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3 years ago

Give reason The SI system of unit is better than the MKS system

Physics

1 answer:

egoroff_w [7]3 years ago

6 0

Answer:

The SI was and is intended to extend and refine the definitions used by the MKS.

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While an emf source supplies energy to a circuit, that energy is dissipated when the current passes through resistance. In deali

fredd [130]

The question is incomplete, the options are;

RI^2

I^2/R

R/I^2

R/V^2

RV^2

V^2/R

VIR

Select all that apply

Answer:

P=RI^2

P=V^2/R

P=VI

Explanation:

Power is the rate at which energy is changing in a circuit. It is shown by the formulas outlined above from the group of answer choices. Since the current (I), voltage (V), and resistance (R) were mentioned in the question, any of three three formulas could be used to obtain the power drawn by the conductor.

7 0

3 years ago

A 0.20 kg baseball is traveling at 40 m/s toward the batter. The ball is hit by the bat with a force of 200N, and is

Paraphin [41]

Answer:

Time, t = 0.015 seconds.

Explanation:

Given the following data;

Mass, m = 0.2kg

Force, F = 200N

Initial velocity, u = 40m/s

Final velocity, v = 25m/s

To find the time;

Ft = m(v - u)

Time, t = m(v - u)/f

Substituting into the equation, we have;

Time, t = 0.2(25 - 40)/200

Time, t = 0.2(-15)/200

Time, t = 3/200

Time, t = 0.015 seconds.

Note: We ignored the negative sign because time can't be negative.

8 0

3 years ago

A capacitor with initial charge q0 is discharged through a resistor. a) In terms of the time constant τ, how long is required fo

-BARSIC- [3]

Answer:

It would take $\tau(\ln 9 - \ln 8)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{8}{9} \, q_0$ .

It would take $\tau(\ln 9 - \ln 7)$ time for the capacitor to discharge from $q_0$ to $\displaystyle \frac{7}{9}\, q_0$ .

Note that $\ln 9 = 2\,\ln 3$ , and that $\ln 8 = 3\, \ln 2$ .

Explanation:

In an RC circuit, a capacitor is connected directly to a resistor. Let the time constant of this circuit is $\tau$ , and the initial charge of the capacitor be $q_0$ . Then at time $t$ , the charge stored in the capacitor would be: