HNesmith16

04.08.2019 • 

Physics

In a simple electric circuit, ohm's law states that v=irv=ir, where vv is the voltage in volts, ii is the current in amperes, and rr is the resistance in ohms. assume that, as the battery wears out, the voltage decreases at 0.030.03 volts per second and, as the resistor heats up, the resistance is increasing at 0.040.04 ohms per second. when the resistance is 200200 ohms and the current is 0.040.04 amperes, at what rate is the current changing?

Ответ:

smallblueberry

16.03.2022

Physics

---

We take the derivative of Ohm's law with respect to time: V = IR
Using the product rule:
dV/dt = I(dR/dt) + R(dI/dt)
We are given that voltage is decreasing at 0.03 V/s, resistance is increasing at 0.04 ohm/s, resistance itself is 200 ohms, and current is 0.04 A. Substituting:
-0.03 V/s = (0.04 A)(0.04 ohm/s) + (200 ohms)(dI/dt)
dI/dt = -0.000158 = -1.58 x 10^-4 A/s

---

0,0(0 оценок)

Ответ:

GrainLeaf

11.12.2020

Physics

---

245 Hz

Explanation:

Speed of sound in air = 343 m/s = v

Frequency of trumpet = f = 262 Hz

Speed of bike = 11.5 m/s =

Applying doppler effect

For the trombone moving towards it

So, frequency of the trombonist is 245 Hz

---

0,0(0 оценок)