Answer:
2.68 hours
Explanation:
A.) Suppose the wind blows out from the west (with the air moving east). The pilot should then head her plane to northwest direction to move directly north.
B.) Given that plane flies at a speed of 102 km/h in still air. And the wind blows out from the west (with the air moving east) at a speed of 46 km/h.
The plan resultant speed can be calculated by using pythagorean theorem.
Resultant Speed = Sqrt( 102^2 + 46^2 )
Resultant Speed = Sqrt( 12520)
Resultant speed = 111.89 km/h
From the definition of speed,
Speed = distance/time
Where distance = 300 km
Substitute the resultant speed and the distance into the formula.
111.89 = 300/time
Time = 300/111.89
Time = 2.68 hours
Therefore, it take her 2.68 hours to reach a point 300 km directly north of her srarting point
The correct option is (B) <span>Aluminum is a metal and is shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.
Since Aluminium is in group 13, and all the elements in group 13 are either metals or metalloids(Boron). Hence we are left with option (B) and (D). Boron is the only metalloid in group 13 and aluminium is a metal(not a metalloid); therefore, we are left with only one option which is Option (B). And Aluminium is </span>shiny, malleable, ductile, conducts heat and electricity, forms basic oxides, and forms cations in aqueous solution.<span>
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Since the ball was not moving before it let Aiden's hand, the formula used to calculate the acceleration is
, where a is acceleration, v is velocity and t is the time. We put them in the formula and get
The acceleration is 490 m/s^2
Answer:
Charge on each is 2 x 10⁻¹⁰.
Explanation:
We know that Force between two point charges is given b the Coulomb's law as:
F = kq₁q₂/r^2
k = 9 x 10^9
r = 3.00 cm
= 0.03 m
q₁ = q₂
F = 4.00 x 10^-7
Rearranging the formula, we get:
F = k q²/r²
q² = Fr²/k
q² = 4 x 10⁻⁷ x 0.03²/(9x10⁹)
q² = 4 x 10⁻²⁰
q = 2 x 10⁻¹⁰
As there is force of repulsion between the charges, the charges must be both positive or both negative.
A type O star is likely to appear blue.
