All categories

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

You might be interested in

A What CE describes the way energy is stored in a sandwich

SVETLANKA909090 [29]

What is Potential Energy? You probably already know that without eating, your body becomes weak from lack of energy. Take a few bites of a turkey sandwich, and moments later, you feel much better. That's because food molecules contain potential energy, or stored energy, that can do work in the future. Hope it helps

8 0

2 years ago

A boy standing on a 19.6 meter tall bridge sees a motorboat approaching the bridge at a constant speed. When the boat is 27 mete

azamat

Answer:

A. 12 m/s

Explanation:

Let’s remember that the definition of velocity is the variation of position of an object respect with to time. We know that the boy dropped the stone when the boat was 27 meters from the bridge and the stone hit the water 3 meters in front of the boat. So, the Boat must have traveled x=27 m-3m=24 m. The next step is calculating the amount of time that took the boat to make that travel; coincidentally, it is the same time that takes the stone to reach the water.

The equation that describes the motion of the stone is:

y = y_0 + v_0 * t+1/2 * a * t^2

The boy drops the stone from rest, so we can say that v_0=0. We can fixate the reference line on top of the bridge, so y_0=0 as well. The equation will be then:

-19,6 m = -1/2 * 9,8 m/s^2 * t^2

t^2= -(19,6 m)/(-4,9 m/s^2) = 4,012 s^2

t=√(4,012 s^2) = 2,003 s

Knowing the time that takes the stone to reach the water, that is the same that time that the boat uses to travel the 24 meters. The velocity of the boat is:

v = ∆x/∆t = (27 m-3 m)/(2,003 s-0s) = 11,9816 m/s ≈ 12 m/s

Have a nice day! :D

8 0

2 years ago

Over a period of more than 30 years, albert klein of california drove 2.5 × 106 km in one automobile. consider two charges, q1 =

Semenov [28]

For q3 to be in equilibrium the total force acting on it has to be zero.
Let's say that total distance traveled by car is L (this is just for the convenience).
We can set up a system of equations to find an answer. Let's say that from q1 to q3 the distance is r_1 and from q3 to q2 the distance is r_2, we know that this distance has to be equal to:
$r_1+r_2=L km$
The second equation is going to the total force acting on the charge q3:
$F_{q3}=F_{q3q1}+F_{q3q2}=0\\ 0=k_c\frac{q_1q_3}{r_1^2}+k_c\frac{q_3q_2}{r^2}$
k_c is the Coulomb's constant. Since left-hand side is zero we just divide whole equation with k_c to get rid of it:
$0=\frac{q_1q_3}{r_1^2}+\frac{q_3q_2}{r^2}$
Let's solve this for r_1^2:
$0=\frac{8}{r_1^2}+\frac{24}{r^2}\\ \frac{1}{r_1^2}=-\frac{3}{r^2}\\ r_1^2=-\frac{r^2}{3};r_2=L-r_1\\ r_1^2=\frac{(L-r_1)^2}{3}\\ r_1^2=\frac{L^2-2Lr_1+r_1^2}{3}\\ 3r_1^2=L^2-2Lr_1+r_1^2\\ 2r_1^2+2Lr_1-L^2=0$
Now we have a quadratic equation with following parameter:
$a=2\\ b=2L\\ c=-L^2$
We know that two solutions are:
$r_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{4L^2+8L^2}}{4}\\ r_{1,\:2}=\frac{-2L\pm \sqrt{12L^2}}{4}\\$
We need a positive solution. When we plug in all the numbers we get:
$r_1=0.915\cdot 10^6$km$

6 0

2 years ago

A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle ϕ1 with

strojnjashka [21]

Answer:

$x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}$

Explanation:

Let 'F₁' and 'F₂' be the forces applied by left and right wires on the bar as shown in the diagram below.

