Answer:
Explanation:
Important here is to know that due north is a 90 degree angle, due east is a 0 degree angle, and due south is a 270 degree angle. Then we find the x and y components of each part of this journey using the sin and cos of the angles multiplied by each magnitude:

Add them all together to get the x component of the resultant vector, V:

Do the same to find the y components of the part of this journey:

Add them together to get the y component of the resultant vector, V:

One thing of import to note is that both of these components are positive, so the resultant angle lies in QI.
We find the final magnitude:
and, rounding to 2 sig dig's as needed:
1.0 × 10² m; now for the direction:
58°
Answer;
A. orographic lifting
Explanation;
Orographic lifting is a process that takes place when an air mass is forced from a low elevation to a higher elevation as it moves over rising terrain.
This can be explained by; When air is blocked by mountains, it cannot go through these mountains, As it ascends or moves up the mountain, the air then cools as it rises and when it cools to its saturation point, the water vapor condenses and cloud forms.
These clouds formed are known as orographic clouds, that develop in response to the lifting forced by the topography the earth.
Answer:
1. Dheere Dheere (slowly slowly)
2. Har (every)
3. Kal (tomorrow)
4. Mat (don't)
5. Andar (inside)
sorry I wasn't able to write in hindi
Answer:

Explanation:
GIVEN
diameter = 15 fm =
m
we use here energy conservation

there will be some initial kinetic energy but after collision kinetic energy will zero

on solving these equations we get kinetic energy initial
J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the
nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V

Answer
given,
For helium
Volume,V = 46 L
Pressure,P = 1 atm
Temperature,T = 25°C = 273 +25 = 298 K
R=0.0821 L . atm /mole.K
n₁ = ?
number of moles
we know
P V = n R T

n₁ = 1.89 moles
For oxygen
Volume,V = 12 L
Pressure,P = 1 atm
Temperature,T = 25°C = 273 +25 = 298 K
R=0.0821 L . atm /mole.K
n₂ = ?
number of moles
we know
P V = n R T

n₂ = 0.49 moles
Total volume of tank = 5 L
temperature of tank = 298 K
Partial pressure of helium


P₁ = 9.25 atm
Partial pressure of oxygen


P₂ = 2.39 atm
total pressure
P = P₁ + P₂
P = 9.25 + 2.39
P = 11.64 atm