What happens to the resistance of two resistors when connected in series circuit

Wire sizes are often reported using the AWG (American Wire Gauge) system in which smaller diameter wires are said to be of highe

NISA [10]

Answer:

Explanation:

24 - gauge wire , diameter = .51 mm .

Resistivity of copper ρ = 1.72 x 10⁻⁸ ohm-m

R = ρ l / s

1.72x 10⁻⁸ / [3.14 x( .51/2)² x 10⁻⁶ ]

= 8.42 x 10⁻² ohm

= .084 ohm

B ) Current required through this wire

= 12 / .084 A

= 142.85 A

C )

Let required length be l

resistance = .084 l

2 = 12 / .084 l

l = 12 / (2 x .084)

= 71.42 m

6 0

3 years ago

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1. What is the formula for the period of a pendulum and what is the main determining factor in its period?

Brut [27]

Answer:

$T=2\pi \sqrt{\frac{L}{g}}$

Explanation:

A simple pendulum is a system consisting of a mass attached to a string, and oscillating in a periodic motion, back and forth, along an equilibrium position.

The period of a pendulum is the time it takes for the pendulum to complete one oscillation.

The period of a pendulum is given by the equation

$T=2\pi \sqrt{\frac{L}{g}}$

where

L is the length of the pendulum

g is the acceleration due to gravity

From the formula, we see that the period of a pendulum does not depend on the mass.

Therefore, the only 2 factors affecting the period of a pendulum are:

- The length of the pendulum: the longer it is, the longer the period of oscillation

- The acceleration due to gravity: the greater it is, the shorter the period of the pendulum

7 0

3 years ago

According to evolutionary psychologists, our predisposition to overconsume fatty, high calorie foods illustrates that we are bio

sashaice [31]

The answer you're looking for is: reproductive success.

Hopefully this has helped! :)

3 0

3 years ago

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A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo

Rasek [7]

Answer:

$t=12.25\ seconds$

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

$x=v_ocos\theta t$

$\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}$

Where vo is the magnitude of the initial velocity, $\theta$ is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

$\displaystyle t=\frac{v_osin\theta}{g}$

There are two times where the value of y is $y_o$ when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making $y=y_o$