Answer:
Explanation:
<u>Instant Acceleration</u>
The kinetic magnitudes are usually related as scalar or vector equations. By doing so, we are assuming the acceleration is constant over time. But when the acceleration is variable, the relations are in the form of calculus equations, specifically using derivatives and/or integrals.
Let f(t) be the distance traveled by an object as a function of the time t. The instant speed v(t) is defined as:
And the acceleration is
Or equivalently
The given height of a projectile is
Let's compute the speed
And the acceleration
It's a constant value regardless of the time t, thus
Answer:
Velocity has both speed and direction. Speed is constant. ... Speed is measured over time.
Explanation:
We don't have enough information in the picture to answer that question.
The disembodied right hand could just as well grab the wire in the other direction, with the thumb pointing to the right.
Maybe if we knew WHY the hand is holding the wire in this direction, and what other electrical phenomenon may be involved, we might be able to say something about the current in the wire.
(Actually, we don't even know if it's a wire. It might be a soda straw, a coat hanger, or a pool cue.)
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
