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Vesnalui [34]
3 years ago
9

The final stages of a star's life depend on its mas. true or false?​

Physics
1 answer:
Diano4ka-milaya [45]3 years ago
5 0

Answer:

TRUE

Explanation:

Low mass stars last lots longer.

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_____ contains chemical energy.<br> A. Heat<br> B. Light<br> C. Natural gas
FinnZ [79.3K]

The answer is C. Natural gas because both A and B are their own forms of energy.

5 0
3 years ago
Read 2 more answers
At the end of a race a runner decelerates from a velocity of 8.90 m/s at a rate of 1.70 m/s2. (a) How far in meters does she tra
Ad libitum [116K]

Answer:

x=22.33m

Explanation:

Kinematics equation for constant deceleration:

x =v_{o}*t - 1/2*at^{2}=8.9*6.3-1/2*1.70*6.3^{2}=22.33m

7 0
3 years ago
An astronaut has a momentum of 280 kg and travels 10 m/s. what is the mass of the astronaut?
Kamila [148]

Answer:

The answer is

<h2>28 kg</h2>

Explanation:

The mass of an object given it's momentum and velocity / speed can be found by using the formula

m =  \frac{p}{v}  \\

where

m is the mass

p is the momentum

v is the speed or velocity

From the question

p = 280 kg/ms

v = 10 m/s

The mass of the object is

m =  \frac{280}{10}  = 28 \\

We have the final answer as

<h3>28 kg</h3>

Hope this helps you

3 0
3 years ago
A solid disk of radius 1.4 cm and mass 430 g is attached by a wire to one of its circular faces. It is twisted through an angle
Neporo4naja [7]

Answer:

    f= 4,186  10²  Hz

Explanation:

El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por

             w = √ k/I

donde ka es constante de torsion de hilo e I es el momento de inercia del disco

El  momento de inercia de indican que giran un eje que pasa                 por enronqueces

           I= ½ M R2  

reduzcamos las cantidades al sistema SI

         R= 1,4 cm = 0,014  m

         M= 430 g = 0,430 kg

substituimos

           w= √ (2 k/M R2)

calculemos  

           w = RA ( 2 370 / (0,430  0,014 2)

           w = 2,963 103 rad/s

la velocidad angular esta relacionada con la frecuencia por

            w =2pi f

            f= w/2π

            f= 2,963 10³/ (2π)

            f= 4,186  10²  Hz

5 0
3 years ago
A very small sphere with positive charge 5.00uC is released from rest at a point 1.20cm from a very long line of uniform linear
JulijaS [17]
Let us situate this on the x axis, and let our uniform line of charge be positioned on the interval <span>(−L,0]</span> for some large number L. The voltage V as a function of x on the interval <span>(0,∞)</span> is given by integrating the contributions from each bit of charge. Let the charge density be λ. Thus, for an infinitesimal length element <span>d<span>x′</span></span>, we have <span>λ=<span><span>dq</span><span>d<span>x′</span></span></span></span>.<span>V(x)=<span>1/<span>4π<span>ϵ0</span></span></span><span>∫line</span><span><span>dq/</span>r</span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>∫<span>−L</span>0</span><span><span>d<span>x/</span></span><span>x−<span>x′</span></span></span>=<span>λ/<span>4π<span>ϵ0</span></span></span><span>(ln|x+L|−ln|x|)</span></span>
5 0
3 years ago
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