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sweet-ann [11.9K]
2 years ago
5

When Earth’s Northern Hemisphere is tilted toward the Sun during June, some would argue that the cause of our seasons is that th

e Northern Hemisphere is physically closer to the Sun than the Southern Hemisphere, and this is the primary reason the Northern Hemisphere is warmer. What argument or line of evidence could contradict this idea?
Physics
1 answer:
Ludmilka [50]2 years ago
4 0

Answer:

Distance of Earth from the Sun has nothing to do with the seasons only the tilt is responsible for the change in seasons.

Explanation:

The Earth's tilt does cause the seasons but the distance from the sun and has nothing to do with the change in seasons. In June, when the Northern Hemisphere is tilted in the direction of the Sun during the Northern Hemisphere summer the Earth is actually farthest from the Sun. In January, when the Southern Hemisphere is tilted in the direction of the Sun during the Northern Hemisphere winter the Earth is actually closest to the Sun. This is caused due to the elliptical orbit of the Earth. So, distance of Earth from the Sun has nothing to do with the seasons.

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Marrrta [24]

Answer:

Position xf is farther away from the sensor than x0, and ax is negative

Explanation:

                     Area of trapezoidal are=

                            =\frac{1}{2} *(1.5+0.75)+\frac{1}{2} (1+0.75)(-0.5)

                            =\frac{1}{2} *(2.25-1.75*0.5)

                            =0.6875 m

As the area is positive therefore displacement from xo is positive

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                                  ax=\frac{-0.5-0}{3} =-\frac{1}{6} ms2

3 0
2 years ago
The burj Khalifa in Dubai is the worlds tallest building. It rises to an amazing 828M above the ground and if you were to get to
antoniya [11.8K]
I don't understand what you mean by "depth" of the steps.  The flat part of the step has a front-to-back dimension, and the 'riser' has a height.  I don't care about the horizontal dimension of the step because it doesn't add anything to the climber's potential energy.  And if the riser of each step is 20cm high, then 3,234 of them only take him (3,234 x 0.2) = 646.8 meters up off the ground.  So something is definitely fishy about the steps.

Fortunately, we don't need to worry at all about the steps in order to derive a first approximation to the answer ... one that's certainly good enough for high school Physics.

In order to lift his bulk 828 meters from the street to the top of the Burj, the climber has to provide a force of 800 newtons, and maintain it through a distance of 828 meters.  The work [s]he does is (force) x (distance) = <em>662,400 joules. </em>
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2 years ago
Specific agricultural uses of water are all of the following except _____. evaporation growing crops raising livestock cleaning
Tamiku [17]
Evaporation.............
5 0
2 years ago
Read 2 more answers
4. A trolley of mass 2kg rests next to a trolley of mass 3 kg on a flat
nydimaria [60]

Answer:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. The total momentum of the trolleys after separation is zero

c. The momentum of the 2 kg trolley after separation is 12 kg·m/s

d. The momentum of the 3 kg trolley is -12 kg·m/s

e. The velocity of the 3 kg trolley = -4 m/s

Explanation:

a. The total momentum of the trolleys which are at rest before the separation is zero

b. By the principle of the conservation of linear momentum, the total momentum of the trolleys after separation = The total momentum of the trolleys before separation = 0

c. The momentum of the 2 kg trolley after separation = Mass × Velocity = 2 kg × 6 m/s = 12 kg·m/s

d. Given that the total momentum of the trolleys after separation is zero, the momentum of the 3 kg trolley is equal and opposite to the momentum of the 2 kg trolley = -12 kg·m/s

e. The momentum of the 3 kg trolley = Mass of the 3 kg Trolley × Velocity of the 3 kg trolley

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3 0
2 years ago
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Mekhanik [1.2K]
The initial volume of the gas is
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The pressure is constant, and it is
p=400 kPa = 400 \cdot 10^3 Pa

Therefore the work done by the gas is
W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J
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The heat energy given to the gas is
Q=+100 J

And the change in internal energy of the gas can be found by using the first law of thermodynamics:
\Delta U = Q-W = 100 J - (-160 J)=+260 J
where the positive sign means the internal energy of the gas has increased.
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2 years ago
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