Answer:
Explanation:
String theory proposes that the fundamental constituents of the universe are one-dimensional “strings” rather than point-like particles. String theory also requires six or seven extra dimensions of space, and it contains ways of relating large extra dimensions to small ones. In statistical mechanics, entropy is an extensive property of a thermodynamic system. It quantifies the number Ω of microscopic configurations that are consistent with the macroscopic quantities that characterize the system theyre related It later developed into superstring theory, which posits a connection called supersymmetry between bosons and the class of particles called fermions. Five consistent versions of superstring theory were developed before it was conjectured in the mid-1990s that they were all different limiting cases of a single theory in 11 dimensions known as M-theory. In late 1997, theorists discovered an important relationship called the AdS/CFT correspondence, which relates string theory to another type of physical theory called a quantum field theory.
Answer:
The ratio of their orbital speeds are 5:4.
Explanation:
Given that,
Mass of A = 5 m
Mass of B = 7 m
Radius of A = 4 r
Radius of B = 7 r
The orbital speed of satellite A,
......(I)
The orbital speed of satellite B,
......(I)
We need to calculate the ratio of their orbital speeds
Using equation (I) and (II)

Put the value into the formula


Hence, The ratio of their orbital speeds are 5:4.
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
= / R
² = mR
ω
²
ω
² = GM
= / R
³
ω
= √(GM
= / R
³)
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
Answer: D. The elements have the same number of valence electrons
Explanation: The chemical reactivity of elements is governed by the valence electrons present in the element.
The elements present in the same group or vertical column have similar valence configurations and thus behave similarly in chemical reactions or have similar bonding properties.
For Example: Both fluorine and chlorine belong to same family or group and both have 7 electrons in their valence shell and thus accept single electron to attain noble gas configuration.




thus both would bond with a cation bearing a single positive charge.
Answer:
The mass and velocity for kinetic energy. Potential Energy: How high an object is and the mass in kilograms or it is the weight in and how high an object is. There are two formulas to calculate potential energy, but the one with grams is used more often.
Explanation:
Hope this helps!