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weeeeeb [17]
3 years ago
5

(Thought question) At noon On June 21 will the shadow length of 42 degrees north latitude be shorter or longer than it is on Mar

ch 21? And, what is your reason for saying it will be either shorter or longer?
Physics
1 answer:
g100num [7]3 years ago
5 0

Answer: Shorter

Explanation: Shadow is formed when an light source is obstructed by an opaque object. The closer the source, shorter is the length of the shadow. In fact, when the source is exactly overhead, no shadow of the object is formed.

June 21 marks the Summer solstice which means the Sun passes directly overhead Tropic of cancer (23.5° N) at noon. March 21 marks the equinox which means sun passes directly overhead equator (0°).

Shadow length of an object at 42° Northern latitude will be shorter on June 21 because the Sun will be closer to this latitude as compared to March 21.

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bonufazy [111]

Переходи на сайте irkmix.top и получай много эмоций из Russia/

7 0
3 years ago
A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
MAVERICK [17]

Answer:

t₂ = 3.89 s

Explanation:

given,

speed of car  = 23 m/s

speed of motorcycle = 23 m/s

after time of 4 s distance between them is equal to = 53 m

motorcycle accelerates at = 7 m/s

time taken to catch up with car = ?

let t₂ be the time in which motorcycle catches car.

distance traveled by car in t₂ s

d = 23 t₂ + 53

distance traveled by motorcycle

using equation of motion

s = ut + \dfrac{1}{2}at^2

s = 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2

now, equating both the distances

23t_2 + 53= 23 t_2 + \dfrac{1}{2}\times 7 \times t_2^2

t_2^2 = 15.143

    t₂ = 3.89 s

time taken by the motorcycle to catch the car is equal to 3.89 s

4 0
3 years ago
How can understanding velocity help to prevent a mid-air collision
dsp73

Answer:

To maintain enough time to prevent a collision, a system operating in air traffic where aircraft speed does not

fall below 100 km/h (most medium-sized UAVs and GA aircraft) will need to be able to detect obstacles which

subtend an arc-width of as small as 0.125 mra

7 0
3 years ago
What would happen to the two balls if one of them were kept positively charged and the charge on the other ball were slowly incr
mariarad [96]

Answer:

the balls would move closer to each other

Explanation:

8 0
3 years ago
A 2.6 kg mass attached to a light string rotates on a horizontal,
Ainat [17]

The maximum speed the mass can have before it breaks is 2.27 m/s.

The given parameters:

  • <em>maximum mass the string can support before breaking, m = 17.9 kg</em>
  • <em>radius of the circle, r = 0.525 m</em>

The maximum speed the mass can have before it breaks is calculated as follows;

T = ma_c\\\\Mg = \frac{Mv^2}{r} \\\\v^2 = rg\\\\v = \sqrt{rg} \\\\v_{max} = \sqrt{0.525 \times 9.8} \\\\v_{max} = 2.27 \ m/s

Thus, the maximum speed the mass can have before it breaks is 2.27 m/s.

Learn more about maximum speed of horizontal circle here:brainly.com/question/21971127

8 0
3 years ago
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