A. The horizontal velocity is
vx = dx/dt = π - 4πsin (4πt + π/2)
vx = π - 4π sin (0 + π/2)
vx = π - 4π (1)
vx = -3π
b. vy = 4π cos (4πt + π/2)
vy = 0
c. m = sin(4πt + π/2) / [<span>πt + cos(4πt + π/2)]
d. m = </span>sin(4π/6 + π/2) / [π/6 + cos(4π/6 + π/2)]
e. t = -1.0
f. t = -0.35
g. Solve for t
vx = π - 4πsin (4πt + π/2) = 0
Then substitute back to solve for vxmax
h. Solve for t
vy = 4π cos (4πt + π/2) = 0
The substitute back to solve for vymax
i. s(t) = [<span>x(t)^2 + y</span>(t)^2]^(1/2)
h. s'(t) = d [x(t)^2 + y(t)^2]^(1/2) / dt
k and l. Solve for the values of t
d [x(t)^2 + y(t)^2]^(1/2) / dt = 0
And substitute to determine the maximum and minimum speeds.
Answer: Newton's second law of motion is F = ma or force is equal to the mass the reason this applies to Newton's third law is because When the game starts, both the sides are pulling the rope and neither side is moving. The force on the rope is the same on each side This is Newton's second law of motion which equates force as mass times acceleration.
Explanation:
Answer:
Explanation:
The 3 kg box will have a downward force of its weight (Mg) and a balancing upward force of the normal force from the table. It will have a horizontal force T pointed towards the hanging mass side from the connecting cable.
The 2 kg hanging box will have a downward force of its weight (mg) and an upward force T from the connecting cable
Refer to the diagram shown below.
The following discussion assumes a simplistic analysis that ignores air resistance and variations in the terrain that the missile travels over.
Let the launch velocity be V₀ at an angle of θ relative to the horizontal.
The horizontal component of velocity is V₀ cosθ.
If the time of flight is
, then
where r = the range of the missile.
Also, the time, t, when the missile is at ground level is given by
where g = acceleration due to gravity.
t = 0 corresponds to when the missile is launched. Therefore
Therefore
Typically, θ=45° to achieve maximum range, so that
This analysis is more applicable to a scud missile rather than a powered, guided missile.
Answer:
Usually, θ=45°
Answer:
The current drawn by the motor from the line is 4.68 A.
Explanation:
Given that,
Internal resistance of the dc motor, r = 3.2 ohms
Voltage, V = 120 V
Emf in the motor,
We need to find the current drawn by the motor from the line. A dc motor with its rotor and field coils connected in series, applying loop rule we get :
I is current drawn by the motor
So, the current drawn by the motor from the line is 4.68 A. Hence, this is the required solution.