The true velocity is 250 km/h.
Explanation:
As the aircraft is moving with speed of 300 km/h and the wind velocity is in east direction with the magnitude of 60 km/h. The resultant velocity or the true velocity can be found by the vector addition of the velocities.
If the true velocity is represented as V and the velocity of aircraft is represented as vₐ and the velocity of wind as vₓ, then
V² = vₐ² + vₓ² - vₐ.vₓ cosθ
So, the angle formed between them is 30°
Then, V² = (300)² + (60)² - 2×(300×60) cos 30°
V²=90000+3600-31177=62423
V =250 km/h
So, the true velocity is having magnitude of 250 km/h.
Thus, the true velocity is 250 km/h.
Oxygen, Carbon dioxide and other small uncharged molecules can move through the cell membrane through a method called osmosis
osmosis is when there is a semipermeable membrane and molecules pass through it, where there is less concentration of molecules.
Answer:
- The emission wavelength for λ(3,4) is 1875.24 nm
- The emission wavelength for λ(4,6) is 2625.34 nm
- The emission wavelength for λ(4,7) is 2165.69 nm
- The emission wavelength for λ(4,8) is 1944.70 nm
- The emission wavelength for λ(4,9) is 1817.54 nm
Explanation:
Using Rydberg equation
where;
R is Rydberg equation = 1.097 x 10⁷ m⁻¹
For λ(3,4)
For λ(4,6)
For λ(4,7)
For λ(4,8)
For λ(4,9)
Answer:
190 m/s
Explanation:
Acceleration = Force/ mass
a = F/m........................ Equation 1
Where F = force, m = mass of the space probe.
Given: m = 20 kg,
F(t) = 6.00t²
when t = 10 s.
F = 6.00(10)²
F = 600 N
Substituting these values into equation 1
a = 600/20²
a = 30 m/s.
But.
a = (v-u)/t
Where a = acceleration of the body, v = final velocity of the body, u = initial velocity of the body, t = time.
making v the subject of the equation,
v = u+at..................... Equation 2
Given: a = 30 m/s², u = 40 m/s, t = 5.00 s.
v = 40 + 30(5)
v = 40+150
v = 190 m/s
Thus the velocity of the space probe = 190 m/s