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3 years ago

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

1 answer:

7 0

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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Four examples of compounds which are classed as carbohydrate

Solnce55 [7]

Answer:

Carbohydrates are divided into four types: monosaccharides, disaccharides, oligosaccharides, and polysaccharides. Monosaccharides consist of a simple sugar; that is, they have the chemical formula C 6 H 12 O 6. Disaccharides are two simple sugars. Oligosaccharides are three to six monosaccharide units, and polysaccharides are more than six.

6 0

3 years ago

If a force of 32000N exerted pressure of 160N/m² , find the area on which the force acts.

Alex17521 [72]

Answer: 200m^2

Explanation:

160N=32000N/x

x*160N=32000

x=200m^2

4 0

1 year ago

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ

Ray Of Light [21]

Answer:

period of oscillations is 0.695 second

Explanation:

given data

mass m = 0.350 kg

spring stretches x = 12 cm = 0.12 m

to find out

period of oscillations

solution

we know here that force

force = k × x .........1

so force = mg = 0.35 (9.8) = 3.43 N

3.43 = k × 0.12

k = 28.58 N/m

so period of oscillations is

period of oscillations = 2π × $\sqrt{\frac{m}{k} }$ ................2

put here value

period of oscillations = 2π × $\sqrt{\frac{0.35}{28.58} }$

period of oscillations = 0.6953

so period of oscillations is 0.695 second

4 0

3 years ago

Certain neutron stars (extremely dense stars) are believed to be rotating at about 10 rev/s. If such a star has a radius of 18 k

Aleks [24]

Answer:

mass of the neutron star =3.45185×10^26 Kg

Explanation:

When the neutron star rotates rapidly, a material on its surface to remain in place, the magnitude of the gravitational acceleration on the central material must be equal to magnitude of the centripetal acc. of the rotating star.

That is

$\frac{GM_{ns}}{R^2}= \omega^2 R$

M_ns = mass odf the netron star.

G= gravitational constant = 6.67×10^{-11}

R= radius of the star = 18×10^3 m

ω = 10 rev/sec = 20π rads/sec

therefore,

$M_{ns}= \frac{\omega^2R^3}{G} = \frac{4\pi^2\times(18\times10^3)^3}{6.67\times10^{-11}}$

= 3.45185... E26 Kg

= 3.45185×10^26 Kg

4 0

2 years ago

Longitudional waves travel through a series of ________ and ___________.

nekit [7.7K]

Answer:

compressions; rarefactions

Explanation:

4 0

2 years ago