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3 years ago

How is the lifetime of a star related to its mass?

1 answer:

6 0

Stars having less mass collapses early than those with more mass. This can be explained by Einstein's equation E=mc².
According to this equation, mass of stars is converted into light due to thermonuclear reactions occuring in the core of star which acts as engine of the stars. This thermonuclear reactions keeps star alive. Thermonuclear reactions occurs slowly in massive stars hence massive stars live more than light stars.

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what are the common elements that make up series and parallel circuits , and how are circuits diagrammed?

Ahat [919]

<h2>Answer:</h2>

A series circuit occurs when the elements are connected along a simple path so the same current flows through all the elements. On the other hand, a parallel circuit occurs when there are two or more paths for the electricity to flow. The diagram are shown in the Figure below. We have chosen a source and resistors to illustrate this problem.

3 0

3 years ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s

for their combination, the rotational momentum is

$(I_{1} +I_{2} )w$

where the subscripts 1 and 2 indicates the values first and second flywheels

$(I_{1} +I_{2} )w$ = (6.655 + 62.72)ω

where ω here is their final angular momentum together

==> 69.375ω

Equating the two rotational momenta, we have

211.76 = 69.375ω

ω = 211.76/69.375 = 3.05 rad/s

Therefore, the energy stored in the first flywheel in this situation is

E = $Iw^{2}$ = 6.655 x $3.05^{2}$ = 61.908 J

c) one third of the initial energy of the flywheel is

6738.27/3 = 2246.09 J

For the car, the kinetic energy = $\frac{1}{2}mv_{car} ^{2}$

where m is the mass of the car

$v_{car}$ is the velocity of the car

Equating the energy

2246.09 = $\frac{1}{2}mv_{car} ^{2}$

making m the subject of the formula

mass of the car m = $\frac{4492.18}{v_{car} ^{2} }$

3 0

3 years ago

A key falls from a bridge that is 45 m above the water. the key falls straight down and lands in a model boat traveling at a con

erastova [34]

Let the key is free falling, therefore from equation of motion

$h = ut +\frac{1}{2}gt^2.$ .

Take initial velocity, u=0, so

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2$ .

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }$

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

$d= v \times t$

From above substituting t,

$d = v \times \sqrt{\frac{2h}{g} }$ .

Now substituting all the given values and g = 9.8 m/s^2, we get

$d = 3.5 \ m/s \times \sqrt{\frac{2 \times 45 m}{9.8 m/s^2} } = 10.60 m$ .

Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.

7 0

3 years ago

A 70.0 kg sprinter starts a race with an acceleration of 1.60 m/s^2, What is the net external force (in N) on him? (Enter the ma

Ierofanga [76]

Answer:

External force on him will be 112 N

Explanation:

We have given the mass of the sprinter m =70 kg

Acceleration of the sprinter $a=1.6m/sec^2$

We have to find the net external force

According to second law of motion force = mass ×acceleration

Force is dependent on the mass and acceleration

So $F=70\times 1.6=112 N$

So external force will be 112 N

6 0

3 years ago

What experimental evidence led scientists to change from the previous model to this one?

san4es73 [151]

The colors of light emitted from heated atoms had very specific energies.

6 0

3 years ago