Answer:
The coefficient of static friction between your partner and the floor is 0.55
Explanation:
Given:
Mass 
Kg
Frictional force 
N
From the formula of frictional force,
Where 
coefficient of static friction, 
Put the above values and find the coefficient of static friction.
Therefore, the coefficient of static friction between your partner and the floor is 0.55
Answer:
a. 192 m/s
b. -17,760 kPa
Explanation:
First let's write the flow rate of the liquid, using the following equation:
Q = A*v
Where Q is the flow rate, A is the cross section area of the pipe (A = pi * radius^2) and v is the speed of the liquid. The flow rate in both parts of the pipe (larger radius and smaller radius) needs to be the same, so we have:
a.
A1*v1 = A2*v2
pi * 0.02^2 * 12 = pi * 0.005^2 * v2
v2 = 0.02^2 * 12 / 0.005^2
v2 = 192 m/s
b.
To find the pressure of the other side, we need to use the Bernoulli equation: (600 kPa = 600000 N/m2)
P1 + d1*v1^2/2 = P2 + d1*v2^2/2
Where d1 is the density of the liquid (for water, we have d1 = 1000 kg/m3)
600000 + 1000*12^2/2 = P2 + 1000*192^2/2
P2 = 600000 + 72000 - 1000*192^2/2
P2 = -17760000 N/m2 = -17,760 kPa
The speed in the smaller part of the pipe is too high, the negative pressure in the second part means that the inicial pressure is not enough to maintain this output speed.
The formula for momentum is p=mv (mass multiplied by velocity), so in this problem, p=50(.5)=25=p or in other words, momentum.
In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
= (5 * 10) Kg m/s
= 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".
