To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.
The altitude is,
And the velocity can be written as,
From the properties of standard atmosphere at altitude z = 20km temperature is
Velocity of sound at this altitude is
Then the Mach number
So front stagnation temperature
Therefore the temperature at its front stagnation point is 689.87K
Answer:
d₂ = 1.466 m
Explanation:
In this case we must use the rotational equilibrium equations
Στ = 0
τ = F r
we must set a reference system, we use with origin at the easel B and an axis parallel to the plank
, we will use that the counterclockwise ratio is positive
+ W d₁ - w_cat d₂ = 0
d₂ = W / w d₁
d₂ = M /m d₁
d₂ = 5.00 /2.9 0.850
d₂ = 1.466 m
because poor and people of back places may not have enough knowledge about contraceptive devices and may not have enough money to buy it.That's why poverty and backwardness are considered to be a cause of population growth
Answer:
A. The electron will begin to move along the axis, towards the centre and the instantaneous velocity because the force acting on it depends largely on acceleration and x until it reaches maximum velocity at centre.
B. Veloctiy (Vb) = 1.66m/s
Explanation:
Given the following data
x(a) = 0.3m
x(b) = 0
q = 1.6×10^-19
Q = 24nc
r = 0.15m
Required: the motion of the electron and the velocity (Vb)
1. At point A the electron will begin to move along the axis from point A to point B, the magnitude of the electric field will change while moving which depends on that and this will produce instantaneous force which will later change and the acceleration will change too while moving, the velocity would reach maximum value at point B
2. Potential energy and kinetic energy are given by
U(a) + K(a) = U(b) + K(b). . .1
Initial P.E and K.E are given as
U(a) = kQ/√x²(a) + a2
By substitution, we have
U(a) = 9×10^9 × (-1.9×10^-19)×24×10^-9/√(0.15)²+(0.3²)
U(a) = -1.03×10^-16
Final P.E and K.E are given as
U(b) = KQ/√x²(b) + a2
By substitution, we have
U(b) = 9×10^9×(-1.9×10^-19)×24×10^-9/√(0.15)²+(0)²
U(b) = -2.3×10^16
3. By substitution into equation 1 becomes
-1.03×10^-6 - 2.3×10^-16 + MV²(b)/2
V(b) = √2×1.27×10^-16/9.1×10^31
V(b) = 1.66×10^7m/s
Answer:
F_total = 10000 N
Explanation:
For this exercise we use that the force is a vector and the best way to do the sum is that since the two tugs pull the boat with the same intensity and angle, the sum of the component of the force perpendicular to the movement becomes zero.
The components parallel to the movement of the tugs is
∑ F = 2 Fcos θ
let's calculate
F_total = 2 10000 cos 60
F_total = 10000 N
