Answer:
v₁f = 0.5714 m/s (→)
v₂f = 2.5714 m/s (→)
e = 1
It was a perfectly elastic collision.
Explanation:
m₁ = m
m₂ = 6m₁ = 6m
v₁i = 4 m/s
v₂i = 2 m/s
v₁f = ((m₁ – m₂) / (m₁ + m₂)) v₁i + ((2m₂) / (m₁ + m₂)) v₂i
v₁f = ((m – 6m) / (m + 6m)) * (4) + ((2*6m) / (m + 6m)) * (2)
v₁f = 0.5714 m/s (→)
v₂f = ((2m₁) / (m₁ + m₂)) v₁i + ((m₂ – m₁) / (m₁ + m₂)) v₂i
v₂f = ((2m) / (m + 6m)) * (4) + ((6m -m) / (m + 6m)) * (2)
v₂f = 2.5714 m/s (→)
e = - (v₁f - v₂f) / (v₁i - v₂i) ⇒ e = - (0.5714 - 2.5714) / (4 - 2) = 1
It was a perfectly elastic collision.
Some of these frictions depend on the Pressure, temperature of atmosphere.
Static Friction: This is the friction force when two objects in contact are not moving relative to each other. This friction is higher than kinetic friction.
Kinetic or Dynamic friction: this the friction force opposing the motion of objects, when two objects in contact are in motion relative to each other. It is less than the static friction. The two surfaces are rubbing against each other as they move.
Rolling friction: This is the friction when two objects are in contact and one object is rolling over the other - like a wheel on a road. The point of contact appears as stationary. The rolling friction is very less compared to static friction & dynamic friction.
Lubricated friction: this is the friction between two solid surfaces in contact with a layer of lubricant fluid flowing in between them. This friction is the least.
Fluid friction - viscosity : this is friction between two adjacent layers that are moving relative to each other at different speeds in a fluid. This is not high.
Internal friction: when an object is compressed and forced to deform, like in a piece of rubber, there is friction between the layers, that opposes this deformation.
Skin friction is the friction that opposes movement of a fluid across a solid surface. This is also called drag. When a coin is dropped in water, there is a friction called drag on the coin. Same is the case when a ball is thrown, a drag is experienced by the ball due to the drag of air.
Answer:
Find the letter below.
Explanation:
1237 Richmond Avenue,
Houston, Texas,
U.S.A.
3rd May, 2021.
The Principal,
Carnegie Vanguard High School,
Houston, Texas,
U.S.A.
Dear Sir,
APPRECIATION FOR THE ADMISSION OFFER IN YOUR SCHOOL
I humbly write this letter to express my sincere appreciation for the offer of admission in your highly esteemed institution of learning. The admission letter states that I was admitted to pursue a programme in Computer Science and this is a source of great delight for me. The opportunity is a unique one because I stand a great chance of acquiring deeper knowledge of programming software that will prove invaluable to me in my future career.
Over the years, I have been particularly thrilled with all the inventions that have been made possible with information technology and programming of many useful software. These inventions have made life a lot easier, hence my interest in getting the required knowledge that will enable me to contribute my quota to the society.
This programme will also afford me the opportunity to relate with seasoned professionals in the field of Information Technology. When in contact with these professionals, I hope to ask them several questions on programming, the technicalities involved in it, and what is obtainable in the real-world scenario. Armed with these pieces of information, I believe that the programme will shape me to be an excellent professional that will be useful to the society.
Once again, thank you for this very special opportunity. I hope to fully maximize the privilege.
Yours Sincerely,
Jessica Liam.
Explanation:
Ubuntu is somewhat a South African concept that involves charity, sympathy, and mainly underlines the concept of universal brotherhood. Hence this concept can help fight social challenges such as racism, crime, violence and many more. It can contribute to maintaining peace and harmony in the country at large
Answer:
2.9 cm
Explanation:
Assuming that the rear wheel has a radius of 0.330 m
Given that
r(a) = 12 cm -> 0.12 m
w(a) = 0.6 rev/s -> 3.77 rad/s
v = 5 m/s
r(w) = 0.330 m
The speed on any point on the rim at the sprocket in the front is
v(a) = w(a).r(a) = 3.77 * 0.12 = 0.4524 m/s
Also,
v(a) = speed at any point on the chain
v(b) = speed at any point on the rim of the rear sprocket
v(a) = v(b)
where v(b) = w(b).r(b)
Recall that the speed at any point on the rear wheel is v, where
v = w(b).r(w)
5 = w(b) * 0.330
w(b) = 5/0.330
w(b) = 15.15 rad/s
On substituting this in the equation, we have
v(b) = w(b).r(b).
Remember also, that v(a) = v(b), so
0.4524 = 15.15 * r(b)
r(b) = 0.4524 / 15.15
r(b) = 0.029 m -> 2.9 cm
Therefore, the radius of the rear sprocket needed is 2.9 cm
