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ozzi
3 years ago
7

What causes fusion to stop within all stars?

Physics
1 answer:
puteri [66]3 years ago
4 0
Hello!

When hydrogen runs out in a star, it can no longer perform nuclear fusion where hydrogen nuclei combine to create helium. This will stop fusion.

Hope this helps!
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a racing car undergoes a uniform acceleration of 4.00m/s2. if the net force causing the acceleration is 3.00 times 10^3 N, what
yKpoI14uk [10]

Answer: 750Kg

Explanation:

Recall that force is the product of the mass M, of an object moving at a uniform acceleration.

i.e Force = Mass x Acceleration

In this case, Mass = ?

Force = 3.00 x 10^3 N = (3.00 x 1000N)

= 3000N

Uniform acceleration = 4.00m/s^2

Force = Mass x Acceleration

3000N = Mass x 4.00m/s^2

Mass = (3000N/4.00m/s^2)

Mass = 750Kg (The SI unit of mass is kilograms)

Thus, the mass of the car is 750Kg

4 0
3 years ago
Which statement is true about alkali metals? A. Some of them explode when exposed to water. B. They are in group 18 of the perio
WINSTONCH [101]
The Answer To This Is A , The Most Reactive Of The Alkali Explode Due To Reactions To Water

5 0
4 years ago
Read 2 more answers
A series rlc circuit is in resonance with 600 hz. by what factor must the capacitance be multiplied change the resonance frequen
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Fvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
6 0
4 years ago
Calculate the mass of an object that has a momentum of 100kg x m/sec and velocity of 4 m/sec
lidiya [134]
P= mv

where p is momentum
m is mass
v is velocity

so it's given p= 100kgm/sec
v= 4m/s

so putting in the formula

100= m × 4

m = 25kg
3 0
3 years ago
Sam, whose mass is 75 kg, straps on his skis and starts down a 50-m-high, 20 frictionless slope. A strong headwind exerts a hori
LenaWriter [7]

Answer:

Explanation:

Sam mass=75kg

Height is 50m

20° frictionless slope

Horizontal force on Sam is 200N

According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.

Therefore

Wg - Ww =∆K.E

Note initial the body was at rest at top of the slope.

Then, ∆K.E is K.E(final) - K.E(initial)

K.E Is given as ½mv²

Since initial velocity is zero then, K.E(initial ) is zero

Therefore, ∆K.E=½mVf²

Wg is work done by gravity and it is given by using P.E formulas

Wg=mgh

Wg=75×9.8×50

Wg=36750J

Ww is work done by wind and it's is given by using formulae for work

Work=force × distance

Ww=horizontal force × horizontal distance

Using Trig.

TanX=opposite/adjacent

Tan20=h/x

x=h/tan20

x=50/tan20

x=137.37m

Then,

Ww=F×x

Ww=200×137.37

We=27474J

Now applying the formula

Wg - Ww =∆K.E

36750 - 27474 =½×75×Vf²

9276=37.5Vf²

Vf²=9275/37.5

Vf²= 247.36

Vf=√247.36

Vf=15.73m/s

3 0
4 years ago
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