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3 years ago

What is the “lag of seasons”?

Physics

1 answer:

Degger [83]3 years ago

3 0

The bottom figure shows the temperatures recorded at the location during this period. The important point is that time of maximum temperature is usually later than the time of maximum insolation. This time difference constitutes the lag of seasons. Usage: Use "Play" to start and "Stop" to halt.

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A student bangs a brick at the head of a table. Three students are positioned equal distance from the head with their hands on t

iris [78.8K]

Explanation:

that the people closer too the head of the table will feel more vibrations than the people at the end of the table. since the vibrations will slow down as they travel farther down the table

Hope this helps!!

5 0

3 years ago

At the top of a pole vault, an athlete actually can do work pushing on the pole before releasing it. Suppose the pushing force t

Salsk061 [2.6K]

Answer:

The work done on the athlete is approximately 2.09 J

Explanation:

From the definition of the work done by a variable force:

$\displaystyle{\int_{x_i}^{x_j}F(x)dx}$

and substituting with the function of our problem:

$\displaystyle{\int_{0}^{0.19}(140x-190x^2)dx\approx2.09\mathrm{J}}$

5 0

3 years ago

An unstable atomic nucleus has a mass of 17.010-27kg, and starts out at rest. When it decays, it the original nucleus disintegra

slega [8]

Answer:

Part a)

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

$E = 4.4 \times 10^{-13} J$

Explanation:

As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same

So we will have

$m_1v_1 + m_2v_2 + m_3v_3 = 0$

$(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0$

$(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0$

$v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s$

Part b)

By equation of kinetic energy we have

$E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2$

$E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}$

$E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}$

$E = 4.4 \times 10^{-13} J$

8 0

3 years ago

A negatively charged rod is placed near two neutral metal spheres, as shown below. Which statements describe the charging method

grin007 [14]

<h3>Answer;</h3>

The spheres develop opposite charges. 
Electrons move from Sphere A to Sphere B. 
The spheres are charged through induction.

<h3>Explanation;</h3>

When a negatively charged rod is placed near two neutral metal spheres, the spheres will develop opposite charges, because the neutral metal spheres have both negative and positive charges. From the basic law of electrostatics unlike charges attracts and like charges repel.
Thus, the sphere will develop opposite charges, electrons will move from Sphere A to sphere B, hence we say that the spheres will be charged by induction such that sphere A will acquire a positive charge while sphere B will acquire negative charge.

3 0

3 years ago

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Soapy water between surfaces

Inessa [10]

Answer:

A.increases friction

Explanation:

6 0

2 years ago

Read 2 more answers