Answer:
3.974 Joule
Explanation:
Diameter of ring = 7.7 cm
a = Distance from the center = d/2 = 3.85 cm = 0.0385 m
Q = Charge = 5 mC
q = Charge to move = 3.4 mC
k = Coulomb constant = 9×10⁹ Nm²/C²
Work done will be equal to Potential energy when mass is at center
∴ Work to move a tiny 3.4 mC charge from very far away to the center of the ring is 3.974 Joule
Answer:
7066kg/m³
Explanation:
The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:
Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)
= 7066kg/m³
Explanation:
Answer:
The depth of focus achievable with those lenses is small.
Explanation:
A larger aperture makes it much harder to focus on more than one object. When using a telephoto lens (the ones the question is referring to), the depth of focus is very small. For example, using a telephoto lens to take a photo of a runner might get the runner in focus, but certainly not the track, or the audience behind them. If you look at photos, especially older photos, of Olympians in almost any sport you can see this.
Hope this helps!
Answer:
The force exerted is 318.86 N
Explanation:
The force exerted by such a stream is calculated by multiplying the mass flow rate of water by the velocity of the stream of water.
mass flow rate = 21.4 kg/s
velocity = 14.9 m/s
Force exerted = 21.4 kg/s × 14.9 m/s = 318.86 kgm/s^2 = 318.86 N
Answer:
Explanation:
a) I = ½mR² = ½(19)(0.15²) = 0.21375 kg•m²
b) τ = Fnet(r) = (25 - 12)(0.15) = 1.95 N•m
c) CCW
d) a = τ/I = 1.95 / 0.21375 = 9.12280701... = 9.1 rad/s²
e) CCW
