The molarity of 10% CaCl2 is 0.9%
concentration of the given salt CaCl₂ = 10%
Density of a solution = 1.0835 g/cm³
Volume = m / d
= 100 / 1.0835
= 92.29 litres
Density = mass / volume
1.0835 × 92.29 = mass
mass = 99.99 gram
Thus the molarity can be calculated by = moles of solute / volume of solution multiplied by 100
= 0.9008/ 92.29 X 100 %
= 0.009 X 100 %
= 0.9 %
The molarity of 10% CaCl2 is 0.9%
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Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N
Answer:
Part a)

Part b)



Part c)



Explanation:
Part a)
As we know that charge density is the ratio of total charge and total volume
So here the volume of the charge ball is given as



now the charge density of the ball is given as

Part b)
Now the charge enclosed by the surface is given as

at radius of 5 cm


at radius of 10 cm


at radius of 20 cm

Part c)
As we know that electric field is given as

so we have electric field at r = 5 cm


electric field at r = 10 cm


electric field at r = 20 cm


Answer:
A. 
B. 
C. 
Explanation:
The capacitance of a capacitor is its ability to store charges. For parallel-plate capacitors, this ability depends the material between the plates, the common plate area and the plate separation. The relationship is

is the capacitance,
is the common plate area,
is the plate separation and
is the permittivity of the material between the plates.
For air or free space,
is
called the permittivity of free space. In general,
where
is the relative permittivity or dielectric constant of the material between the plates. It is a factor that determines the strength of the material compared to air. In fact, for air or vacuum,
.
The energy stored in a capacitor is the average of the product of its charge and voltage.

Its charge,
, is related to its capacitance by
(this is the electrical definition of capacitance, a ratio of the charge to its voltage; the previous formula is the geometric definition). Substituting this in the formula for
,

A. Substituting for
in
,

B. When the distance is
,


C. When the distance is restored but with a dielectric material of dielectric constant,
, inserted, we have

Answer:
Explanation:
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