Answer:
The three best answers are:
Electric fields can be positive or negative (depending on source)
Electric fields can attrace or repel objets (assuming the object is charged)
Electric forces are strongest near the charged object E ∝ 1 / R^2
Answer:
a = 7.35 ft / s²
Explanation:
For this exercise we must use the kinematics relations
x = v₀ t + ½ a t²
as the runner leaves the starting line his initial velocity is zero
x = ½ a t²
a =
let's reduce the distance to foot
x = 60 yd (3ft / 1yd) = 180 ft
let's calculate
a = 2 180 / 7²
a = 7.35 ft / s²
Answer:
I = 1.06886 N s
Explanation:
The expression for momentum is
I = F t = Δp
therefore the momentum is a vector quantity, for which we define a reference system parallel to the floor
Let's find the components of the initial velocity
sin 28.2 = v_y / v
cos 28.2= vₓ / v
v_y = v sin 282
vₓ = v cos 28.2
v_y = 42.8 sin 28.2 = 20.225 m / s
vₓ = 42.8 cos 28.2 = 37.72 m / s
since the ball is heading to the ground, the vertical velocity is negative and the horizontal velocity is positive, it can also be calculated by making
θ = -28.2
v_y = -20.55 m / s
v_x = 37.72 m / s
X axis
Iₓ = Δpₓ = 
since the ball moves in the x-axis without changing the velocity, the change in moment must be zero
Δpₓ = m
- m v₀ₓ = 0
v_{fx} = v₀ₓ
therefore
Iₓ = 0
Y axis
I_y = Δp_y = p_{fy} -p_{oy}
when the ball reaches the floor its vertical speed is downwards and when it leaves the floor its speed has the same modulus but the direction is upwards
v_{fy} = - v_{oy}
Δp_y = 2 m v_{oy}
Δp_y = 2 0.0260 (20.55)
= 1.0686 N s
the total impulse is
I = Iₓ i ^ + I_y j ^
I = 1.06886 j^ N s
1) 30 volts is correct
2) Rt = (4x6)/(4+6)=2.4 Ω
3) P = I^2 x R = 36 W
we use AC because it's easy to transmit via 3 phase on cables and easy to step up and down