Answer:
a)11.25 J
b)Number of revolution = 1
Explanation:
Given that
Radius ,r= 0.8 m
m= 0.3 kg
Initial speed ,u= 10 m/s
final speed ,v= 5 m/s
a)
Initial energy


KEi= 15 J
Final kinetic energy


KEf=3.75 J
The energy transformed from mechanical to internal = 15 - 3.75 J = 11.25 J
b)
The minimum value to complete the circular arc

Now by putting the values

V= 2.82 m/s
So kinetic energy KE


KE=1.19 J
ΔKE= KEi - KE
ΔKE= 15- 1.19 J
ΔKE=13.80 J
The minimum energy required to complete 2 revolutions = 2 x 11.25 J
= 22.5 J
Here 22.5 J is greater than 13.8 J.So the particle will complete only one revolution.
Number of revolution = 1
Answer:
The sum of all forces for the two objects with force of friction F and tension T are:
(i) m₁a₁ = F
(ii) m₂a₂ = T - F
1) no sliding infers: a₁ = a₂= a
The two equations become:
m₂a = T - m₁a
Solving for a:
a = T / (m₁+m₂) = 2.1 m/s²
2) Using equation(i):
F = m₁a = 51.1 N
3) The maximum friction is given by:
F = μsm₁g
Using equation(i) to find a₁ = a₂ = a:
a₁ = μs*g
Using equation(ii)
T = m₁μsg + m₂μsg = (m₁ + m₂)μsg = 851.6 N
4) The kinetic friction is given by: F = μkm₁g
Using equation (i) and the kinetic friction:
a₁ = μkg = 6.1 m/s²
5) Using equation(ii) and the kinetic friction:
m₂a₂ = T - μkm₁g
a₂ = (T - μkm₁g)/m₂ = 12.1 m/s²
I believe the answer here is <span>C).seedless, seed (nonflowering), and seed (flowering).</span><span>
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We know that
g = LcosΘ
<span>where g, L and Θ are centripetal gravity length, and angle of object
</span><span>ω² = g/LcosΘ </span>
<span>ω = √(g / LcosΘ) </span>