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3 years ago

When the northern hemisphere is tilted toward the sun that part of the earth is closer to the sun and thus hotter? True or false

Physics

1 answer:

Ratling [72]3 years ago

8 0

<u>False</u>

<u>Explanation:</u>

Most of us have a belief that Earth is very close to the sun in summer season and so it is hotter. Also, since Earth is far away from the sun in winter season, it is colder. Though this idea looks true, it is incorrect.

The orbit of the Earth isn't a perfect circle. It is quite tilted. During some months of the year, Earth is very close to the sun than at other times. We have winter when the Earth is very close to the sun and summer when it is far away in the Northern Hemisphere. So when we compare the distance of the Sun from the Earth, this change in Earth's distance throughout the year does not affect our weather much.

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Which of the following is a false statement about dispersion forces? View Available Hint(s) Which of the following is a false st

icang [17]

Explanation :

Dispersion forces are also known as London dispersion forces. It is the weakest force. Also, it is the part of the Van der Waals forces.

(1) This force is exhibited by all atoms and molecules.

(2) These forces are the result of the fluctuations in the electron distribution within molecules or atoms. Due to these fluctuations, the electric field is created. The magnitude of this force is explained in terms of Hamaker constant 'A'.

(3) Dispersion forces result from the formation of instantaneous dipoles in a molecule or atom. When electrons are more concentrated in a place, instantaneous dipoles formed.

(4) Dispersion force magnitude depends on the amount of surface area available for interactions. If the area increases, the size of the atom also increase. As a result, stronger dispersion forces.

So, the false statement is "Dispersion forces always have a greater magnitude in molecules with a greater molar mass".

4 0

3 years ago

Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.04 m away from a waterfall 0.585 m in heigh

S_A_V [24]

Answer:

V₀ = 5.47 m/s

Explanation:

The jumping motion of the Salmon can be modelled as the projectile motion. So, we use the formula for the range of projectile motion here:

R = V₀² Sin 2θ/g

where,

R = Range of Projectile = 3.04 m

θ = Launch Angle = 41.7°

V₀ = Minimum Launch Speed = ?

g = 9.81 m/s²

Therefore,

3.04 m = V₀² [Sin2(41.7°)]/(9.81 m/s²)

V₀² = 3.04 m/(0.10126 s²/m)

V₀ = √30.02 m²/s²

6 0

2 years ago

A cyclotron is to accelerate protons to an energy of 5.4 MeV. The superconduction electromagnet of the cyclotron produces a 2.9T

mart [117]

<h3><u>Answer;</u></h3>

Radius = 0.0818 m

Angular velocity = 2.775 × 10^7 rad/sec

<h3><u>Explanation;</u></h3>

The mass of proton m=1.6748 × 10^-27 kg;

Charge of electron e= 1.602 × 10^-19 C;

kinetic energy E= 2.7 MeV

= 2.7 × 10^6 × 1.602 × 10^-19 J;

= 4.32 × 10^-13 Joules

But; K.E =0.5m*v^2,

Hence v=√(2K.E/m)

Velocity = 2.27 × 10^7 m/s

Angular velocity, ω = v/r

Therefore; V = ωr

Hence; V = √(2K.E/m) = ωr

r= √(2E/m)/w = √E*√(2*m)/(eB)

= √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)

but E = 4.32 × 10^-13 Joules

r = 0.0818 m

Angular speed

Angular velocity, ω = v/r , where r is the radius and v is the velocity

Therefore;

Angular velocity = 2.27 × 10^7 / 0.0818 m

= 2.775 × 10^7 rad /sec

3 0

2 years ago

Please explain your answer when you found the answer (Only question number 9)

Zinaida [17]

<h3>With a uniform acceleration of 2 m/s ²</h3>

5 0

2 years ago

If the load on Pulley E is 20N, how much effort is needed to life it?

Mariana [72]

It would need to be over 20 because if the load of the Pulley E is 20 and the effort is 20, then they will be equal and the Pulley would not move, so your answer is at least 20

7 0

3 years ago