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Zolol [24]
2 years ago
6

We know that the earth's axis is tilted 23 ½ degrees. On or about June 21 or 22 each year, the summer solstice occurs for those

of us who live in the Northern Hemisphere. On this particular day, the sun's rays are striking earth directly, or at 90 degrees, at a latitude approximately 23 ½ degrees north of the equator. If you were at 23 1/2 degrees north latitude on that day, and you looked up at noon, the sun would be directly overhead.
Imagine you have taken a flight north to New York City. Describe the sun rays at exactly noon on the same day of the year.

A) The sun's rays would also be directly overhead. The sun's rays strike earth at 90 degrees throughout the hemisphere.

B) The sun's rays would also be directly overhead but only for one hour, twelve noon until one o'clock. After that the earth's movement distorts the angle of the sun's rays.

C) The sun's rays will be directly overhead in the Northern Hemisphere during the fall equinox. The movement of earth changes the angle of incidence of the sun's rays throughout the year.

D) The sun's rays will never be directly overhead. The latitude of 23 ½ degrees north is known as the Tropic of Cancer. Above this imaginary line the sun's rays hit earth with decreased angles.
Physics
2 answers:
Dima020 [189]2 years ago
8 0
<span>D) The sun's rays will never be directly overhead. The latitude of 23 ½ degrees north is known as the Tropic of Cancer. Above this imaginary line the sun's rays hit earth with decreased angles.</span>
arlik [135]2 years ago
8 0

Answer:

Option (D)

Explanation:

The path in which the sun moves across the sky usually varies from one season to another. This path is directly related to the position of sunrise and sunset, as the position of sunrise and sunset changes with the changing position of the sun. Even though the sun appears to be at the highest position during the winter solstice where the sun is at the maximum height and receives the highest amount of daylight, the sun is never directly overhead in the state of New York.

Thus, the rays of the sun never falls directly overhead in this region. At this condition, i.e above the tropic of cancer, the sun's rays hit the earth with gradually decreased angles.

Hence, the correct answer is option (D).

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Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.
m_a_m_a [10]

Answer:

F = 10.788 N

Explanation:

Given that,

Charge 1, q_1=6\ \mu C=6\times 10^{-6}\ C

Charge 2, q_2=2\ \mu C=2\times 10^{-6}\ C

Distance between charges, d = 0.1 m

We know that there is a force between charges. It is called electrostatic force. It is given by :

F=\dfrac{kq_1q_2}{d^2 }\\\\F=\dfrac{8.99\times 10^9\times 6\times 10^{-6}\times 2\times 10^{-6}}{(0.1)^2 }\\\\F=10.788\ N

So, the force applied between charges is 10.788 N.

3 0
2 years ago
Read 2 more answers
Most of the nutrients in the rainforest ecosystem are in the _____.
joja [24]
<span>The answer should be the vegitation. </span>
4 0
2 years ago
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A 125-g metal block at a temperature of 93.2 °C was immersed in 100. g of water at 18.3 °C. Given the specific heat of the metal
Nataly_w [17]

Answer:

34.17°C

Explanation:

Given:

mass of metal block = 125 g

initial temperature T_i = 93.2°C

We know

Q = m c \Delta T   ..................(1)

Q= Quantity of heat

m = mass of the substance

c = specific heat capacity

c = 4.19 for H₂O in J/g^{\circ}C

\Delta T = change in temperature

Now

The heat lost by metal = The heat gained by the metal

Heat lost by metal = 125\times 0.9\times (93.2-T_f)

Heat gained by the water = 100\times 4.184\times(T_f -18.3)

thus, we have

125\times 0.9\times (93.2-T_f) = 100\times 4.184\times(T_f -18.3)

10485-112.5T_f = 418.4T_f - 7656.72

⇒ T_f = 34.17^oC

Therefore, the final temperature will be = 34.17°C

6 0
2 years ago
Nicki rides her bike at a constant speed for 6 km. That part of her ride takes her 1 h. She then rides her bike at a constant sp
Savatey [412]

km x h = km/h

First trial: 6 x 1 = 6km/h

Second trial: 9 x 2 = 18km/h

6 + 18 = <u>24km/h</u> (Total)

Or

6 + 9 = 15 km

2 + 1 = 3h

15 + 3 = 18

15 x 2 = 30

3 x 2 = 6

30 - 6 = <u>24km/h</u>

8 0
2 years ago
A 50-kg load is suspended from a steel wire of diameter 1.0 mm and length 11.2 m. By what distance will the wire stretch? Young'
lbvjy [14]

Answer:

3.5 cm

Explanation:

mass, m = 50 kg

diameter = 1 mm

radius, r = half of diameter = 0.5 mm = 0.5 x 10^-3 m

L = 11.2 m

Y = 2 x 10^11 Pa

Area of crossection of wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3  

                                              = 7.85 x 10^-7 m^2

Let the wire is stretch by ΔL.

The formula for Young's modulus is given by

Y =\frac{mgL}{A\Delta L}

\Delta L =\frac{mgL}{A\times Y}

ΔL = 0.035 m = 3.5 cm

Thus, the length of the wire stretch by 3.5 cm.

5 0
2 years ago
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