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3 years ago

8

Each of the following figures shows Earth with a different axis tilt. (Assume that Earth's rotation period is unchanged.) Each a

lso shows a person located in Florida (not to scale). Rank the figures based on how much time the person spends in daylight during a 24-hour period, from most to least.
- tilt = 23.5 [Florida:summer]
- tilt = 0
- tilt = 23.5
- tilt = 45
- tilt = 90

1 answer:

mart [117]3 years ago

3 0

Answer:

0º - 23.5º - 45º - 90º

Explanation:

The inclination angle of the Earth's axis with respect to its orbit around the Sun, determines that the duration enters the day and the night is different.

If it is not inclined (angle 0º) the two and niches have the same hardness

If the angle were very high (about 90º) the night would be very long, about half of the orbit period about six months.

Based on this presentation we can answer the question

The Sumo exposition from highest to lowest is

0º - 23.5º - 45º - 90º

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A charge of -3.40 nC is placed at the origin of an xy-coordinate system, and a charge of 2.45 nC is placed on the y axis at y =

gavmur [86]

Answer:

The magnitude of this force is $1.866\times10^{-4}\ N$

The direction of this force is 55.69°.

Explanation:

Given that,

Charge at origin $q_{1}= -3.40\ nC$

Charge at y axis $q_{2}= 2.45\ nC$

Distance on y axis = 4.25 cm

Third charge $q_{3}= 5.00\ nC$

Distance on x axis = 2.90 cm

(a). We need to calculate the force F₁₃

Using formula of force

$F_{13}=\dfrac{kq_{1}q_{3}}{r^2}$

Put the value into the formula

$F_{13}=\dfrac{9\times10^{9}\times(-3.40\times10^{-9})\times5.00\times10^{-9}}{(2.90\times10^{-2})^2}$

$F_{13}=-0.00018192\ N$

$F_{13}=-1.82\times10^{-4}\ N$

We need to calculate the force F₁₂

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{r^2}$

Put the value into the formula

$F_{12}=\dfrac{9\times10^{9}\times(-3.40\times10^{-9})\times2.45\times10^{-9}}{(4.25\times10^{-2})^2}$

$F_{12}=-0.00004150\ N$

$F_{12}=-4.15\times10^{-5}\ N$

The magnitude of this force is

The total force exerted on this charge by the other two charges.

$F=\sqrt{F_{13}^2+F_{12}^2+2F_{13}F_{12}\cos \theta}$

Here, $\theta=90$

$F=\sqrt{F_{13}^2+F_{12}^2}$

Put the value into the formula

$F=\sqrt{(-1.82\times10^{-4})^2+(-4.15\times10^{-5})^2}$

$F=0.0001866\ N$

$F=1.866\times10^{-4}\ N$

(c). We need to calculate the direction of this force

Using formula of direction

$\tan\theta=\dfrac{y}{x}$

Put the value into the formula

$\theta=\tan^{-1}(\dfrac{4.25}{2.90})$

$\theta=55.69^{\circ}$

Hence, The magnitude of this force is $1.866\times10^{-4}\ N$

The direction of this force is 55.69°.

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3 years ago

Which of the following is true about ALL energy conversions?

iogann1982 [59]

Answer:

d I think

Explanation:

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3 years ago

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A 15.00 kg particle starts from the origin at time zero. Its velocity as a function of time is given by = 8t2î + 5tĵ where is in

Elden [556K]

Answer:

Explanation:

I will assume the equation reads:

v = 8t²î + 5tĵ

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$x = \int\limits^t_0 {v} \, dt = \int\limits^t_0 {8t^{2}\hat i + 5t\hat j} \, dt = \frac{8}{3} t^{3} \hat i + \frac{5}{2}t^{2}\hat j |^t_0 = \frac{8}{3} t^{3} \hat i + \frac{5}{2}t^{2}\hat j - \frac{8}{3} \hat i - \frac{5}{2} \hat j\\ x = \frac{8}{3} (t^{3} - 1 )\hat i + \frac{5}{2} (t^{2} - 1 )\hat j$

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4 years ago

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vlada-n [284]

Answer:

3rd

Explanation:

3rd

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3 years ago

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velikii [3]

Answer:

$V = \frac{V_0x}{d}$

Explanation:

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I attached an image that could help to understand the representation of the field. The formula used to calculate it is given by,

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If we want to consider the change in Voltage with respect to the position then it would be,

$E= -\frac{dV}{dx}$

According to the information provided, the potential is $V_0$ and there is a distance d, therefore

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Substituting (2)

$V = -\int \frac{V_0}{d} dx$

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Where x is the height from the grounded plate.

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3 years ago