Answer:
B) The total number of atoms does not change, so mass is conserved in the reaction.
Explanation:
According to the <u><em>Law of Conservation of Mass</em></u>, no matter can ever be created or destroyed. The number of atoms does not change, thus the mass does not change.
The electric field between plates is 4000V/m.
An electric field (sometimes E-field) is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them. It also refers to the physical field for a system of charged particles.
The value of the electric field has dimensions of force per unit charge. In the metre-kilogram-second and SI systems, the appropriate units are newtons per coulomb, equivalent to volts per metre.
The voltage between points A and B is
V=E.d
E =V/d (uniform E- field only)
where d is the distance from A to B, or the distance between the plates.
Given:
distance d = 3 cm
voltage V = 120 V
Electric field E = V/d
E = 120 V / 3cm
E = 40 V / 1 cm [ 1 cm = 1/100 m ]
E = 4000 V/m.
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Answer:
The required angular speed the neutron star is 10992.32 rad/s
Explanation:
Given the data in the question;
mass of the sun M
= 1.99 × 10³⁰ kg
Mass of the neutron star
M
= 2( M )
M = 2( 1.99 × 10³⁰ kg )
M = ( 3.98 × 10³⁰ kg )
Radius of neutron star R = 13.0 km = 13 × 10³ m
Now, let mass of a small object on the neutron star be m
angular speed be ω.
During rotational motion, the gravitational force on the object supplies the necessary centripetal force.
GmM = / R² = mRω²
ω² = GM = / R³
ω = √(GM = / R³)
we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²
we substitute
ω = √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)
ω = √( 2.65466 × 10²⁰ / 2.197 × 10¹²
ω = √ 120831133.3636777
ω = 10992.32 rad/s
Therefore, The required angular speed the neutron star is 10992.32 rad/s
Explanation:
W = PE
W = mgh
1500 J = (20 kg) (9.8 m/s²) h
h = 7.65 m
Round as needed.
