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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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A 6.60-kg block slides with an initial speed of 1.56 m/s up a ramp inclined at an angle of 28.4° with the horizontal. The coeffi

Vlad [161]

Answer:

The distance travel by block before coming to rest is 0.122 m

Explanation:

Given:

Mass of block $m = 6.60$ kg

Initial speed of block $v _{i} = 1.56$ $\frac{m}{s}$

Final speed of block $v_{f} = 0$ $\frac{m}{s}$

Coefficient of kinetic friction $\mu _{k} = 0.62$

Ramp inclined at angle $\theta =$ 28.4°

Using conservation of energy,

Work done by frictional force is equal to change in energy,

$\mu _{k} mgd \cos 28.4 = \Delta K - \Delta U$

Where $\Delta U = mg d\sin 28.4$

$\mu _{k} mgd \cos 28.4 = \frac{1}{2}mv_{i} ^{2} - mgd\sin 28.4$

$\mu _{k} mgd \cos 28.4 +mgd\sin 28.4 = \frac{1}{2}mv_{i} ^{2}$

$d(6.60 \times 9.8 \times 0.62 \times 0.879 + 6.60 \times 9.8 \times 0.475) = \frac{1}{2} \times 6.60 \times (1.56)^{2}$

$d = 0.122$ m

Therefore, the distance travel by block before coming to rest is 0.122 m

7 0

2 years ago

A plane has an average air speed (this is the speed the plane moves through air) of 750 mph. The plane flies a route of 5000 mil

Digiron [165]

Answer:

6 hours 15 minutes

Explanation:

On the trip from L.A. to London, the plane travels at 750 mph against a headwind of 50 mph, and that makes the net 700 mph (in aviation speak, 750 is the airspeed, while 700 is the groundspeed). 5000 miles divided by 700 mph results in about 7.14 hours, or about 7 hours and 9 minutes. On the return trip, ASSUMING THE SAME WIND, the plane travels at 750 mph, but this time the wind of 50 mph is a tail wind. So the net (groundspeed) is 800 mph. Traveling 5000 miles at 800 mph only takes 6.25 hours, or 6 hours and 15 minutes.

Outbound flight 7 hours 9 minutes

Return flight 6 hours 15 minutes

6 0

1 year ago

Two vertical springs have identical spring constants, but one has a ball of mass m hanging from it and the other has a ball of m

OverLord2011 [107]

To solve this problem we will start from the definition of energy of a spring mass system based on the simple harmonic movement. Using the relationship of equality and balance between both systems we will find the relationship of the amplitudes in terms of angular velocities. Using the equivalent expressions of angular velocity we will find the final ratio. This is,

The energy of the system having mass m is,

$E_m = \frac{1}{2} m\omega_1^2A_1^2$