Answer: e = 3.333% or 33 1/3%
^
repeated
Identify the given information.
Work input (Wi)= 600 K
Work output (Wo) = 200 K
Choose which formula you should use. In this case, you are finding efficiency.
e = (Wo/Wi) x 100%
Substitute and solve.
e = (200 K/600 K) X 100%
e = 0.333 X 100%
e = 33.333% or 33 1/3%
Answer:
v2^2 - v1^2 = 2 g s fundamental formula
v2 = v1 + 2 g = v1 + 19.8 increase in velocity in 2 sec
v1^2 + 39.6 v1 + 392 - v1^2 = 2 * 9.8 * 123.1 = 2412.76
v1 = (2412.76 - 392) / 39.6 = 51.03
v2 = 51.03 + 19.6 = 70.63
T = 70.63 / .8 = 7.207 sec time to fall height of tower
S = 1/2 g T^2 = 4.9 * 7.207^2 = 254.5 m
(Note v2^2 - v1^2 = 70.63^2 - 51.03^2 = 2385 m
2385 / (2 * 9.8) = 122 m (close to 123.1 as was given
Answer:
q = 7.4 10⁻¹⁰ C
Explanation:
a) The magnetic force is given by the expression
F = q v x B
Where the blacks indicate vectors, q is the electric charge, v at particle velocity and B the magnitude of the magnetic field. If the velocity is perpendicular to the magnetic field, the sine is 1
F = q v B
Let's calculate the charge
q = F / vB
q = 1.00 10⁻¹² / 30.0 B
For the magnetic field of the earth we have a value between 25μT and 65μT, an intermediate value would be 45 μT, let's use this value.
q = 1 10⁻¹² / (30 45 10⁻⁶)
q = 7.4 10⁻¹⁰ C
b) In laboratories and modern electronics, currents of up to 1 10⁻⁶ A can be achieved without much difficulty, in advanced and research laboratories currents of up to 1 10⁻¹² can be managed. Load values (coulomb) cannot they are widely used today for work, but 1 mA = 3.6C, so we see that getting loads with the value of 10⁻¹⁰ C implies very small current less than 1 10⁻¹³ A, which only in laboratories of Very specialized can be created. Consequently, from the above it would be difficult to find loads lower than the calculated
The electrostatic charge is the one created by the friction between two surfaces, it is an indicated charge, in this case it would be possible to have better wing loads found from 10⁻¹⁰C
