Given that,
Distance in south-west direction = 250 km
Projected angle to east = 60°
East component = ?
since,
cos ∅ = base/hypotenuse
base= hyp * cos ∅
East component = 250 * cos 60°
East component = 125 km
Answer:
Relative population is 2.94 x 10⁻¹⁰.
Explanation:
Let N₁ and N₂ be the number of atoms at ground and first excited state of helium respectively and E₁ and E₂ be the ground and first excited state energy of helium respectively.
The ratio of population of atoms as a function of energy and temperature is known as Boltzmann Equation. The equation is:
= 
= 
Here g₁ and g₂ be the degeneracy at two levels, K is Boltzmann constant and T is equilibrium temperature.
Put 1 for g₁, 3 for g₂, -19.82 ev for (E₁ - E₂) and 8.6x10⁵ ev/K for K and 10000 k for T in the above equation.
= 
= 3.4 x 10⁹
= 2.94 x 10⁻¹⁰
Answer:
17 m/s south
Explanation:
= Mass of dog = 10 kg
= Mass of skateboard = 2 kg
v = Combined velocity = 2 m/s
= Velocity of dog = 1 m/s
= Velocity of skateboard
In this system the linear momentum is conserved

The velocity of the skateboard will be 17 m/s south as the north is taken as positive
Answer: -2 km
Explanation:
If we imagine Jin's movement to be the hypothenuse of a right triangle, then the southern component of Jin's movement corresponds to the side of the triangle opposite to the angle of 30 degrees. Therefore, the magnitude of this southern component is given by
However, the angle of 30 degrees is south of east: this means that the direction of this southern component is south, and since we generally take north as positive direction, we must add a negative sign, so the correct answer is
-2 km
Answer:
h = 2 R (1 +μ)
Explanation:
This exercise must be solved in parts, first let us know how fast you must reach the curl to stay in the
let's use the mechanical energy conservation agreement
starting point. Lower, just at the curl
Em₀ = K = ½ m v₁²
final point. Highest point of the curl
= U = m g y
Find the height y = 2R
Em₀ = Em_{f}
½ m v₁² = m g 2R
v₁ = √ 4 gR
Any speed greater than this the body remains in the loop.
In the second part we look for the speed that must have when arriving at the part with friction, we use Newton's second law
X axis
-fr = m a (1)
Y Axis
N - W = 0
N = mg
the friction force has the formula
fr = μ N
fr = μ m g
we substitute 1
- μ mg = m a
a = - μ g
having the acceleration, we can use the kinematic relations
v² = v₀² - 2 a x
v₀² = v² + 2 a x
the length of this zone is x = 2R
let's calculate
v₀ = √ (4 gR + 2 μ g 2R)
v₀ = √4gR( 1 + μ)
this is the speed so you must reach the area with fricticon
finally have the third part we use energy conservation
starting point. Highest on the ramp without rubbing
Em₀ = U = m g h
final point. Just before reaching the area with rubbing
= K = ½ m v₀²
Em₀ = Em_{f}
mgh = ½ m 4gR(1 + μ)
h = ½ 4R (1+ μ)
h = 2 R (1 +μ)