This is a statement but yes a star forms inside nebulae which are gigantic clouds of gas. stars form inside as the gases own gravity pulls it together after which it becomes large enough to perform fusion and become a star.

3 0

3 years ago

Importance of electric circuit

Feliz [49]

Answer:

It is important because it carries useful energy through your house that you can use for a variety of tasks.

Explanation:

Hope this helped !

8 0

3 years ago

2. A person began running due east and covered 15 kilometers in 2.0 hours. What is the average velocity of the person?

Angelina_Jolie [31]

Answer:

7.5 km/h (2.1 m/s) due east

Explanation:

The average velocity of the person is given by:

$v=\frac{d}{t}$

where

d is the displacement

t is the time taken

In this problem,

d = 15 km is the displacement

t = 2.0 h is the time elapsed

so the average velocity is

$v=\frac{15 km}{2.0 h}=7.5 km/h$

and the direction is the same as the displacement (east).

We can also convert the velocity into SI units (m/s). We have:

d = 15 km = 15,000 m

t = 2.0 h * 3600 s/h = 7200 s

$v=\frac{15,000 m}{7200 s}=2.1 m/s$

4 0

3 years ago

Read 2 more answers

A 2.0 µF capacitor is charged through a 50,000 ohm resistor. How long does it take for the capacitor to reach 90% of full charge

Nesterboy [21]

Answer:

0.23 s

Explanation:

First of all, let's find the time constant of the circuit:

$\tau=RC$

where

$R=50,000 \Omega$ is the resistance

$C=2.0\mu F=2.0\cdot 10^{-6}F$ is the capacitance

Substituting,

$\tau=(50,000 \Omega)(2.0\cdot 10^{-6}F)=0.1 s$

The charge on a charging capacitor is given by

$Q(t)=Q_0 (1-e^{-t/\tau} )$ (1)

where

$Q_0$ is the full charge

we want to find the time t at which the capacitor reaches 90% of the full charge, so the time t at which

$Q(t)=0.90 Q_0$

Substituting this into eq.(1) we find

$0.90 Q_0 = Q_0 (1-e^{-t/\tau})\\0.90=1-e^{-t/\tau}\\e^{-t/\tau}=1-0.90=0.10\\-\frac{t}{\tau}=ln(0.10)\\t=-\tau ln(0.10)=(0.1 s)ln(0.10)=0.23 s$

4 0

3 years ago