Answer:
e = 0.46 m
Explanation:
From the laws of friction, frictional force, F is proportional to normal reaction, R.
F₁ = μR
where μ is coefficient of friction; R = mg and g = 9.8 ms⁻²
Also, from Hooke's law, extension, e, in an elastic spring is proportional to applied force.
F₂ = Ke
where K is force constant of the spring
Since the box is just about to move, the coefficient of friction involved is static friction.
The force on the spring equals the frictional force experienced by the box the box; F₁ = F₂
Ke = μR
e = μR/K
where μ = 0.65; R = 18 kg * 9.8 ms⁻²; K = 250 N/m
e = (0.65 * 18 * 9.8)/250
e = 0.46 m
Answer:
1.312 x 10⁻¹² J/nucleon
Explanation:
mass of ¹³⁶Ba = 135.905 amu
¹³⁶Ba contain 56 proton and 80 neutron
mass of proton = 1.00728 amu
mass of neutron = 1.00867 amu
mass of ¹³⁶Ba = 56 x 1.00728 amu + 80 x 1.00867 amu
= 137.10128 amu
mass defect = 137.10128 - 135.905
= 1.19628 amu
mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg
= 1.9858 x 10⁻²⁷ Kg
speed of light = 3 x 10⁸ m/s
binding energy,
E = mass defect x c²
E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²
E = 17.87 x 10⁻¹¹ J/atom
now,
binding energy per nucleon =
= 0.1312 x 10⁻¹¹ J/nucleon
= 1.312 x 10⁻¹² J/nucleon
Answer:
They are reflected back at the same angle they came in
Explanation:
When parallel light rays hit a convex mirror they reflect outwards and travel directly away from an imaginary focal point (F). Each individual ray is still reflecting at the same angle as it hits that small part of the surface.
Answer:
Final angular velocity = 2.15rad/s
Explanation:
The final angular velocity can be determined using the equation:
L=Ii × wi
L= (Ii + II) × wf
Where I is the moment of inertia of the 2nd cylinder and wf is the final angular velocity.
Given:
Ii=12.6kgm^2
I=41.8kgm^2
Wi=9.28rad/s
Ii ×wi= (Ii + I)×wf
12.6×9.28 = (12.6+41.8)×wf
116.93= 54.4wf
Wf=116.93/54.4
Wf= 2.15rad/s
Answer is A) Fulcrum
The fixed point that a lever rotates around is called the fulcrum.