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:
Explanation:
Given that,
Force is downward I.e negative y-axis
F = -2 × 10^-14 •j N
Magnetic field is westward, +x direction
B = 8.3 × 10^-2 •i T
Charge of an electron
q = 1.6 × 10^-19C
Velocity and it direction?
Force in a magnetic field is given as
F = q(V×B)
Angle between V and B is 270, check attachment
The cross product of velocity and magnetic field
F =qVB•Sin270
2 × 10^-14 = 1.6 × 10^-19 × V × 8.3 × 10^-2
Then,
v = 2 × 10^-14 / (1.6 × 10^-19 × 8.3 × 10^-2)
v = 1.51 × 10^6 m/s
Direction of the force
Let x be the direction of v
-F•j = v•x × B•i
From cross product
We know that
i×j = k, j×i = -k
j×k =i, k×j = -i
k×i = j, i×k = -j OR -k×i = -j
Comparing -k×i = -j to given problem
We notice that
-F•j = q ( -V•k × B×i)
So, the direction of V is negative z- direction
V = -1.51 × 10^6 •k m/s
Answer:
I think it's B
Explanation:
Because hese receptors are found in muscles, tendons, and joints.
M = 7.0 kg, the mass of the groceries
h = 1.2 m, the elevation of the bag of groceries
The bag of groceries moves a constant velocity over the 2.7-m room.
At constant velocity, there is no applied force, and the kinetic energy remains constant.
At an elevation of 1.2 m, there is an increase in PE (potential energy) given by
V = m*g*h
= (7.0 kg)*(9.8 m/s²)*(1.2 m)
= 82.32 J
The change in PE is equal to the work done.
Answer: 82.3 J
