Answer:
Explanation:
In this problem, the capacitors C1 and C2 are connected in series to the battery.
The voltage of the battery, which is variable, is V.
The voltge across the capacitor C2 is V2: the graph of V2 versus V shows that a straight line with a slope of 0.45.
This means that the relationship between V and V2 is
(1)
So, 45% of the voltage is on the capacitor C2; and therefore, 55% of the voltage must be on capacitor C1:
(2)
When two capacitors are connected in series, they have the same charge on their plates, so
Using the relationship between charge (Q), capacitance (C) and voltage (V) for a capacitor:
We can rewrite the equation as
Or
Substituting (1) and (2),
And since
We find:
To find the velocity of the bullet just before reaching the ground, we use the formula.
Vy = Vosen (30 °) + gt (1).
For that we need to know the time it takes for the bullet to reach the ground from the moment it is fired.
For that we use the formula:
Xy = Xo + Vosen (30 °) * t + 0.5gt ^ 2 (2)
where:
Vy = Component of the speed in the y direction (vertical)
Vo = initial velocity of the bullet at the exit of the cannon
g = acceleration of gravity
Xy = vertical component of the position.
Xo = initial position.
t = time.
To find the time it takes for the bullet to reach the ground, we make Xy = 0 and clear t.
then it would be:
0 = 10 + 80sen (30 °) * t +0.5 * (- 9.8) t ^ 2.
The solution to that equation is:
t = 8.4 seconds.
We substitute that time in equation (1) and clear Vy.
We have left:
Vy = 80 * sin (30 °) + 9.8 * (8.4)
Vy = 42.38 m / s
Answer:
3.4093
Explanation:
NPSHa = hatm + hel + hf +hva
the elevation head is the hel
friction loss head is hf
NPSHa is the head of vapour pressure of fluid
atmospheric pressure head is hatm
log₁₀P* =
log₁₀Pv =
= 4.07827 - 1343.943/333.377
=4.07827 - 4.0313009
= 0.0469691
we take the log
p* = 1.114218
we convert this value to get 111421.8
hvap = 111421.8 * 1/776.14 * 1/9.81
= 14.63
hatm = 1.1 *101325/1 * 1/9.81 *1/776.14
=14.64
hf = 7000/1 * 1/776.14 * 1/9.81
= 0.9193
NPSHa = 2.5
hel = 0.9193 + 2.5 + 14.63 - 14.64
hel = 3.4093
The electric field at a distance of 7.00 cm is 8.60×10^−2C/m2
The charge per unit area of wall is
σ = (8.60×10^−6C/cm2)( 100cm /m)^2 = 8.60×10^−2C/m2
The electric field at a distance of 2.00 cm is then
E = σ /7∈0 = (8.60×10^−2C/m^2) / 7(8.85×10−12C2/N⋅m2)
= 1.38×10^9N/C away from the wall.
Electric field, an electric property associated with each point in space when charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.
Learn more about electric field here:
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There is a correlation between a main sequence star's mass and its luminosity.