Answer:
hope this helps!
Explanation:
Volume of the air bubble, V1=1.0cm3=1.0×10−6m3
Bubble rises to height, d=40m
Temperature at a depth of 40 m, T1=12oC=285K
Temperature at the surface of the lake, T2=35oC=308K
The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa
The pressure at the depth of 40 m: P1=1atm+dρg
Where,
ρ is the density of water =103kg/m3
g is the acceleration due to gravity =9.8m/s2
∴P1=1.103×105+40×103×9.8=493300Pa
We have T1P1V1=T2P2V2
Where, V2 is the volume of the air bubble when it reaches the surface.
V2=
Answer:
678.2 km/h and 80.54° north of east
Explanation:
From the question,
Using pythagoras theorem,
a² = b²+c²..................... Equation 1
Where a = resultant velocity
Given: b = 600 km/h, c = 100 km/h
Substitute these values into equation 1
R² = 600²+100²
R² = 360000+10000
R² = 460000
R = √460000
R = 678.2 km/h.
And the direction is
tanθ = 600/100
tanθ = 6
tanθ = 6
θ = tan⁻¹(6)
θ = 80.54°.
Hence the resultant velocity of the aircraft is 678.2 km/h and 80.54° north of east
Answer:
At the midpoint of the line joining the two equal charges
Explanation:
The midpoint of the line joining the two equal charges has equal distance from each charge. Since the charges are at x = a and x = -a, at the midpoint of the line joining the two equal charges, the magnitude of the electric field experienced as a result of each charge is the same. But these two fields are in opposite direction, hence the net resulting electric field would be equal to zero.
Answer:
C. A spring is stretched.
Explanation:
Kinetic energy is the energy of motion, potential energy is stored energy. The motion of stretching the spring is kinetic, the energy it has before it's released is potential.
Answer:
Explanation:
Resistivity and resistance are proportional and depends of the length and the cross-sectional area of the wire:
furthermore, the density is the mass divided by the volume, and the volume can be written as the area multiplyed by the length:
Now you have tw equations and two variables, so you can solve for each of them.
first, solve for A in both equations and replace them:
now replace this into any of the previous equiations:
If you assume the wire has circular cross-sectional area, then the area is:
solving for d:
replacing A and simplifying:
![d=2 \sqrt[4]{\frac{m\rho}{\rho_m R \pi^2} }](https://tex.z-dn.net/?f=d%3D2%20%5Csqrt%5B4%5D%7B%5Cfrac%7Bm%5Crho%7D%7B%5Crho_m%20R%20%5Cpi%5E2%7D%20%7D)
