Answer: 1.14 N
Explanation :
As any body submerged in a fluid, it receives an upward force equal to the weight of the fluid removed by the body, which can be expressed as follows:
Fb = δair . Vb . g = 1.29 kg/m3 . 4/3 π (0.294)3 m3. 9.8 m/s2
Fb = 1.34 N
In the downward direction, we have 2 external forces acting upon the balloon: gravity and the tension in the line, which sum must be equal to the buoyant force, as the balloon is at rest.
We can get the gravity force as follows:
Fg = (mb +mhe) g
The mass of helium can be calculated as the product of the density of the helium times the volume of the balloon (assumed to be a perfect sphere), as follows:
MHe = δHe . 4/3 π (0.294)3 m3 = 0.019 kg
Fg = (0.012 kg + 0.019 kg) . 9.8 m/s2 = 0.2 N
Equating both sides of Newton´s 2nd Law in the vertical direction:
T + Fg = Fb
T = Fb – Fg = 1.34 N – 0.2 N = 1.14 N
Answer:
From smallest ratio to the largest ratio:
Coasting Universe - Critical Universe - Recollapsing Universe(From left to right)
Explanation:
The coasting universe is one that expands at a constant rate given by the Hubble constant throughout all of cosmic time. It has a ratio of actual density to critical density that is less than 1
The critical universe is one that is at balance with no expansion .I.e. the actual density and the critical density are equal, which makes the ratio of actual density to critical density to be equal to 1
Recollapsing Universe: The expansion of the universe reverses in the future and the universe eventually recollapses. The recollapsing universe has the ratio of the actual density to the critical density to be greater than 1
Answer:
Decrease
Explanation:
I = Current = 3.7 A
e = Charge of electron = 
n = Conduction electron density in copper = 
= Drift velocity of electrons
r = Radius = 1.23 mm
Current is given by
The drift speed of the electrons is
From the equation we can see the following
So, if the number of conduction electrons per atom is higher than that of copper the drift velocity will decrease.
6 meters is left because you subtract 12 meters from 6
