2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s

a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)


2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
Power grid
All the poles and wires you see along the highway and in front of your house are called the electrical transmission and distribution system. Today, generating stations all across the country are connected to each other through the electrical system (sometimes called the "power grid").
Answer:
x=0.154kg
Explanation:
(x*L)+(0.5kg*4200*50)+(x*4200*(-50)=0
(x*333 000J/kg*c)+(0.5kg*4200J/kg*C*(-40C))+(x*4200J/kg*C*50C)=0
The correct answer would be left
Answer:
Explanation:
Given
Both cars mass is m
and solving problem in Vertical and horizontal direction
considering + y and +x to be positive and u be the final velocity of system
Conserving Momentum in Vertical direction

------1
Conserving momentum in x direction
-----2
squaring and adding 1 &2



