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4 years ago

A main sequence star cools and expands. It must be entering which stage?

1 answer:

4 0

If a main sequence star is cooling and expanding, it is entering the red giant stage. This means that the star has burned up all of it's hydrogen and is now starting to burn its helium making it cooler.

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(a) A load of coal is dropped (straight down) from a bunker into a railroad hopper car of inertia 3.0 × 104 kg coasting at 0.50

Firlakuza [10]

Answer:

a) m=20000Kg

b) v=0.214m/s

Explanation:

We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, $p=m_Av_A=m_Bv_B=m_Cv_C$ , were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as $m_h$ , and the mass of the first and second coals as $m_1$ and $m_2$ respectively

We start with the transition between parts A and B, so we have:

$m_Av_A=m_Bv_B$

Which means

$m_hv_A=(m_h+m_1)v_B$

And since we want the mass of the first coal thrown ( $m_1$ ) we do:

$m_hv_A=m_hv_B+m_1v_B$

$m_hv_A-m_hv_B=m_1v_B$

$m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}$

Substituting values we obtain

$m_1=\frac{(3\times10^4Kg)(0.5m/s-0.3m/s)}{0.3m/s}=20000Kg=2\times10^4Kg$

For the transition between parts B and C, we can write:

$m_Bv_B=m_Cv_C$

Which means

$(m_h+m_1)v_B=(m_h+m_1+m_2)v_C$

Since we want the new final speed of the car ( $v_C$ ) we do:

$v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}$

Substituting values we obtain

$v_C=\frac{(3\times10^4Kg+2\times10^4Kg)(0.3m/s)}{(3\times10^4Kg+2\times10^4Kg+2\times10^4Kg)}=0.214m/s$

5 0

3 years ago

a bowling ball of mass 7.3 kg and radius 9.6 cm rolls without slipping down a lane at 3.1 m/s . part a calculate its total kinet

erma4kov [3.2K]

49 J is the total kinetic energy. If a bowling ball of mass 7.3 kg and radius 9.6 cm rolls without slipping down a lane at 3.1 m/s. Kinetic energy is the energy an bowling ball has because of its motion.

Given: m = 7.3 Kg ; r = 9.4 cm = 0.094 m ; v = 3.1 m

Now total kinetic energy in this case is given by KE = Kinetic energy due to rotation + Kinetic energy due to translation

i,e KE = 1/2*m*v2 + 1/2*I*ω2 where I is the moment of inertia of the bowling ball about it's center and ω is the angular velocity

Now for pure rotation (without slipping) v = rω

also for the ball (solid sphere) I = 2/5*m*r2

Hence our kinetic energy becomes

KE = 1/2*m*v2 + 1/5*m*v2 = 7/10*m*v2

so KE = 0.7*7.3*(3.1)2 = 49.10 J = 49 J

Learn more about kinetic energy here

brainly.com/question/12669551

#SPJ4

5 0

1 year ago

A 5.0 kg block hangs from a spring with spring constant 2000 N/m. The block is pulled down 5.0 cm from the equilibrium position

Svet_ta [14]

Answer:

Explanation:i think this would help u

7 0

3 years ago

Julietta and Jackson are playing miniature golf. Julietta's ball rolls into a long. Straight upward incline with a speed of 2.95

Verdich [7]

Answer:

The length of the incline is 3.504 meters.

Explanation:

Let suppose that Julietta's ball decelerates uniformly, then we determine the length of the incline is determined by the following equation of motion:

$\Delta s = v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (Eq. 1)