Answer:
v₀ₓ = 63.5 m/s
v₀y = 54.2 m/s
Explanation:
First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:
K.E = (0.5)(mv₀²)
where,
K.E = initial kinetic energy of projectile = 1430 J
m = mass of projectile = 0.41 kg
v₀ = launch velocity of projectile = ?
Therefore,
1430 J = (0.5)(0.41)v₀²
v₀ = √(6975.6 m²/s²)
v₀ = 83.5 m/s
Now, we find the launching angle, by using formula for maximum height of projectile:
h = v₀² Sin²θ/2g
where,
h = height of projectile = 150 m
g = 9.8 m/s²
θ = launch angle
Therefore,
150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)
Sin θ = √(0.4216)
θ = Sin⁻¹ (0.6493)
θ = 40.5°
Now, we find the components of launch velocity:
x- component = v₀ₓ = v₀Cosθ = (83.5 m/s) Cos(40.5°)
<u>v₀ₓ = 63.5 m/s</u>
y- component = v₀y = v₀Sinθ = (83.5 m/s) Sin(40.5°)
<u>v₀y = 54.2 m/s</u>
As we know,
so, let's solve for charge (q) :
time = 5 minutes = 5 × 60 seconds = 300 seconds.
hence, the charge = 60 coulombs (C)
Answer:
Magnitude of the force is 4350N
Explanation:
As the woman accelerates at a distance of 22 m to go from rest to 62.5 m / s, we can use the kinematics to find the acceleration
v² = v₀² + 2 a x
v₀ = 0
a = v² / 2x
a = 62.5²/(2 × 22)
a = 88.78m/s²
the time you need to get this speed
v = v₀ + a t
t = v / a
t = 62.5 / 88.78
t = 0.704s
Let's caculate the magnitude of the force
F = ma
= 49 × 88.78
= 4350.22
≅ 4350N
Magnitude of the force is 4350N
t = 1,025 s
a = 55.43 m / s²
Flow of electric charge in a wire requires " ELECTRONS "
