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2 years ago

Which is one factor that contributes to the formation of polar, temperate, and tropical zones?

Physics

1 answer:

Tju [1.3M]2 years ago

4 0

Answer:

The angle of the Sun's rays:

At about June 21 the sun is directly overhead at about 23 deg north of the equator

On about Dec 21 the sun is directly overhead at about 23 deg south of the equator

It is this tilt 23 deg to the ecliptic (path of earth about sun) that causes the major temperature changes on earth

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If a 1-megaton nuclear weapon is exploded

Elenna [48]

Answer:

1,780,000 N

Explanation:

0.2 atm × (1.013×10⁵ Pa/atm) = 20,260 Pa

Force = pressure × area

F = 20,260 Pa × (3.89 m × 22.6 m)

F = 1,780,000 N

4 0

2 years ago

You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi

Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0

3 years ago

A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.

ValentinkaMS [17]

A) $2.03 m/s^2$

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

$mg sin \theta - \mu_k R = ma$ (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

$\theta=22.5^{\circ}$

$\mu_k = 0.19$ is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

$R-mg cos \theta =0$ (2)

From (2) we find

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_k mg cos \theta = ma$

Solving for a,

$a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2$

B) 5.94 m/s

We can solve this part by using the suvat equation

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

$a=2.03 m/s^2$

s = 8.70 m

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s$

6 0

3 years ago

Which of the following is NOT considered electromagnetic radiation? x-rays radio waves cosmic rays ultraviolet light

OlgaM077 [116]

Answer:

Cosmic ray is not considered electromagnetic radiation.

Explanation:

Electromagnetic radiation refers to

It is Radiation that has both electric and magnetic fields and travels in waves.
Electromagnetic radiation can vary in strength from low energy to high energy.

X rays are the rays produced when a negatively charged electrode is heated by electricity and electrons are released, thereby producing energy. It is a type of radiation called EM waves.

Radio wave are wave from the portion of the electromagnetic spectrum at lower frequencies than microwaves. It is an EM wave.

Ultraviolet wave Invisible rays that are part of the energy that comes from the sun. It is an example of EM wave.

Cosmic ray is a high-speed particle either an atomic nucleus or an electron that travels through space.

Cosmic ray is not an electromagnetic wave.

Hence

Cosmic ray is not considered electromagnetic radiation

learn more about electromagnetic radiation here:

<u>brainly.com/question/13695751</u>

#SPJ4

4 0

1 year ago

Concrete colums are constructed with reinforcing steel in them to make them stronger and more ductile. The reinforcing bars are

Sergio039 [100]

Answer:

$21678.47223\ lbf-in^2$

$383.1109\ lbf-in^2$

Explanation:

d = Diameter of column = 0.5 inch

$A_c$ = Area of concrete = $119.4\ in^2$

The strain in the system is conserved

$\dfrac{F_sL}{A_sE_s}=\dfrac{F_cL}{A_cE_c}\\\Rightarrow F_c=\dfrac{F_sA_cE_c}{A_sE_s}\\\Rightarrow F_c=\dfrac{F_s \times 119.4\times 4.1\times 10^6}{8\times \dfrac{\pi \dfrac{1}{2^2}}{4}\times 29\times 10^6}\\\Rightarrow F_c=10.74658F_s$