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3 years ago

When a star begins to run out of fuel, what two types of stars can it become

Physics

1 answer:

Nikolay [14]3 years ago

7 0

Answer:

A red giant or red super giant then a white dwarf

Explanation:

This also depends on what the star classification was to begin with.

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Suppose you are on a cart, initially at rest, which rides on a frictionless horizontal track. You throw a ball at a vertical sur

Len [333]

Answer:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

Explanation:

If we assume that we have the situation in the figure attached.

For this case we assume that the momentum changes are equal in magnitude and opposite in direction, so then we satisfy this:

$F_c t_ c = -F_b t_b$

And the forces are equal but in the opposite direction. So then we can write by general rule:

$m_c \Delta V_{c} = -m_b \Delta V_b$

Or equivalently:

$m_c \Delta V_{c} +m_b \Delta V_b =0$

Where: $V_c$ represent the speed of the car and $V_b$ the speed of the ball

$m_c$ represent the mass of the car

$m_b$ represent the mass of the ball

Since the ball is moving to the left and we assume that the total momentum not changes then the car need to move to the right in order to satisfy the equation and satisfy the balance.

By conservation of the momentum the car will move to the right since the ball is moves to the left.

So then the correct option for this case is :

A.Yes, and it moves to the right.

3 0

3 years ago

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0

3 years ago

Near the surface of Earth an electric field points radially downward and has a magnitude of approximately 100 N/C. What charge (

pentagon [3]

Answer:

$q=2.997\times 10^{-4}$ C

Sign-Negative

Explanation:

We are given that

Electric field = $E=100NC^{-1}$ (Radially downward)

Acceleration= $0.19 ms^{-2}$ (Upward)

Mass of charge=3 g= $3\times 10^{-3}$ kg

1kg=1000g

We have to find the magnitude and sign of charge would have to be placed on a penny .

By newton's second law

$\sum F_y=ma$

$\sum F_y=qE-mg$

Substitute the values then we get

$qE-mg=ma$

Substitute the values then we get

$q(100)-3\times 10^{-3}(9.8)=3\times 10^{-3}(0.19)$

$100q-29.4\times 10^{-3}=0.57\times 10^{-3}$

$100q=0.57\times 10^{-3}+29.4\times 10^{-3}=29.97\times 10^{-3}$

$q=\frac{29.97\times 10^{-3}}{100}$

$q=2.997\times 10^{-4}$ C

Sign of charge =Negative

Because electric force acting in opposite direction of electric field therefore,charge on penny will be negative.

4 0

3 years ago

Which line represents a stationary object?

sineoko [7]

Answer:

I think its B but I may be wrong

3 0

3 years ago

Read 2 more answers

When must a psychological researcher debrief human test subjects?

myrzilka [38]

<span>Psychological researchers must debrief human test subjects </span><span>at the end of every experiment.

The current code of ethics in p</span>sychological research states that researchers absolutely must debrief human test subjects at the end of every study regardless or whether or not harm or deception was involved.

Debriefing a subject after a study is an essential opportunity for the researcher to explain the purpose and aim of the study to the subject, make sure the subject is not harmed or mentally disturbed, clarify why deception was used (if deception was involved) and overall, to clarify any questions or doubts the subject might have.

3 0

3 years ago