Answer:
v = 1.08 m/s
Explanation:
What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?
The decrease in PE is
d = 80.0cm * 1 / 1000m = 0.80m
h = 0.80 m /2 = 0.40 m
ΔPE = m*g*h
ΔPE = (0.0500 - 0.0200)kg * 9.8m/s² * 0.400 m
ΔPE = 0.1176 J
The moment of inertia of the assembly is
I = 1/12*m*L² + (m1 + m2)*(L/2)²
I = 1/12*0.390kg*(0.800m)² + 0.0700kg*(0.400m)²
I = 0.032 kg·m²
KE = ½Iω²
0.1176 J = ½ * 0.032kg·m² * ω²
ω = 2.71 rad/s
v = ωr = 2.71 rad/s * 0.400m
The linear velocity
v = 1.08 m/s
I think it will go down like decrease minus or whatever you call it
Answer:
Time= 6.12*10^4s
mass flow rate m=0.98kg/s
Explanation:
Given
Volume= 60m^3
diamter= 2.5cm= 0.025m
radius= 0.0125m
area A= πr^2
area A= 3.142*0.0125^2
area A= 4.9*10^-4m^2
the velocity of the flow 2m/s
<u>volume flow rate </u>
V=vA
V=2* 4.9*10^-4
V=9.82*10^-4 m^3/s
<u>Time taken to fill the pool</u>
time= volume/volume flow rate
time= 60/9.82*10^-4
time= 6.12*10^4s
<u>Mass flow rate </u>
m= density *volume flow rate
Assuming the density of water to be 997kg/m^3
m= 997*9.82*10^-4
m=0.98kg/s
The change in surface area of Gaussian surface with radius (r) is 8πr.
<h3>
Electric field from Coulomb's law</h3>
The electric field experienced by a charge is calculated as follows;
where;
- E is the electric field
- Q is the charge
- r is the radius
The electric field reduces by a factor of 
<h3>Surface area of a Gaussian surface;</h3>
The surface area of a sphere is given as;
<h3>Change in area with r</h3>
Thus, the change in surface area of Gaussian surface with radius (r) is 8πr.
Learn more about area of Gaussian surfaces here: brainly.com/question/17060446
