A glass pipe system has a very corrosive liquid flowing in it (think hydrofluoric acid, say). The liquid will destroy flow meters, but you need to know the flow rate. One way of measuring the flow rate is to add a fluorescent dye to the liquid at a known concentration, and then downstream activate the dye by UV light and then measure the dye concentration by emitted light. If the dye is added at 1.00 g/s, and the dye concentration downstream is 0.050% by mass, what is the unknown flow rate in kg/h
glass
We are given an object that is speeding up on a level ground.
Let's remember that the gravitational energy depends on the change in height, therefore, if the object is not changing its height it means that the gravitational energy remains constant.
The kinetic energy depends on the velocity. If the velocity is increasing this means that the kinetic energy is also increasing.
Now, every change in velocity requires acceleration and acceleration requires a force. The force and the distance that the object moves are equivalent to the work that is transferred to the object and therefore, the change in kinetic energy. This means that the total energy of the system increases as work is transferred to the mass.
We have that the total energy of the system increases in the form of kinetic energy and that the gravitational potential energy remains constant. Therefore, the diagrams should look like pie charts that grow but the area of the segment of the potential energy stays the same. It should look similar to the following.
I wanna say B Red-billed oxpeckers eat ticks off of impalas
The calculated coefficient of kinetic friction is 0.33125.'
The rate of kinetic friction the friction force to normal force ratio experienced by a body moving on a dry, uneven surface is known as k. The friction coefficient is the ratio of the normal force pressing two surfaces together to the frictional force preventing motion between them. Typically, it is represented by the Greek letter mu (). In terms of math, is equal to F/N, where F stands for frictional force and N for normal force.
given mass of the block=10 kg
spring constant k= 2250 Nm
now according to principal of conservation of energy we observe,
the energy possessed by the block initially is reduced by the friction between the points B and C and rest is used up in work done by the spring.
mgh= μ (mgl) +1/2 kx²
10 x 10 x 3= μ(600) +(1125) (0.09)
μ(600) =300 - 101.25
μ = 198.75÷600
μ =0.33125
The complete question is- A 10.0−kg block is released from rest at point A in Fig The track is frictionless except for the portion between point B and C, which has a length of 6.00m the block travels down the track, hits a spring of force constant 2250N/m, and compresses the spring 0.300m form its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between point Band (C)
Learn more about kinetic friction here-
brainly.com/question/13754413
#SPJ4
