Jonathan is studying solutions in his chemistry class. The teacher instructs the students to measure 2 tablespoons of salt into
1 answer:
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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
Answer:
The y-component of the car's position vector is 670m/s.
The x-component of the acceleration vector is -3, and the y-component is 40.
Explanation:
The displacement vector of the car with velocity
is the integral of the velocity.
Integrating we get the displacement vector
:
Now if the initial position if the car is
then the displacement of the car at time is
Now at , we have
The y-component of the car's position vector is 670m/s.
The acceleration vector is the derivative of the velocity vector:
and at it is
The x-component of the acceleration vector is -3, and the y-component is 40.