<span>the speed of something in a given direction. so i think none of these</span>
Answer:
F' = (3/2)F
Explanation:
the formula for the electric field strength is given as follows:
E = F/q
where,
E = Electric Field Strength
F = Force due to the electric field
q = magnitude of charge experiencing the force
Therefore,
F = E q ---------------- equation (1)
Now, if we half the electric field strength and make the magnitude of charge triple its initial value. Then the force will become:
F' = (E/2)(3 q)
F' = (3/2)(E q)
using equation (1)
<u>F' = (3/2)F</u>
Answer:
Its diameter increases as it flows down from the pipe. Assuming laminar flow for the water, then Bernoulli's equation can be applied.
P1-P2 + (rho)g(h1 - h2) + 1/2(rho)(v1² - v2²) = 0
Explanation:
P1 = P2 = atmospheric pressure so, P1 - P2 = 0
h1 is greater than h2 so h1-h2 is positive. Rearranging the equation above 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho = v2²
From the continuity equation for fluids
A1v1 = A2v2
v2 = A1v1/A2
Substituting into the equation above
(A1v1/A2)² = 2{ (rho)g(h1-h2) + 1/2(rho)v1²}/rho
Making A2² the subject of the formula,
A2² = (A1v1)²× rho/(2{ (rho)g(h1-h2) + 1/2(rho)v1²}
The denominator will be greater than the numerator and as a result the diameter of the flowing stream decreases.
Thank you for reading.
Answer:
The peak-to-peak ripple voltage = 2V
Explanation:
120V and 60 Hz is the input of an unfiltered full-wave rectifier
Peak value of output voltage = 15V
load connected = 1.0kV
dc output voltage = 14V
dc value of the output voltage of capacitor-input filter
where
V(dc value of output voltage) represent V₀
V(peak value of output voltage) represent V₁
V₀ = 1 - (
)V₁
make C the subject of formula
V₀/V₁ = 1 - (1 / 2fRC)
1 / 2fRC = 1 - (v₀/V₁)
C = 2fR ((1 - (v₀/V₁))⁻¹
Substitute for,
f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V
C = 2 * 240 * 1 (( 1 - (14/15))⁻¹
C = 62.2μf
The peak-to-peak ripple voltage
= (1 / fRC)V₁
= 1 / ( (120 * 1 * 62.2) )15V
= 2V
The peak-to-peak ripple voltage = 2V