Answer:
(a) 19.62 N
(b) Box moves down the slope
(c) 24.43 N
Explanation:
(a)
2 Kg box causes tension
![T=mgwhere m is mass, g is gravitational force taken as 9.81T=2*9.81 =19.62 N (b) Block mass of 4 Kg [tex]T'-mg sin \theta=0](https://tex.z-dn.net/?f=T%3Dmg%3C%2Fp%3E%3Cp%3Ewhere%20m%20is%20mass%2C%20g%20is%20gravitational%20force%20taken%20as%209.81%3C%2Fp%3E%3Cp%3E%3C%2Fp%3E%3Cp%3ET%3D2%2A9.81%20%3D19.62%20N%20%20%3C%2Fp%3E%3Cp%3E%28b%29%20%20%3C%2Fp%3E%3Cp%3EBlock%20mass%20of%204%20Kg%20%20%3C%2Fp%3E%3Cp%3E%5Btex%5DT%27-mg%20sin%20%5Ctheta%3D0)
hence 
where m is mass and g is gravitational force
T'=4*9.81 sin 35= 22.5071 N
Since T' is greater than 
, then the box moves down the slope
(c)
Acceleration a= 
When moving, the box will exert force T"= 
T"= 4*9.81 sin 35 +(4*0.48)= 24.43 N
From the items on this list, the only one that allows calculation
of the mechanical advantage is 'B' ... the lengths from the fulcrum
to the effort and the resistance.
The MA can also be calculated when you know the two forces ...
the effort and the resistance ... when the lever is just balanced.
Answer:
Explanation:
To solve this problem, we can use the following suvat equation:
where
is the vertical displacement of the frog
is the initial vertical velocity
t is the time
a is the acceleration
We have chosen this formula because apart from , all the other quantities are known. In fact:
is the vertical displacement
t = 2 s is the total time of flight
is the acceleration due to gravity (negative because it is downward)
Therefore, solving for , we find the initial velocity of the frog:
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.
