All categories

3 years ago

What happens during the fall in September?

Physics

2 answers:

san4es73 [151]3 years ago

4 0

Answer:

Explanation:

8090 [49]3 years ago

3 0

Answer:

During the fall in September, the Sun passes from northern hemisphere towards equator. ... On this day, Sunlight directly falls on the equator i.e. the sun crosses celestial equator. Day time is equal to night time in the two hemispheres

Explanation:

make me brainiest if it helps you

You might be interested in

Laskar, J.: 1990, The chaotic motion of the solar system. A numerical estimate of the size of the chaotic zones, Icarus, 88, 266

balandron [24]

The chaotic nature of the Solar System excluding Pluto was established by the numerical computation of the maximum Lyapunov exponent of its secular system over 200 myr.

<h3>What is chaotic motion of the solar system ?</h3>

There has been an increase in awareness of chaotic dynamics in the solar system over the past 20 years. The orbits of tiny objects in the solar system, such as asteroids, comets, and interplanetary dust, are now known to be chaotic and to experience significant variations across geological time periods.

a completely unpredictable orbit, or one where significant changes in the orbit can result from even small changes in the position and/or velocity of the orbiting entity.

Learn more about Chaotic motion here:

brainly.com/question/13717859

#SPJ4

7 0

1 year ago

The kinetic energy of a rocket is increased by a factor of eight after its engines are fired, whereas its total mass is reduced

Rudik [331]

The momentum increases by a factor of 2

Explanation:

We can solve this problem by rewriting the momentum of the rocket in terms of the kinetic energy and the mass.

The kinetic energy of the rocket is:

$K=\frac{1}{2}mv^2$ (1)

where

m is the mass

v is the velocity

The momentum of the rocket is

$p=mv$ (2)

From eq.(1) we get

$v=\sqrt{\frac{2K}{m}}$

and substituting into (2),

$p=\sqrt{2mK}$

Now in this problem we have:

- The kinetic energy of the rocket is increased by a factor 8:

$K' = 8K$

- The mass is reduced by half:

$m'=\frac{m}{2}$

Substituting, we find the new momentum:

$p'=\sqrt{2(\frac{m}{2}(8K)}=\sqrt{4(2mK)}=2\sqrt{2mK}=2p$

So, the momentum increases by a factor of 2.

Learn more about momentum and kinetic energy:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

brainly.com/question/6536722

#LearnwithBrainly

4 0

3 years ago

An empty truck traveling at 10 km/h has kinetic energy. How much kinetic energy does it have when loaded so that its mass and it

Katena32 [7]

Answer:

8 time increase in K.E.

Explanation:

Consider Mass of truck = m kg and speed = v m/s then

K.E. = 1/2 ×mv²

If mass and speed both are doubled i.e let m₀ = 2m and v₀ = 2v then

(K.E.)₀ = 1/2 ×2m(2v)²

(K.E.)₀ = 8 (1/2 × mv²) = 8 × K.E.

7 0

3 years ago

What mistake did Farah make in this experiment? Farah conducted the following experiment to check whether fabrics of different c

Cerrena [4.2K]

She hung up the pieces of cotton BEFORE putting the fan on, also the water evaporating can take longer or shorter even if they are all the same :))

3 0

3 years ago

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun.

lord [1]

Answer:

(a) $F_{sm} = 4.327\times 10^{20}\ N$

(b) $F_{em} = 1.983\times 10^{20}\ N$

Solution:

As per the question:

Mass of Earth, $M_{e} = 5.972\times 10^{24}\ kg$

Mass of Moon, $M_{m} = 7.34\times 10^{22}\ kg$

Mass of Sun, $M_{s} = 1.989\times 10^{30}\ kg$

Distance between the earth and the moon, $R_{em} = 3.84\times 10^{8}\ m$

Distance between the earth and the sun, $R_{es} = 1.5\times 10^{11}\ m$

Distance between the sun and the moon, $R_{sm} = 1.5\times 10^{11}\ m$

Now,

We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:

$F_{G} = \frac{Gmm'_{2}}{r^{2}}$ (1)

Now,

(a) The force exerted by the Sun on the Moon is given by eqn (1):

$F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}$

$F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}$

$F_{sm} = 4.327\times 10^{20}\ N$

(b) The force exerted by the Earth on the Moon is given by eqn (1):

$F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}$

$F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}$

$F_{em} = 1.983\times 10^{20}\ N$

$F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}$

$F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}$

$F_{se} = 3.521\times 10^{20}\ N$

7 0

3 years ago