Explanation:
In order to find out if the keys will reach John or not, we can use the formula of projectile motion to find the maximum height reached by the keys:
H = V²Sin²θ/2g
where,
V = Launch Speed = 18 m/s
θ = Launch Angle = 40°
g = 9.8 m/s²
Therefore,
H = (18 m/s)²[Sin 40°]²/(2)(9.8 m/s²)
H = 6.83 m
Hence, the maximum height that can be reached by the projectile or the keys is greater than the height of John's Balcony(5.33 m).
Therefore, the keys will make it back to John.
Answer:
They both have the same angular speed.
Explanation:
The mathematical formula for angular speed is:
where is angular speed, is a constant, and is the period (the time it takes the marry-go-round to complete a lap).
What we can see from the formula is that, since the does not change its value, the angular speed depends only on the period T.
In this case for both the children closer to the outher edge and for the children closer to the center, the time to complete a lap is the same, because the time does not depend on where they are sitting in the marry go round. This means that the period for both is the same.
Thus, since the period for both is the same, the angular speed given by
will also be the same
I think it’s speed will increase, if I understood it correctly
Answer:
it creates a gas called carbon dioxide. The gas begins to expand in the bottle and starts to inflate the balloon
Explanation:
Why does this happen? well, The faster-moving particles inside the bottle start to move faster and faster and soon they expand to fill the balloon.
Answer:
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference
Explanation:
For this exercise let's use the electric field expression
E = k q / r²
where k is the Coulomb constant that is equal to 9 109 N m² /C², q the charge and r the distance to the point of interest positive test charge, in this case the distance to the bee
let's calculate the field for each charge
Q = 24 pC = 24 10⁻¹² C
E₁ = 9 10⁹ 24 10⁻¹² / 0.20²
E₁ = 5.4 N / C
Q = 32 pC = 32 10⁻¹² C
E₂ = 9 10⁹ 32 10⁻¹² / 0.2²
E₂ = 7.2 N / C
let's find the difference between these two fields
ΔE = E₂ -E₁
ΔE = 7.2 - 5.4
ΔE = 1.8 N / C
the minimum detection field is
E_minimum = 0.77 N / C
ΔE> E_minimo
We see that the field difference between these two flowers is greater than the minimum field, so the bee knows if it has been recently visited, so the answer is if it can detect the difference