Virtual upright and the same size
Answer:
13.309 m/s²
Explanation:
Length from shoulder to hand, l = 30 cm = 0.3 m
initial velocity, u = 1 m/s
final velocity, v = 2.5 m/s
time, t = 3 s
Let the tangential acceleration is a.
by using first equation of motion
v = u + at
2.5 = 1 + 3 a
a = 0.5 m/s²
Let the centripetal acceleration is a'.
a' = v'²/l
a' = 2 x 2 / 0.3
a' = 13.3 m/s²
The tangential acceleration and the centripetal acceleration are both perpendicular to each other. So, the net acceleration is given by
A = 13.309 m/s²
Answer:
Kf > Ka = Kb > Kc > Kd > Ke
Explanation:
We can apply
E₀ = E₁
where
E₀: Mechanical energy at the beginning of the motion (top of the incline)
E₁: Mechanical energy at the end (bottom of the incline)
then
K₀ + U₀ = K₁ + U₁
If v₀ = 0 ⇒ K₀
and h₁ = 0 ⇒ U₁ = 0
we get
U₀ = K₁
U₀ = m*g*h₀ = K₁
we apply the same equation in each case
a) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
b) U₀ = K₁ = m*g*h₀ = 70 Kg*9.81 m/s²*8m = 5493.60 J
c) U₀ = K₁ = m*g*h₀ = 35 Kg*9.81 m/s²*4m = 1373.40 J
d) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*16m = 1098.72 J
e) U₀ = K₁ = m*g*h₀ = 7 Kg*9.81 m/s²*4m = 274.68 J
f) U₀ = K₁ = m*g*h₀ = 105 Kg*9.81 m/s²*6m = 6180.30 J
finally, we can say that
Kf > Ka = Kb > Kc > Kd > Ke
Answer:
it makes our work easy and
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it multiply our force applied
it complete our work with high efficiency
by using this,less effort is required for the work.
Answer:
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