Answer:
Solution given;
load =600N
effort=?
load distance: 0.6m
effort distance;2.6m
we have
load *load distance :effort *effort distance
0.6*600=effort *2.6
360/2.6=effort
effort:138.46N
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Answer:Twice
Explanation:
It is given that car is raised on a lift with respect to ground.
Suppose car is raised by a height of h m from ground having mass m so gain in Potential Energy by car is
Gain in 
Now if it were raised twice as high,i.e. 2 h height from ground then gain in Gravitational Potential Energy
It is clear that 
Thus in second case gain in Potential Energy is twice as compared to first case
Answer: 14.16
Explanation:
Given
d = 38cm
r = d/2 = 38/2 = 19cm = 0.19m
K.E = 510J
m = 10kg
I = 1/2mr²
I = 1/2*10*0.19²
I = 0.18kgm²
When it has 510J of Kinetic Energy then,
510J = 1/2Iω²
ω² = 1020/I
ω² = 1020/0.18
ω² = 5666.67
ω = √5666.67 = 75.28 rad/s
Velocity is the block, v = ωr
V = 75.28 * 0.19
V = 14.30m/s
The "effective mass" M of the system is
M = (14.0 + ½*10.0) kg = 19.0 kg
The motive force would be
F = ma
F = 14 * 9.8
F = 137.2N
so that the acceleration would be
a = F/m
a = 137.2/19
a = 7.22m/s²
Finally, using equation of motion.
V² = u² + 2as
14.3² = 0 + 2*7.22*s
204.49 = 14.44s
s = 204.49/14.44
s = 14.16m
