Dispersion im pretty sure
three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650
Answer:
T = 1.2 s
T = 15.1 m = 15 m
Explanation:
This is a case of projectile motion:
TOTAL TIME OF FLIGHT:
The formula for total time of flight in projectile motion is:
T = 2 V₀ Sinθ/g
where,
T = Total Time of Flight = ?
V₀ = Launch Speed = 13.9 m/s
θ = Launch Angle = 25°
g = 9.8 m/s²
Therefore,
T = (2)(13.9 m/s)(Sin 25°)/(9.8 m/s²)
<u>T = 1.2 s</u>
<u></u>
RANGE OF BALL:
The formula for range in projectile motion is:
R = V₀² Sin2θ/g
where,
R = Horizontal Distance Covered by ball = ?
Therefore,
T = (13.9 m/s)²(Sin 2*25°)/(9.8 m/s²)
<u>T = 15.1 m = 15 m</u>
Answer:
Θ
Θ
Θ = 
Explanation:
Applying the law of conservation of momentum, we have:
Δ
Θ (Equation 1)
Δ
Θ (Equation 2)
From Equation 1:
Θ
From Equation 2:
sinΘ = 
Replacing Equation 3 in Equation 4:
Θ (Equation 5)
And we found Θ from the Equation 5:
tanΘ=
Θ=
The distance covered by Arthur is
3t
The distance covered by Betty from where Arthur started is
100 - 2t
They meet each other when this distance is equal
3t = 100 - 2t
t = 20 s
Since the dog has a constant speed of 5m/s, the distance that Spot has run is
5 m/s (20s) = 100 m
