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3 years ago

5

The diagram below shows the relative positions of Earth and the Sun at a certain time of the year. Based on the diagram, which s

eason is occurring in the Southern Hemisphere of Earth?
Question 4 options:

Fall

Winter

Spring

Summer

1 answer:

AlladinOne [14]3 years ago

7 0

Answer:

summer

Explanation:

Notice the higher density of the rays of the sun hitting straight the latitudes below the equator.

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Why does gravity on Earth have a stronger attraction with you than the sun has with you? Plsssssss help I really need this rn

n200080 [17]

Answer:

Although the Earth has a lesser mass than the Sun, it is far closer to you, allowing for a stronger pull.

Explanation:

5 0

3 years ago

You observe a distant galaxy. You find that a spectral line of hydrogen that is shifted from its normal location in the visible

dimulka [17.4K]

Answer:

The galaxy is moving away from us

Explanation:

Galaxy refers to a system of millions or billions of stars accompanied by gas and dust such that they are held together as a result of gravitational attraction.

When we observe a distant galaxy, we find that a spectral line of hydrogen that is shifted from its normal location in the visible part of the spectrum into the infrared part of the spectrum.

It means that the galaxy is moving away from us.

6 0

3 years ago

A whale swims due east for a distance of 6.9 km, turns around and goes due west for 1.8 km, and finally turns around again and h

xeze [42]

The question is incomplete. Here is the complete question:

A whale swims due east for a distance of 6.9 km, turns around and goes due west for 1.8 km and finally turns around again and heads 3.7 km due east. (a) What is the total distance traveled by the whale? (b) What are the magnitude and direction of the displacement of the whale?

Answer:

(a) Distance = 12.4 km

(b) Displacement = 8.8 km due east

Explanation:

Consider east direction as positive and west direction as negative.

Given:

The motion is along the east-west line.

The whale first swims 6.9 km due east, then 1.8 km due west and again 3.7 km due east.

(a)

Distance traveled by the whale is equal to the sum of the lengths of all the distances traveled. Therefore,

Distance traveled = 6.9 km + 1.8 km + 3.7 km = 12.4 km

Therefore, the distance traveled by the whale is 12.4 km.

(b)

Displacement of the whale is given by considering the sign of each of the individual displacements. Therefore,

Displacement of the whale = (+6.9 km) + (-1.8 km) + (+3.7 km)

Displacement of the whale = 6.9 km - 1.8 km + 3.7 km = 8.8 km

The answer is positive. So, the direction is due east.

Therefore, the displacement of the whale is <u>8.8 km due east.</u>

6 0

3 years ago

A pendulum is made by letting a 4 kg mass swing at the end of a string that has a length of 1.5 meter. The maximum angle that th

olga nikolaevna [1]

Answer:

Approximately $7.8\; \rm J$ .

Explanation:

The change in the gravitational potential energy of the pendulum is directly related to the change in its height.

Refer to the sketch attached. The pendulum is initially at $\rm P_2$ . Its highest point is at $P_1$ . The length of segment $\rm BP_2$ gives the change in its height.

The lengths of $\rm AP_1$ and $\rm AP_2$ are simply the length of the string, $1.5\; \rm m$ . To find the length of $\rm BP_2$ , start by calculating the length of $\rm AB$ .

$\rm AB$ forms a leg in the right triangle $\rm \triangle AP_1B$ . Besides, it is adjacent to the $30^\circ$ angle $\rm P_1\hat{A}B$ . Its length would be:

$\rm AB = 1.5 \times \cos(30^\circ) \approx 1.30\; \rm m$ .

The length of $\rm BP_2$ would thus be

$\rm BP_2 = AP_2 - AB = 1.5 - 1.30 \approx 0.20\; \rm m$ .

The change in gravitational potential energy can be found with the equation

$\Delta \mathrm{GPE} = m \cdot g \cdot \Delta h$ . In this equation,

$m$ is the mass of the object,
$g \approx 9.81\; \rm N \cdot kg^{-1}$ near the surface of the earth, and
$\Delta h$ is the change in the object's height.

In this case, $m = 4\; \rm kg$ and $\Delta h \approx 0.20\; \rm m$ . Therefore:

$\Delta \mathrm{GPE} = 4 \times 9.81 \times 0.20 \approx 7.8\; \rm J$ .

6 0

3 years ago

A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a d

zvonat [6]

Answer:

<u /> $D_l=d$ <u />

Explanation:

From the question we are told that:

The Electric field of strength direction =Right

The Velocity of The First Electron=V_0

The Velocity of The Second Electron=V_0

Therefore

$V_{e1}=V_{e2}$

Generally, the equation for the Horizontal Displacement of electron is mathematically given by

$D=\frac{at^2}{2}$

Where

Acceleration is given as

$a=\frac{V_o}{2d}$

And

Time

$T=\frac{d}{v_0}$

Therefore horizontal displacement towards the left is

$D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}$

<u /> $D_l=d$ <u />

5 0

3 years ago