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3 years ago

12

How to calculate the total current of a circuit which has both parallel and series connection

1 answer:

tankabanditka [31]3 years ago

3 0

1) using kirchhoff voltage and current law.
2) voltage and current divider rule
3) ohm's law

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2. It is to bring down the foot forcibly to the f a heavy step) with or without weight. 2 Ormain lateral position - This is done

irga5000 [103]

Im confused bc im not sure if this is a full question

5 0

3 years ago

A car is traveling at 20 meters/second and is brought to rest by applying brakes over a period of 4 seconds. What is its average

frez [133]

(u) = 20 m/s
(v) = 0 m/s
<span> (t) = 4 s
</span>
<span>0 = 20 + a(4)

</span><span>4 x a = -20
</span>
so, the answer is <span>-5 m/s^2. or -5 meter per second</span>

8 0

3 years ago

Read 2 more answers

A golf ball has a mass of 46 grams, and when struck by a driver reaches a speed of 42 m/s in a time of 0.00050 seconds. what was

Fittoniya [83]

Impulse in physics is the integral of force, F, with respect with time, t. This value is a vector quantity since force is a vector quantity as well. It can be calculated from the product of force and time. We do as follows:

Impulse = Ft
= m(a)(t)
= m(v/t)t
= 0.046 (42/0.0005) (0.0005)
= 1.932 N-s

4 0

3 years ago

A 290-turn solenoid having a length of 32 cm and a diameter of 11 cm carries a current of 0.30 A. Calculate the magnitude of the

iris [78.8K]

The magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

To find the answer, we need to know about the magnetic field inside the solenoid.

<h3>What's the expression of magnetic field inside a solenoid?</h3>

Mathematically, the expression of magnetic field inside the solenoid= μ₀×n×I

n = no. of turns per unit length and I = current through the solenoid

<h3>What's is the magnetic field inside the solenoid here?</h3>

Here, n = 290/32cm or 290/0.32 = 906

I= 0.3 A

So, Magnetic field= 4π×10^(-7)×906×0.3 = 3.4×10^(-4) T.

Thus, we can conclude that the magnitude of the magnetic field inside the solenoid is 3.4×10^(-4) T.

Learn more about the magnetic field inside the solenoid here:

brainly.com/question/22814970

#SPJ4

6 0

1 year ago

Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

shepuryov [24]

Answer:

a. $\Delta s ^2 = 8.0888 \ 10^{17} m^2$

b. $\Delta s ^2 = 3.0234 \ 10^{16} m^2$

c. $\Delta s ^2 = 3.0234 \ 10^{20} m^2$

Explanation:

The spacetime interval $\Delta s^2$ is given by

$\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2$

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

$\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2$ .

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

<h3>a.</h3>

$\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 5,625 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2$

$\Delta (c t) ^ 2 = (899,377,374 \ m)^2$

$\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2$

so

$\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2$

$\Delta s ^2 = 8.0888 \ 10^{17} m^2$

<h3>b.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2$

$\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{16} m^2$

<h3>c.</h3>

$\Delta \vec{x}^2 = (5 \ 10 \ m)^2$

$\Delta \vec{x}^2 = 2,500 m^2$

$\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2$

$\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2$

$\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2$

so

$\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2$

$\Delta s ^2 = 3.0234 \ 10^{20} m^2$

5 0

3 years ago