The isobars in the conventional series that will be needed
to complete the pressure analysis between the lowest and highest values on this
map are: 1008, 1012, 1016, 1020.
To add, an isobar is <span>a line on a map connecting points having the
same atmospheric pressure at a given time or on average over a given period.</span>
Answer:
b) a horse pulls a wagon at a constant velocity
Explanation:
As we know that ,work done is the dot product of force vector and displacement vector.
W= F.d
W=work
F=Force
d=Displacement
We also know that
F = m a
m= mass ,a = acceleration
When velocity is constant then rate of change in the velocity will be zero,then we can say that acceleration will be zero.
When a= 0 Then F= 0
W= F.d ( F=0)
W = 0
Therefore option b i correct because horse is going with constant velocity.
Two parallel plates having charges of equal magnitude but opposite sign are separated by 11.0 cm. Each plate has a surface charge density of 49.0 nC/m2. A proton is released from rest at the positive plate.
(a) Determine the magnitude of the electric field between the plates from the charge density.
b) Determine the potential difference between the plates.
Answer:
a
The Electric Field between the two plate is
b
The potential difference between the plate is V =
Explanation:
From the we are given that
The separation between the plate is
The surface charge density is
Generally Electric field between the plate is mathematically given as
Note that is the permitivity of free space and its value is
Now substituting values we have
Generally Potential difference between the plate is mathematically given as
Where E is the electric field which is
Substituting value we have
V =
Answer: 1350kg/m3
Explanation:
Db = Mb/Vb
900= Mb/Vb
(where Db is the density of the block of wood = 900kg/m3, Mb is its mass and Vb is its volume
Df = Mf/Vf (where Df, Mf and Vf are the density, mass and velocity of the fluid respectively
note that the same mass is involved because the block of wood can only displace its own mass of the fluid
It displaces two-thirds of its own volume in the fluid.
2/3Vf = Vb
Vf = 2Vb/3
Db = Mb/Vb
900 = 2Mb/3Vf
2700Vf = 2Mb
1350Vf = Mb
since Mb = Mf
Df = 1350Vf/Vf
Df= 1350Kg/m3