When thinking about an electric circuit, you usually focus not on the motion of individual charges, but rather on the continuous current (charge per unit time) flowing through the circuit. Thus, rather than considering the work done on a particular charge, it is useful to compute the work done per unit time on the charge flowing through the circuit, or in other words, the power. Find the electrical power P delivered to the resistor via the work done on the individual charges passing through it. (Again, this power ultimately appears in the form of heat).

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Ответ:

GaudySky

24.07.2022

Physics

v = 26.94 m/s

Explanation:

given,

angle of inclination = 30°

mass of the sleigh = 200 kg

coefficient of kinetic friction = 0.2

height of inclination = 10 m

pull force be = 5000 N

now,.

$T - f_s - mg sin \theta = m a$

$T - \mu N - mg sin \theta = m a$

$T - \mu mg - mg sin \theta = m a$

$a = \dfrac{T}{m} - \mu g - g sin \theta$

$a = \dfrac{5000}{200} - \0.2\times 9.8 - 9.8 \times sin 30^0$

$a = 18.14\ m/s^2$

$L = \dfrac{10}{sin 30}$

L = 20 m

v² = u² + 2 as

v² = 0 + 2 x 18.14 x 20

v = 26.94 m/s

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