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3 years ago

on june 21 some earth locations have 24 hours of daylight these locations are all between the latitudes of

Physics

1 answer:

andrew-mc [135]3 years ago

4 0

66° N and 90° N

the area of the artic circle in the northern hemisphere

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Does the sun sun traces shortest path across local sky on june solstice

viktelen [127]

The June solstice in the Northern hemisphere is the summer solstice. The June Solstice in the Southern hemisphere is the winter solstice. The summer solstice is equivalent to the longest day while the winter solstice is equivalent to the shortest day. Therefore on the local sky, when is the June solstice we have have the longest day (longest path of sun in the sky) in the Northern hemisphere and the shortest day (shortest path of sun in the sky) in the Southern hemisphere.

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3 years ago

A projectile is fired at v 0 = 381.0 m/s at an angle of θ = 73.5 ∘ , with respect to the horizontal. Assume that air friction wi

Ipatiy [6.2K]

Answer:

Explanation:

velocity of projection, vo = 381 m/s

angle of projection, θ = 73.5°

The formula for the range is

$R=\frac{u^{2}Sin2\theta }{g}$

$R=\frac{381^{2}Sin147 }{9.8}$

R = 8067.4 m

Range in shorten by 34.1 %

So, the new range is

R' = 8067.4 - 34.1 x 8067.4/100

R' = 5316.4 m

5 0

3 years ago

Free point cuz i cewl like dat‍♀️‍♂️

DedPeter [7]

Yes thank u teehee

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8 0

2 years ago

Read 2 more answers

How do we represent Earth’s face?

jeka94

Answer:

Human-driven changes in arrive utilize and arrive cover such as deforestation, urbanization, and shifts in vegetation designs moreover change the climate, coming about in changes to the reflectivity of the Soil surface (albedo), emanations from burning timberlands, urban warm island impacts and changes within the normal water cycle.

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3 years ago

A uniformly charged conducting sphere of 1.1 m diameter has a surface charge density of 6.2 µC/m2. (a) Find the net charge on th

ira [324]

Answer:

(a) q = 2.357 x 10⁻⁵ C

(b) Φ = 2.66 x 10⁶ N.m²/C

Explanation:

Given;

diameter of the sphere, d = 1.1 m

radius of the sphere, r = 1.1 / 2 = 0.55 m

surface charge density, σ = 6.2 µC/m²

(a) Net charge on the sphere

q = 4πr²σ

where;

4πr² is surface area of the sphere

q is the net charge on the sphere

σ is the surface charge density

q = 4π(0.55)²(6.2 x 10⁻⁶)

q = 2.357 x 10⁻⁵ C

(b) the total electric flux leaving the surface of the sphere

Φ = q / ε

where;

Φ is the total electric flux leaving the surface of the sphere

ε is the permittivity of free space

Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)

Φ = 2.66 x 10⁶ N.m²/C

8 0

2 years ago