A two-dimensional section of the body is known as an

Why is space black if the sun is shining

faltersainse [42]

Answer:One star can't light up a whole universe

Explanation:It is like saying one light can feel up the whole town which it can't do.

5 0

3 years ago

In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs

bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

$V=\frac{I}{nAq}$

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

$\frac{I}{q} = 1.2*10^{18}$

$A= 1.3*10^{-8}m^2$

$n=6.3*10^{28} e/m^3$

$\omicron{O} = 1.2*10^{-4}(m/s)(N/c)$ Mobility

We can find the drift velocity replacing,

$V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}$

$V= 1.465*10^-3m/s$

The electric field is given by,

$E= \frac{V}{\omicron{O}}$

$E=\frac{1.465*10^-3}{1.2*10^{-4}}$

$E=12.2V/m$

7 0

4 years ago

How does the sun's gravity and the earth inertia keep us orbiting in the solar system

scZoUnD [109]

<span>Inertia keeps us orbiting because any object with mass has the tendency to resist changes to their direction and speed of movement. Combine that with the interaction of the gravitational attraction of the sun, and that is what keeps Earth in orbit. The sunâ€™s gravitational force is one that is proportional to Earthâ€™s mass, and it acts in a way that is almost exactly perpendicular to Earthâ€™s motion. This keeps Earth from spinning into the sun or far away from it.</span>

6 0

4 years ago

In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 5.0-m-diameter hollow steel cylinder with their b

lesya692 [45]

Answer:2.55 rad/s

Explanation:

Given

Diameter of ride=5 m

radius(r)=2.5 m

Static friction coefficient range=0.60-1

Here Frictional force will balance weight

And limiting frictional force is provided by Centripetal force

$f=\mu N=\mu m\omega ^2\cdot r$

weight of object=mg

Equating two

f=mg

$\mu m\omega ^2\cdot r=mg$

$\omega ^2=\frac{g}{\mu r}$

$\omega =\sqrt{\frac{g}{\mu r}}$

$\omega =2.55 rad/sec$

8 0

3 years ago

An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider compl

pickupchik [31]

Answer:

a. 2.1 s

b.0.48 Hz

c. A=24cm

d. 72cm/s

Explanation:

An air-track glider attached to a spring oscillates between the 10.0 cm mark and the 57.0 cm mark on the track. The glider completes 15.0 oscillations in 31.0 s.What are the (a) period, (b) frequency, (c) amplitude, and (d) maximum speed of the glider?

What are the period,

period is the time taken for a wave particle to make one complete oscillation

a) 31 / 15 = 2.066 seconds

= 2.1 s

(b) frequency : this the number of oscillation made in one seconds.

it is also the inverse of the period.

= oscillations / time

= 15/31= 0.48 Hz

(c) amplitude : maximum displacement from the origin

amplitude = 1/2 of the difference of oscillation marks

= 1/2(57-10) = 47/2cm

23.5cm

A=24cm

(d) maximum speed of the glider?

V=ωA

angular frequency *Amplitude

V=a*pi*f*amplitude

2π x frequency x amplitude = maximum speed

= 2π x .48 x 24

=72.38 cm/s

72cm/s

5 0

3 years ago