Your answer would be C. Hope this helps!!
D , since Voltage is one joule per coulomb
Answer:
Explanation:
Given the equation modelled by the height of the train given as:
s(t) = 18t²-2t³ for for 0 ≤ t ≤ 9
a) Velocity is the rate of change of displacement.
Velocity = dS(t)/dt
V = dS(t)/dt = 36t - 6t² miles
Velocity at t = 3hrs is determiner by substituting t = 3 into the velocity function.
V = 36(3) -6(3)²
V= 108 - 72
Velocity = 36mi/hr
b) for Velocity at time = 7hrs
V(7) = 36(7) - 6(7)²
V(7) = 252 - 294
V(7) = -42mi/hr
The velocity at t = 7hrs is -42mi/hr
c) Acceleration is the rate of change of velocity.
a(t) = dV(t)/dt
Given v(t) = 36t - 6t²
a(t) = 36 - 12t
Acceleration at t=1 is given as:
a(1) = 36 -12(1)
a(1) = 24mi/hr²
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
