Answer:
i = 0.477 10⁴ B
the current flows in the counterclockwise
Explanation:
For this exercise let's use the Ampere law
∫ B . ds = μ₀ I
Where the path is closed
Let's start by locating the current vines that are parallel to the z-axis, so it must be exterminated along the x-axis and as the specific direction is not indicated, suppose it extends along the y-axis.
From BiotSavart's law, the field must be perpendicular to the direction of the current, so the magnetic field must go in the x direction.
We apply the law of Ampere the segment parallel to the x-axis is the one that contributes to the integral, since the other two have an angle of 90º with the magnetic field
Segment on the y axis
L₀ = (y2-y1)
L₀ = 3-0 = 3 cm
Segment on the point x = 2 cm
L₁ = 3-0
L₁ = 3cm
B L = μ₀ I
B 2L = μ₀ I
i = 2 L B /μ₀
i= 2 0.03 / 4π 10⁻⁷ B
i = 4.77 10⁴ B
The current is perpendicular to the magnetic field whereby the current flows in the counterclockwise
Mechanical advantage is the amount by which a machine multiplies input force. It is a measure of the force amplification using a tool or machine system.
Answer:
The wave speed is calculated below:
Explanation:
Given,
number of waves passed per minute = 8
time period = 1 minute = 60 s
distance between successive wave crests = 20 m
waves passing interval per second =

Now,
wave speed = 20 m ×

=
m/s
= 2.67 m/s
Hence the wave speed is 2.67 m/s.
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :

1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²