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katrin [286]
3 years ago
5

When triggers nuclear fusion in stars

Physics
1 answer:
Brut [27]3 years ago
5 0
Fusion in stars are usually the result of gravity. Once a mass of hydrogen accumulates enough mass, the gravity of all that mass compresses the core of the star to the point that the hydrogen atoms there begin fusing into helium. The <span>process then cascades outward, and the end result is a star. Hopes ths Helps. Also mark me brainliest since I was first

</span>
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A 2.35 kg ball is attached to a ceiling by a3.53 m long string. The height of the room is5.03 m .The acceleration of gravity is
Studentka2010 [4]

Answer:-81.29 J

Explanation:

Given

mass of ball m=2.35 kg

Length of string L=3.53 m

height of Room h=5.03 m

Gravitational Potential Energy is given by

P.E.=mgh

where h=distance between datum and object

here Reference is ceiling

therefore h=-3.53 m

Potential Energy of ball w.r.t ceiling

P.E.=2.35\times 9.8\times (-3.53)=-81.29 J

i.e. 81.29 J of Energy is required to lift a ball of mass 2.35 kg to the ceiling    

3 0
3 years ago
Define the term mid-ocean ridge in simple, easy-to-understand words. Thanks :)
Likurg_2 [28]
According to wikipedia <em>a mid-ocean ridge is an underwater mountains system formed by tectonic plates.

</em>Happy studying!<em>
</em>
6 0
4 years ago
A tall cylinder with a cross-sectional area 13.0 cm2 is partially filled with mercury; the surface of the mercury is 6.00 cm abo
kvv77 [185]

Answer:

V = 1060.8 cm³

Explanation:

we know that the pressure,

  P = density  x  gravity  x  depth

ρm is the density of mercury

ρw is the density of water

the pressure due to mercury P₁= (ρm) g h₁

the pressure due to water P₂ = (ρw) g h₂

the total pressure

P = P₁ + P₂

P =  (ρm) g h₁  + (ρw) g h₂

but the total pressure is double the pressure due to mercury.  

(ρm) g h₁ =(ρw) g h₂

h_2 = \dfrac{\rho_m\times h_1}{\rho_w}

h_2 =13.6\times 6

h₂ = 81.6 cm

the height of the water is 81.6 cm

the volume

V = height  x area

V = 81.6 x 13

V = 1060.8 cm³

the volume of water must be added to double the gauge pressure is V = 1060.8 cm³

8 0
3 years ago
What additional load will be required to cause the extension of 2.0cm when an elastic wire extend by 1.0cm when a load of 20g ra
xxTIMURxx [149]

Answer:

The additional load is 20g

5 0
3 years ago
A charged particle that is moving in a static uniform magnetic field
Studentka2010 [4]

Answer:

may experience a magnetic force, but its speed will not change.

Explanation:

A charged particle moving in a static uniform magnetic field experiences a magnetic force (called Lorentz force) which is given by:

F=qvB sin \theta

where

q is the charge

v is the speed of the particle

B is the magnetic field intensity

\theta is the angle between the direction of the motion of the charge and the direction of the magnetic field

We can notice that when \theta=0 (so, when the particle is moving parallel to the field), the magnetic force is zero, so the particle does not experience any force. This means that we can immediately exclude the following choice:

- will always experience a magnetic force, regardless of its direction of motion.

Moreover, the magnetic force acts perpendicular to the direction of motion of the particle. This means it will change the direction of motion of the particle, so we can also exclude the following option

- may experience a magnetic force, but its direction of motion will not change.

Finally, the fact that the magnetic force is perpendicular to the direction of motion of the particle also implies that the force does no work on the particle. According to the work-energy theorem (which states that the kinetic energy gained by the particle is equal to the work done on it), this implies that the particle gains no kinetic energy, so its speed does not change. This allows us to exclude also the following choice

- a magnetic force which will cause its speed to change.

Therefore, the correct option is the only one remained:

- may experience a magnetic force, but its speed will not change.

3 0
4 years ago
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