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1 year ago

Which season is signaled by average lower temperature and indirect, angled sunlight?

1 answer:

4 0

When the earth in its translation movement has the earth's axis is further from the sun and the energy per unit area is minimal, we have the winter season

The translation is the movement of the earth around the Sun, this movement establishes the duration of the year, also the earth's axis is inclined about 23.5º in relation to the orbit.

A consequence of this inclination of the Earth's axis is that the solar rays reach the surface of the planet with different inclinations, when they arrive more perpendicularly the energy deposited by the surface unit is greater and therefore the temperature increases, this period is called Summer.

When the rays arrive with the greater inclination, the energy deposited per unit area is minimal, which is why the temperature decreases, this period is called Winter.

In the periods when the axis is almost vertical we have an average absorption of energy, these two periods are called Spring and Autumn.

In conclusion, the winter season occurs when the terrestrial axis is furthest from the sun and the energy per unit of area is minimal.

Learn more about the seasons here:

brainly.com/question/20435793

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All of these are examples of pseudopsychology EXCEPT:

jeyben [28]

Answer:

psychoanalysis

Explanation:

psychoanalysis a system of psychological theory and therapy which aims to treat mental disorders by investigating the interaction of conscious and unconscious elements in the mind and bringing repressed fears and conflicts into the conscious mind by techniques such as dream interpretation and free association.

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2 years ago

Read 2 more answers

Two small aluminum spheres, each of mass 0.0250 kilograms, areseparated by 80.0 centimeters.

sineoko [7]

Answer:

Total number of electrons

$N = 7.25 \times 10^{24}$

electrons removed from each sphere

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = 7.27 \times 10^{-10}$

Explanation:

As we know that moles is defined as

$n =\frac{mass}{molar mass}$

$n = \frac{0.0250}{0.026982}$

$n = 0.93$

so number of atoms of Al in each sphere is given as

$N = 0.93(6.02 \times 10^{23})$

$N = 5.58 \times 10^{23}$

Now number of electrons in each atom is given as

atomic number = number of electrons in each atom = 13

total number of electrons in each sphere is

$N = 13 \times (5.58 \times 10^{23})$

$N = 7.25 \times 10^{24}$

Also we know that force of attraction between them is given as

$F= \frac{kq_1q_2}{r^2}$

$1.00 \times 10^4 = \frac{(9\times 10^9)q^2}{0.80^2}$

$q = 8.4 \times 10^{-4} C$

now we have

$q = Ne$

$8.4 \times 10^{-4} = N(1.6 \times 10^{-19}$

$N = \frac{(8.4 \times 10^{-4})}{1.6 \times 10^{-19}}$

$N = 5.27 \times 10^{15}$

Fraction of electrons transferred is given as

$f = \frac{5.27 \times 10^{15}}{7.25 \times 10^{24}}$

$f = 7.27 \times 10^{-10}$

6 0

2 years ago

A 0.45 Caliber bullet (m = 0.162 kg) leaving the muzzle of a gun at 860

Charra [1.4K]

Answer:139.32kgm/s

Explanation:

Mass(m)=0.162kg

Velocity(v)=860m/s

Momentum=mass x velocity

Momentum=0.162 x 860

Momentum=139.32kgm/s

3 0

2 years ago

A projectile has an initial horizontal velocity of 15 meters per second and an initial vertical velocity of 25 meters per second

Artyom0805 [142]

Answer:

75 m

Explanation:

The horizontal motion of the projectile is a uniform motion with constant speed, since there are no forces acting along the horizontal direction (if we neglect air resistance), so the horizontal acceleration is zero.

The horizontal component of the velocity of the projectile is

$v_x = 15 m/s$

and it is constant during the motion;

the total time of flight is

t = 5 s

Therefore, we can apply the formula of the uniform motion to find the horizontal displacement of the projectile:

$d= v_x t =(15 m/s)(5 s)=75 m$

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2 years ago

3. A 4.1 x 10-15 C charge is able to pick up a bit of paper when it is initially 1.0 cm above the paper. Assume an induced charg

Anni [7]

Answer:

$\mathbf{1.51\times10^{-15}N}$

Explanation:

The computation of the weight of the paper in newtons is shown below:

On the paper, the induced charge is of the same magnitude as on the initial charges and in sign opposite.

Therefore the paper charge is

$q_{paper}=-4.1\times10^{-15}C$

Now the distance from the charge is

$r=1cm=0.01m$

Now, to raise the paper, the weight of the paper acting downwards needs to be managed by the electrostatic force of attraction between both the paper and the charge, i.e.

$mg=\frac{k_{e}q_{1}q_{2}}{r^{2}}$

$\Rightarrow W=mg$

$=\frac{9\times10^{9}\times(4.1\times10^{-15})^{2}}{0.01^{2}}$

$=\mathbf{1.51\times10^{-15}N}$

6 0

2 years ago