Now, the horizontal and vertical components of these forces are:

$F_{1x} = -F_1cos(\phi_1)\\F_{1y}=F_1sin(\phi_1)\\\\F_{2x}=F_2cos(\phi_2)\\F_{2y}=F_2sin(\phi_2)$

As the system is in equilibrium, the net force in x and y directions is 0 and net torque about any point is also 0. Therefore,

$\sum F_x=0\\F_{1x}=F_{2x}\\F_1cos(\phi_1)=F_2cos(\phi_2)\\\frac{F_1}{F_2}=\frac{cos(\phi_2)}{cos(\phi_1)}-------1$

Now, let us find the net torque about a point 'P' that is just above the center of mass at the upper edge of the bar.

At point 'P', there are no torques exerted by the F₁x and F₂x nor the weight of the bar as they all lie along the axis of rotation.

Therefore, the net torque by the forces $F_{1y}\ and\ F_{2y}$ will be zero. This gives,

$-F_{1y}\times x + F_{2y}(L-x) = 0\\F_{1y}\times x=F_{2y}(L-x)\\x=\frac{F_{2y}(L-x)}{F_{1y}}$

But, $F_{1y}=F_1sin(\phi_1)\ and\ F_{2y}=F_2sin(\phi_2)$

Therefore,

$x=\frac{F_2sin(\phi_2)(L-x)}{F_1sin(\phi_1)}\\\textrm{From equation (1),}\frac{F_2}{F_1}=\frac{cos(\phi_1)}{cos(\phi_2)}\\\therefore x=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times (L-x)\\x=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times L-\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}\times x\\\\$

$x(1+\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)})=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}L\\x(1+\frac{cos(\phi_1)}{sin(\phi_1)}\times \frac{sin(\phi_2}{cos(\phi_2)})=\frac{cos(\phi_1)}{cos(\phi_2)}\times \frac{sin(\phi_2}{sin(\phi_1)}L$

We know,

$tan(\phi)=\frac{sin(\phi)}{cos(\phi)}\\\\cot(\phi)=\frac{cos(\phi)}{sin(\phi)}$

∴ $x=\frac{L}{tan(\phi_1)cot(\phi_2)+1}$

6 0

2 years ago

1. Suppose a tank filled with water has a liquid column with a height of 10 meters. If the area is 2 square meters (m²), what's

Anit [1.1K]

Answers:

1. Firstly, we have to define that Pressure $P$ is Force applied $F$ per unit area $A$ . It is mathematically expressed as follows:

$P=\frac{F}{A}$ (1)

The unit of P is Pascal (Pa) which is equivalent to $\frac{kg}{ms^{2}}$ and also equivalent to $\frac{N}{m^{2} }$

There is also another expression of the Pressure in which it is dependent on the density $d$ of the liquid, the height $h$ of the container and the gravity force $g$ :

$P=d*h*g$ (2)

In this problem the liquid is water, and its known density is approximately:

$d=1000kg/m^{3}$

So, we have to substitute the values in equation (2) to obtain the pressure (Being careful with the units):

$P=1000\frac{kg}{m^{3}}*10m*9.8\frac{m}{s^{2}}$

$P=98000Pa$

Then, we have to substitute this value in equation (1) and clear $F$ :

$F=P*A$

Finally:

$F=196000N$

2. For this problem, we will use equation (1) to find the Pressure. We already know the area $A$ and the force exerted by water in the container $F$ :

$P=\frac{F}{A}=\frac{900N}{3m^{2}}$

$P=300Pa$

3. In this case, equation (2) is the perfect way to find the hydrostatic pressure at any point at the bottom of the tank (be careful with the units):

$P=d*h*g$

$P=1000\frac{kg}{m^{3}}*7.5m*9.8\frac{m}{s^{2}}$

$P=73500Pa$

4. In this case, it's important to know that in fluids (in this case the water) the higher the fluid is, the lower the pressure. Then, if $P_{1}$ and $P_{2}$ are the respective pressures at the heights $h_{1}$ and $h_{2}$ , and knowing that the water density and the gravity force in this case are constants, we can use the following expression to solve this problem: