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Akimi4 [234]
3 years ago
13

Imagine two starts are exactly the same size, but one of the stars to greater mass than the other.which star will burn out all o

f its fuel first
Physics
1 answer:
Neko [114]3 years ago
4 0

Answer:

The larger star will burn out faster. as nuclear fuel is burned out due to higher percentage being consumed.

Explanation:

Gravity on the other hand is only affected a little during this process and mass use. Solar mass loss is caused by mass that is turned into energy (E=mc²), but also by the material that is lost to the solar wind.

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Please help!
Kisachek [45]

B. Inelastic collision.

In elastic collision , both momentum and kinetic energy are conserved while in inelastic collision only momentum is conserved. there is some loss of energy in inelastic collision during collision.

During the collision of bat with baseball, some energy gets lost to heat and sound. hence the kinetic energy is not conserved although the momentum is conserved.

3 0
3 years ago
Read 2 more answers
Having greater biodiversity in an ecosystem is beneficial to humans because
11Alexandr11 [23.1K]

Answer:

The answer is A

Explanation:

4 0
3 years ago
Read 2 more answers
Which of the following situations would produce and average velocity of zero
enyata [817]
There are no situations listed. Try adda picture or more text for your question to be answered.
3 0
4 years ago
Building codes usually limit the current carried by a No. 14 copper wire to 15 A. Many household circuits are wired with this si
Sedbober [7]

Answer:

The drift velocity of the electrons in this case is c) 5.52 x 10-4 m/s

Explanation:

Hi

First of all, we need to find the volume occupied by 63.3g of copper, so V=\frac{m}{\rho}= \frac{0.0633Kg}{8,900kg/m^{3}} =7.11 \times 10^{-6} m^{3}, then if we assume each atom of copper contributes with one free electron to the material body n=\frac{N_{A}}{V}=\frac{6.02 \times 10^{23} electrons}{7.11 \times 10^{-6} m^{3}} =8.46 \times 10^{28} electrons/m^{3}.

Finally, we apply v_{d}=\frac{I}{nqA}, where A is area so, A=\pi (\frac{d}{2})^{2}= \pi (\frac{0.0016m}{2})^{2}=2 \times 10^{-6} m^{2}, thus v_{d}=\frac{I}{nqA}=\frac{15C/s}{(8.46 \times 10^{28} m^{-3})(1.60 \times 10^{-19} C)(2 \times 10^{-6} m^{2})}=5.50 \times 10^{-4} m/s.

As we can see c. 5.52 x 10-4 m/s is the nearest one.

3 0
3 years ago
At the instant the traffic light turns green, an automobile that has been waiting at an intersection starts ahead with a constan
Aloiza [94]

Answer

given,

Initial speed of the car = 0 m/s

acceleration of car = 2.4 m/s²

Speed of the truck = 15.5 m/s

to car to overtake the truck distance travel by both should be same

distance traveled by the truck

d = 15.5 t

distance traveled by the car

d = u t + 1/2 gt ²

d = 0.5 x 9.8 t²

now, equating both the equations

15.5 t = 4.9 t²

t = 3.163 s

Time taken by car to overtake truck is 3.163 s

distance travel by the car = 15.5 x 3.163

d = 49 m

b) Speed of the automobile

using equation of motion

    v = u + at

    v = 0 + 2.4 x 3.163

    v = 7.59 m/s

Speed of the automobile while crossing truck is equal to v = 7.59 m/s

4 0
3 years ago
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