6) D) 19.2 m/s
7) D) 4 N
8) 20 Hz, 0.05 s
Explanation:
6)
We can solve this problem by using the continuity equation.
In fact, the flow rate of a liquid through a pipe must be constant, therefore we can write:
[tex]A_1 v_1 = A_2 v_2[/tex]
where
[tex]A_1[/tex] is the cross-sectional area of the pipe in section 1
[tex]A_2[/tex] is the cross-sectional area of the pipe in section 2
[tex]v_1[/tex] is the speed of the liquid in section 1
[tex]v_2[/tex] is the speed of the liquid in section 2
Here we have:
[tex]d_1 = 9.6 cm=0.096m[/tex] is the diameter in section 1, so the area is
[tex]A_1 = \pi (\frac{d_1}{2})^2=\pi (\frac{0.096}{2})^2=7.24\cdot 10^{-3} m^2[/tex]
[tex]d_2= 2.5 cm=0.025m[/tex] is the diameter in section 2, so the area is
[tex]A_2 = \pi (\frac{d_2}{2})^2=\pi (\frac{0.025}{2})^2=0.49\cdot 10^{-3} m^2[/tex]
[tex]v_1=1.3 m/s[/tex]
Solving for v2, we find the speed in the nozzle:
[tex]v_2=\frac{A_1 v_1}{A_2}=\frac{(7.24\cdot 10^{-3})(1.3)}{0.49\cdot 10^{-3}}=19.2 m/s[/tex]
7)
The apparent weight of an object submerged in a liquid is given by
[tex]W'=W-B[/tex]
where
W' is the apparent weight
W is the real weight of the object
B is the buoyant force (which is the weight of the volume of liquid displaced by the object)
In this problem:
W = 6 N is the real weight of the brick, in air
B = 2 N is the buoyant force (the weight of water displaced)
Re-arranging the equation, we find the apparent weight of the brick:
[tex]W'=6 N - 2 N = 4 N[/tex]
8)
The frequency of a periodic motion is the number of oscillations that the system completes in one second.
Mathematically, it is given by:
[tex]f=\frac{N}{t}[/tex]
where
N is the number of oscillations completed
t is the time elapsed
In this problem:
N = 30 is the number of oscillations
t = 1.5 s is the time taken for N oscillations
So, the frequency of the motion is:
[tex]f=\frac{30}{1.5}=20 Hz[/tex]
The period is instead the time taken for one complete oscillation. It is equal to the reciprocal of the frequency:
[tex]T=\frac{1}{f}[/tex]
So in this case,
[tex]T=\frac{1}{20}=0.05 s[/tex]
