Can someone help me asap this is due at 2 50

The hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would pro

Gnesinka [82]

The hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would produce is apparently as a result of Eddington luminosity

<h3>What are stars?</h3>

Stars are a fixed luminous point in the sky which is a large and remote incandescent body

So therefore, the hypothetical upper limit to the mass a star can be before it self-destructs due to the massive amount of fusion it would produce is apparently as a result of Eddington luminosity

Learn more about stars:

brainly.com/question/13018254

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4 0

1 year ago

What does a plant need to take in from environment to live?

Maksim231197 [3]

Answer:

Carbon Dioxide.

Explanation:

8 0

2 years ago

Read 2 more answers

What is a constellation as astronomers define it today? What does it mean when an astronomer says, “I saw a comet in Orion last

MissTica

Explanation:

Constellation: The complete sky has been divided in 88 different areas, in a way we have divided Earth in countries, not necessarily having same shapes and size. These 88 areas are known as constellations. These contains a lot of stars. When we join the brightest stars together we can imagine a shape out of them which is called as Asterism. Most of the people are unaware of this difference. Some of the famous constellations are Orion, Taurus, Gemini, Hydra, Ursa Major etc.

When an astronomer says that there is a comet is in the Orion, he means that a comet is in the boundaries of Orion constellation.

4 0

2 years ago

A racquetball strikes a wall with a speed of 30 m/s and rebounds in the opposite direction with a speed of 26 m/s. The collision

Fudgin [204]

Answer:

The average acceleration of the ball during the collision with the wall is $a=2,800m/s^{2}$

Explanation:

Known Data

We will asume initial speed has a negative direction, $v_{i}=-30m/s$ , final speed has a positive direction, $v_{f}=26m/s$ , $\Delta t=20ms=0.020s$ and mass $m_{b}$ .

Initial momentum

$p_{i}=mv_{i}=(-30m/s)(m_{b})=-30m_{b}\ m/s$

final momentum

$p_{f}=mv_{f}=(26m/s)(m_{b})=26m_{b}\ m/s$

Impulse

$I=\Delta p=p_{f}-p_{i}=26m_{b}\ m/s-(-30m_{b}\ m/s)=56m_{b}\ m/s$

Average Force

$F=\frac{\Delta p}{\Delta t} =\frac{56m_{b}\ m/s}{0.020s} =2800m_{b} \ m/s^{2}$

Average acceleration

$F=ma$ , so $a=\frac{F}{m_{b}}$ .

Therefore, $a=\frac{2800m_{b} \ m/s^{2}}{m_{b}} =2800m/s^{2}$

8 0

3 years ago

A steel cable has a cross-sectional area 4.49 × 10^-3 m^2 and is kept under a tension of 2.96 × 10^4 N. The density of steel is

Lemur [1.5K]

Answer:

The transverse wave will travel with a speed of 25.5 m/s along the cable.

Explanation:

let T = 2.96×10^4 N be the tension in in the steel cable, ρ = 7860 kg/m^3 is the density of the steel and A = 4.49×10^-3 m^2 be the cross-sectional area of the cable.

then, if V is the volume of the cable:

ρ = m/V

m = ρ×V

but V = A×L , where L is the length of the cable.

m = ρ×(A×L)

m/L = ρ×A

then the speed of the wave in the cable is given by:

v = √(T×L/m)

= √(T/A×ρ)

= √[2.96×10^4/(4.49×10^-3×7860)]

= 25.5 m/s

Therefore, the transverse wave will travel with a speed of 25.5 m/s along the cable.

7 0

3 years ago