Answer:
The charge on the drop is
q = 1.741 x 10 ⁻²¹ C
Explanation:
Electric field due to plates
Ef = V/d
Ef = 2033 V / (2.0 * 10^-2 m )
Ef = 101650 V/m
So, we can write
Ef * q = m*g
q = m*g / E
f
The mass can be equal using the density and the volume so:
m = ρ * v
The volume can be find as:
v = 2.298 x 10 ⁻ ¹⁶ m³
q = ρ * v * g / Ef
q = 81 x 10 ³ kg/ m³ * 2.2298 x 10 ⁻ ¹⁶ m³ * 9.8 m/s² / 101650 V/m
The charge on the drop is
q = 1.741 x 10 ⁻²¹ C
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:
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Answer:
The other angle is 120°.
Explanation:
Given that,
Angle = 60
Speed = 5.0
We need to calculate the range
Using formula of range
...(I)
The range for the other angle is
....(II)
Here, distance and speed are same
On comparing both range
Hence, The other angle is 120°
The solution has reacted.
