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.
Answer:
Time period of the osculation will be 0.0671 sec
Explanation:
It is given a vertical spring is stretched by 4 cm
So change in length of the spring x = 4 cm = 0.04 m
Mass which is hung from it m = 12 gram = 0.012 kg
Sprig force will be equal to weight of the mass
So
k = 244.7 N/m
Now new mass is m = 28 gram = 0.028 kg
So time period with new mass will be
C, electrons are negative, and most of the atom’s mass comes from the nucleus of the atom
Answer:
d= 794.4 cmExplanation:
Given that
Speed ,V= 286 km/h
V=79.44 m/s
Given that time ,t= 100 ms
t= 0.1 s
We know that ( if acceleration is zero)
Distance = Speed x time
d= V t
Now by putting the values in the above equation
d = 79.44 x 0.1 m
d= 7.944 m
We know that 1 m = 100 cm
d= 794.4 cm