Answer:
I dont the child mass...you should substitute that value to (m)
then you can get your answer
We have 2 conditions of balance. First of all, the total forces have to have a net sum of 0 since the bridge is balancing. Hence, if we denote by F1 the force of the first pole, F2 the power of the second pole (the one closer to the car), W the weight of the bridge and w the weight of the car, we have that W+w=F1+F2=4.196*10^5 N.
We also have that it does not rotate. Hence, taking as origin of our frame of reference the car, we have that 5*F2+4*W=9*F2 by calculating the distances from our point of reference. Thus yields 5F2+8*10^5=9F2. When we solve the system of equations that is created above (best way here is by substitution), we get that F1=2.07*10^5 N while F2=2.126*10^5 N . Each pole takes up around half the weight but due to the car the pole closer to it has more weight to bear; nevertheless the car does not weigh a lot so the difference is small.
True. Think of a magnet and how they only connect to the opposite charges.
Answer:
ROYGBIV or Roy G. Biv is an acronym for the sequence of hues commonly described as making up a rainbow: red, orange, yellow, green, blue, indigo and violet. The initialism is sometimes referred to in reverse order, as VIBGYOR.
For the part a) we need only the momentum of the box and we have the data to find it.
Momentum is given by,
where clearly, p is the momentum, m the mass of the box and v is the velocity.
Substituting,
For part b) we need an analysis of the situation. We understand that the box on a surface that has no friction will continue to rotate at the same speed previously defined. The box can only stop with friction, so,
<em>It is the same that part a)</em>
