1. Which science will discuss how matter is created throughout the universe?

What yall up to yall wanna hit

gayaneshka [121]

Answer:

umm what?

Explanation:

3 0

3 years ago

Read 2 more answers

A certain capacitor, in series with a resistor, is being charged. At the end of 10 ms its charge is half the final value. The ti

vichka [17]

To solve this problem we will apply the expression of charge per unit of time in a capacitor with a given resistance. Mathematically said expression is given as

$q(t) = e^{-\frac{t}{(R*C)}}$

Here,

q = Charge

t = Time

R = Resistance

C = Capacitance

When the charge reach its half value it has passed 10ms, then the equation is,

$\frac{1}{2}*q_{final} = e^{-\frac{0.01}{(R*C)}}$

$Ln(\frac{1}{2}) = -\frac{0.01}{RC}$

$- RC = \frac{0.01}{Ln(1/2)}$

$RC = 0.014s$

We know that RC is equal to the time constant, then

$T = RC = 0.014s = 14ms$

Therefore the time constant for the process is about 14ms

5 0

3 years ago

A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the gen

Zinaida [17]

Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

$\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s$

$N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns$

Therefore, the number of turns of wire needed is 573.8 turns

4 0

3 years ago

A small metal ball is given a negative charge, then brought near to end a of the rod (figure 1). What happens to end a of the ro

erma4kov [3.2K]

What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.

<h3>Electrostatics</h3>

I have attached the image of the rod.

We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.

This means that their fields will cancel.

Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.

This also applies to a strong conducting rod and therefore it is strongly attracted.

Read more about Electrostatics at; brainly.com/question/18108470

7 0

2 years ago

Fill in the blank .Technically color consists of electromagnetic _____

8_murik_8 [283]

Electromagnetic radiation

5 0

3 years ago