Answer:
v₂ = 5131.42 m/s
Explanation:
given,
radius of the planet = r₁ = 9.00×10⁶ m
mass of the satellite = m₁ = 68 Kg
orbital radius = r₁ = 8 x 10⁷ m
orbital speed = v₁ = 4800 m/s
mass of second satellite = m₂ = 84.0 kg
orbital radius = r₂ = 7.00×10⁷ m
orbital speed of second satellite = v₂ = ?
using orbital speed of satellite

so,

now,


v₂ = 5131.42 m/s
The orbital speed of second satellite is equal to v₂ = 5131.42 m/s
There are other Brainly websites for non-English users.
Omg I love Lewis structures!
You have to first identify the central atom. In this case the central atom is Carbon, and it's surrounded by two oxygen atoms.
Carbon has 4 valence electrons. Oxygen has 6 valence electrons. it will have two double bonds.
I'll take a pic of it.
Answer:
93.54 Hz
Explanation:
✓From the question, Number of harmonic frequency is 4
✓ the frequency (f₄ )= 116.5 Hz
✓harmonic frequency can be calculated using below expresion
fₙ = [ (nv)/4L]..........eqn(1)
v = speed of sound= 343 m/s
n = number of given harmonic frequency
L = Length of the rope
Using above expresion ,and substitute the values at (n=4) which is 4th harmonic frequency to find the " initial Lenght of the rope
fₙ = [ (nv)/4L]
f₄ = 4× 343 /4L
f₄ = 343 /L
L= 343 /f₄
But f₄= 116.5 Hz
L= 343/116.5= 2.944m
Hence, initial Lenght of the rope= 2.944m
We can determine the frequency of new length as ( initial Lenght of the rope + tubing Lenght)
= ( 2.944m + 0.721m )
= 3.667m
Hence, new length= 3.667m
To find the new frequency of the 4th harmonic we will use eqn(2)
f₄ = v/l ...............eqn(2)
From equation (2) If we substitute the values we have
f₄ = (343/3.667)
= 93.54 Hz
Hence, the the new frequency of the 4th harmonic is
93.54 Hz
Answer:
Explanation:
Point P is situated outside sphere . For a non conducting sphere , electric field outside sphere is given by the relation
E = k Q / d²
15000 = 9 x 10⁹ x Q / .5²
Q = 416.67 X 10⁻⁹
Maximum field will be when d = r ( radius )
maximum E = kQ / r²
= 9 x 10⁹ x 416.67 X 10⁻⁹ / .3²
= 41667 N / C