ) = angle(α)

First we are going to write two important equations to solve this problem :

Vy(t) and y(t)

We start by decomposing the speed in the direction ''y''

$sin(\alpha) = \frac{Vyi}{Vi}$

$Vyi = Vi.sin(\alpha ) = 9.2 \frac{m}{s} .sin(46) = 6.62 \frac{m}{s}$

Vy in this problem will follow this equation =

$Vy(t) = Vyi -g.t$

where g is the gravity acceleration

$Vy(t) = Vyi - g.t= 6.62 \frac{m}{s} - (9.8\frac{m}{s^{2} }) .t$

This is equation (1)

For Y(t) :

$Y(t)=Yi+Vyi.t-\frac{g.t^{2} }{2}$

We suppose yi = 0

$Y(t) = Yi +Vyi.t-\frac{g.t^{2} }{2} = 6.62 \frac{m}{s} .t- 4.9\frac{m}{s^{2} } .t^{2}$

This is equation (2)

We need the time in which Vy = 0 m/s so we use (1)

$Vy (t) = 0\\0=6.62 \frac{m}{s} - 9.8 \frac{m}{s^{2} } .t\\t= 0.675 s$

So in t = 0.675 s → Vy = 0. Now we calculate the y in which this happen using (2)

$Y(0.675s) = 6.62\frac{m}{s}.(0.675s)-4.9 \frac{m}{s^{2} } .(0.675s)^{2} \\Y(0.675s) =2.236 m$

2.236 m is the maximum height from the shell (in which Vy=0 m/s)

Let's calculate now the height for t = 0.555 s

$Y(0.555s)= 6.62 \frac{m}{s} .(0.555s)-4.9\frac{m}{s^{2} } .(0.555s)^{2} \\Y(0.555s) = 2.165m$

The height asked is

∆h = 2.236 m - 2.165 m = 0.071 m

6 0

3 years ago

HELP ME! One airline limits the size of carry-on luggage to a volume of 40,000 cm^3. A passenger has a carry-on that has an area

EleoNora [17]

You must times the area by the volume, look at it as if the area is just one of 23 layers that makes up the volume.
1960x23=45080
so no it cannot be carried as it is 5080cm^3 over the limit

8 0

3 years ago

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Paha777 [63]

<span>Using Coulomb's law: k*(-0.3)*(-0.3)/(d^2)=19.2 D is the distance between the two negative charges</span>

8 0

3 years ago