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:
v= 2413.5 m/s
Explanation:
maximum change of speed of rocket
=(initial exhaust velocity)×ln [(initialmass/finalmass)]
let initial mass= m
final mass = m-m(4/5) = m/5
[since the 80% of mass which is fuel is exhausted]
V-0 = 1500 ln (1/0.2)
V= 1500×1.609 = 2413.5 m/s
therefore, its exhaust speed v= 2413.5 m/s
Answer:
t=0.704s
Explanation:
A child is running his 46.1 g toy car down a ramp. The ramp is 1.73 m long and forms a 40.5° angle with the flat ground. How long will it take the car to reach the bottom of the ramp if there is no friction?
from newton equation of motion , we look for the y component of the speed and look for the x component of the speed. we can then find the resultant of the speed

Vy^2=0+2*9.8*1.73sin40.5
Vy^2=22.021
Vy=4.69m/s
Vx^2=u^2+2*9.81*cos40.5
Vy^2=25.81
Vy=5.08m/s
V=(Vy^2+Vx^2)^0.5
V=47.71^0.5
V=6.9m/s
from newtons equation of motion we know that force applied is directly proportional to the rate of change in momentum on a body.
f=force applied
v=velocity final
u=initial velocity
m=mass of the toy, 0.046
f=ma
f=m(v-u)/t
v=u+at
6.9=0+9.8t
t=6.9/9.81
t=0.704s
Answer:
t = 2 s
Explanation:
As we know that fish is pulled upwards with uniform maximum acceleration
then we will have

here we know that maximum possible acceleration of so that string will not break is given as

now we have


now for such acceleration we can use kinematics


t = 2 s
Answer:
Weight W = 392.4 N
Density
= 784.31 
Specific gravity S = 0.78431
Force required F = 10 N
Explanation:
Given data
Mass (m) = 40 kg
Volume (V) = 0.051 
Weight W = m × g
⇒ W = 40 × 9.81
⇒ W = 392.4 N
This is the weight of the methanol.
Density
= 
⇒
= 
⇒
= 784.31 
This is the density of the methanol.
Specific gravity (S) = 
⇒ 
⇒ S = 0.78431
This is the specific gravity of the methanol.
Force needed to accelerate this tank F = ma
⇒ F = 40 × 0.25
⇒ F = 10 N
This is the force required to accelerate the tank.