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Effectus [21]
3 years ago
5

How does occluded fronts form.

Physics
1 answer:
Genrish500 [490]3 years ago
8 0

Answer:

when a cold front catches up to a warm front

Explanation:

cold fronts moving along much faster than warm fronts do. the cold front over takes the warm front and cold air moves into cooling air ahead of the warm front

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During both the vernal equinox and _____________, there is an equal amount of night and day.
choli [55]
During both the Vernal Equinox and Autumnal equinox, there is an equal amount of night and day
Vernal Equinox marks the beginning of Spring and Autumnal Equinox marks the beginning of Fall

hope this helps
5 0
3 years ago
Read 2 more answers
a 1.05 kg water bottle is sitting on the teacher's desk it is .20 m tall and has a radius of .03 m find the force that the water
arsen [322]
The force applied would be 1.05*9.8 = 10.3 N
the pressure is equal to F/a
area will be πr^2 = 0.002826
thus pressure will be = 10.3/0.002826= 3644.72 N/m^2
4 0
3 years ago
Calculate the force of gravity on the 0.60-kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth r
KIM [24]
The force of gravity between two objects is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the mass of the object is m_1=0.60 kg, while the Earth's mass is m_2=5.97 \cdot 10^{24} kg. Their separation is r=1.3 \cdot 10^7 m, therefore the gravitational force exerted on the object is
F=(6.67 \cdot 10^{-11}m^3 kg^{-1} s^{-2}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=1.4 N
5 0
3 years ago
2. A Se lanza un electrón con rapidez inicial v0 = 1.60×106 m/s hacia el interior de un campo uniforme entre las placas paralela
Vanyuwa [196]

Answer:

A)     E = 145.6 N / C , B)  y= 2,8 10-7 m with a downward direction

C) he shape of the trajectory of the two particles is to simulate a parabola,

D)     F_{e} /F_{g} = 10³⁴

Explanation:

A) For this exercise we use Newton's second law to find the acceleration of the electron, where the force is electric

           F = m a  

           - e E = m a

          a = - e E / m

with the field directed downward, the acceleration is in the vertical upward direction.

We look for how much the electron moves with kinematics, in the x direction there is no acceleration,

x axis (parallel to plates)

           x = v₀ t

           t = x / v₀

y axis (perpendicular to plates)

          y = y₀ + v_{oy} t + ½ a t²

Let's take the zero of the system in the middle of the plates y₀ = 0, also the initial vertical velocity is zero (v_{oy} = 0) the width of the plate is known

          y = ½ a t²

we substitute

         y = ½ (e E /m)  (x / v₀)²

         y = ½ e x2 /m v₀²   E

we look for the electric field

        E = 2 m y v₀² / e x²

where to use this expression the length and width of the condenser must be known, suppose that the length is x = l = 1 cm = 1 10⁻² m and the width is y = 0.5 mm = 0.5 10⁻³ m

let's calculate

         E = 2  9.1 10⁻³¹ 0.5 10⁻³ (1.6 10⁶)² / (1.6 10⁻¹⁹ (1 10⁻²)²)

         E = 145.6 N / C

B) The electron is exchanged for a proton

Let's look for the vertical displacement, in this case as the proton has a positive charge it moves towards the bottom of the plates

          y = ½ e x² / m v₀² E

          y = ½ 1.6 10⁻¹⁹ 1 10⁻⁴ / (1.67 10⁻²⁷ (1.6 10⁶)²   145.6

          y = 28.4375 10⁻⁸ m

since the distance between the plates is 0.5 10-3 m, the proton passes the condensate because its deflection is very small

In summary, its displacement is y= 2,8 10-7 m and with a downward direction (the same direction of the electric field)

C) The shape of the trajectory of the two particles is to simulate a parabola, but one for having a negative charge (electron) the force is upwards and the other for having a positive charge (proton) the trajectory is downwards

D) The force of gravity

           F_{g} = G m M / R²

electron

          Between the electron and the positive charges of the conducting plate

           F_{g}= 6.67 10⁻¹¹ 1.67 10⁻²⁷ 9.1 10⁻³¹ / (0.5 10⁻³)²

           F_{g} = 4.1 10⁻⁵¹ N

           

electric force

           F_{e} = -e E

           F_{e} = - 1.6 10⁻¹⁹ 145.6

           F_{e} = 2.3 10⁻¹⁷ N

let's look for the reason between these two forces

         F_{e} / F_{g} = 2.3 10⁻¹⁷ / 4.1 10⁻⁵¹

          F_{e} /F_{g} = 10³⁴

We see that the electric force is many orders of magnitude higher than the gravitational force.

5 0
3 years ago
12
tensa zangetsu [6.8K]

Answer:

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Explanation:

3 0
3 years ago
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