Answer:
Two stars, each of mass M, form a binary system. ... used is the distance between the centers of the planets; here that distance is 2R. ... r appears in the denominator of Newton's law of gravitation, the force of ... The orbital speed of a satellite orbiting the earth in a circular orbit at the ... is undergoing uniform circular motion?
Explanation:
Two stars, each of mass M, form a binary system. ... used is the distance between the centers of the planets; here that distance is 2R. ... r appears in the denominator of Newton's law of gravitation, the force of ... The orbital speed of a satellite orbiting the earth in a circular orbit at the ... is undergoing uniform circular motion?
Move from higher to lower energy levels
Substances that move to the stronger parts of a magnetic field are termed paramagnetic substances; the atomic feature responsible for this property is presence of unpaired electrons in atoms.
<h3>What is a paramagnetic substance?</h3>
A paramagnetic substance is the substance that possess unpaired electrons that are heavily attracted in a magnetic field.
A magnetic field is defined as the field that exists around a magnet that produces a field of force.
Examples of paramagnetic substance include the following:
- aluminum,
- gold,
- copper.
- Chromium, and
- Manganese.
These substances are known as paramagnetic substances because they possess a high number of unpaired electrons.
Other properties of a paramagnetic substance include the following:
- They have a permanent dipole moment or permanent magnetic moment.
- They are weakly magnetized in the direction of the magnetizing field.
- They usually have constant relative permeability (μr) slightly greater than 1.
Therefore, Substances that move to the stronger parts of a magnetic field are termed paramagnetic substances; the atomic feature responsible for this property is presence of unpaired electrons in atoms.
Learn more about magnets here:
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Answer:
a = - 0.248 m/s²
Explanation:
Frictional drag force
F = ½ *(ρ* v² * A * α)
ρ = density of air , ρ = 1.295 kg/m^3
α = drag coef , α = 0.250
v = 100 km/h x 1000m / 3600s
v = 27.77 m/s
A = 2.20m^2
So replacing numeric in the initial equation
F = ½ (1.295kg/m^3)(27.77m/s)²(2.30m^2)(0.26)
F = 298.6 N
Now knowing the force can find the acceleration
a = - F / m
a = - 298.6 N / 1200 kg
a = - 0.248 m/s²
