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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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Consider as a system the gas in a vertical cylinder; the cylinder is fitted with a piston on which a number of small weights are
Amiraneli [1.4K]

Answer:

Explanation:

a ) At constant pressure , work done = P x Δ V

= 200 x 10³ x ( .1 - .04 )

= 12 x 10³ J .

b )

At constant temperature work done

= n RT ln v₂ / v₁

= PV ln v₂ / v₁

= 200 x 10³ x .04 ln .1 / .04

8 x 10³ x .916

= 7.33 x 10³ J .

5 0
1 year ago
Consider a person standing in an elevator that is moving at constant speed upward. The person, of mass m, has two forces acting
larisa [96]

Answer:

The weight of the person has a smaller magnitude.

Explanation:

For an observer in inertial frame of reference for the person in the elevator Newton's Second Law can be written as

Normal reaction acts upwards

Weight acts downwards

\sum F_{v}=ma_{v}\\\\N-mg=m\times a_{v}\\\\m\times a_{v}> 0\\\\\therefore N-mg> 0\\\\\therefore N> mg

Here

N is the normal reaction force

mg is the weight of the person

g is acceleration due to gravity

4 0
2 years ago
Read 2 more answers
A piano wire has a length of 81 cm and a mass of 2.0
choli [55]
<span>Frequency = 394 Hz
 Length of the string L = 81 cm = 0.81 m
 Mass of the string = 0.002 kg
 Tension T = ?
 Wave length of the string is two times the length.
  n x lambda = 2L, we also have lambda = vt = v / f, t is time period and given n = 1.
  Therefore L = v / 2f => v = 2fL
 Deriving form force equation, force here is tension T so
  v = squareroot of (TL/m) hence
   2fL = squareroot of (TL/m) => 4 x f^2 x L^2 = (T x L) / m => T = 4 x f^2 x L x m
 T = 4 x 0.81 x (394)^2 x 0.002 = 4 x 0.81 x 155236 x 0.002
 T = 1005.9 N = 1.006 x 10^3 N</span>
4 0
2 years ago
A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15A. If it is lying flat on the gro
grin007 [14]

Answer:

torque is 1.7 * 10^{-2} Nm

Explanation:

Given data

turns n = 1000 turns

radius r  = 12 cm

current I = 15A

magnitude B = 5.8 x 10^-5 T

angle θ = 25°

to find out

the torque on the loop

solution

we know that torque on the loop is

torque = N* I* A*B* sinθ

here area A = πr² = π(0.12)²

put all value

torque = N* I* A*B* sinθ

torque = 1000* 15* π(0.12)² *5.8 x 10-5 * sin25

torque = 0.0166 N m

torque is 1.7 * 10^{-2} Nm

5 0
2 years ago
A very long, straight wire has charge per unit length 3.50×10^−10 C/m . At what distance from the wire is the electricfield magn
Dafna11 [192]

Answer:

r= 2.17 m

Explanation:

Conceptual Analysis:

The electric field at a distance r from a charge line of infinite length and constant charge per unit length is calculated as follows:

E= 2k*(λ/r) Formula (1)

Where:

E: electric field .( N/C)

k: Coulomb electric constant. (N*m²/C²)

λ: linear charge density. (C/m)

r : distance from the charge line to the surface where E calculates (m)

Known data

E= 2.9  N/C

λ = 3.5*10⁻¹⁰ C/m

k= 8.99 *10⁹ N*m²/C²

Problem development

We replace data in the formula (1):

E= 2*k*(λ/r)

2.9= 2*8.99 *10⁹*(3.5*10⁻¹⁰/r)

r =( 2*8.99 *10⁹*3.5*10⁻¹⁰) / (2.9)

r= 2.17 m

5 0
2 years ago
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