Answer:
d) 1.2 mT
Explanation:
Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.
First of all, we observe that:
- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is
I = 15 A
- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).
Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by
where
is the vacuum permeability
I = 15 A is the current in the conductor
r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field
Substituting, we find:
Answer:
See the answers below.
Explanation:
to solve this problem we must make a free body diagram, with the forces acting on the metal rod.
i)
The center of gravity of the rod is concentrated in half the distance, that is, from the end of the bar to the center there is 40 [cm]. This can be seen in the attached free body diagram.
We have only two equilibrium equations, a summation of forces on the Y-axis equal to zero, and a summation of moments on any point equal to zero.
For the summation of forces we will take the forces upwards as positive and the negative forces downwards.
ΣF = 0
Now we perform a sum of moments equal to zero around the point of attachment of the string with the metal bar. Let's take as a positive the moment of the force that rotates the metal bar counterclockwise.
ii) In the free body diagram we can see that the force acts at 18 [cm] of the string.
ΣM = 0
![(15*9) - (18*W) = 0\\135 = 18*W\\W = 7.5 [N]](https://tex.z-dn.net/?f=%2815%2A9%29%20-%20%2818%2AW%29%20%3D%200%5C%5C135%20%3D%2018%2AW%5C%5CW%20%3D%207.5%20%5BN%5D)
Complete question:
The coordinate of a particle in meters is given by x(t)=1 6t- 3.0t³ , where the time tis in seconds. The
particle is momentarily at rest at t is:
Select one:
a. 9.3s
b. 1.3s
C. 0.75s
d.5.3s
e. 7.3s
Answer:
b. 1.3 s
Explanation:
Given;
position of the particle, x(t)=1 6t- 3.0t³
when the particle is at rest, the velocity is zero.
velocity = dx/dt
dx /dt = 16 - 9t²
16 - 9t² = 0
9t² = 16
t² = 16 /9
t = √(16 / 9)
t = 4/3
t = 1.3 s
Therefore, the particle is momentarily at rest at t = 1.3 s
Explanation:
1. To graphically add vectors, use the tail-to-tip method. Draw the first vector (it doesn't matter which), then draw the second vector where the first vector ends. The resultant vector is from the tail of the first vector to the tip of the second vector.
This graph shows two ways to get the resultant: A + B or B + A.
desmos.com/calculator/bqhcclhhqc
2. To algebraically add vectors, split each vector into x and y components.
Aₓ = 5.0 cos 45 = 3.5
Aᵧ = 5.0 sin 45 = 3.5
Bₓ = 2.0 cos 180 = -2.0
Bᵧ = 5.0 sin 180 = 0
The components of the resultant vector are the sums of the components of A and B.
Cₓ = 3.5 + -2.0 = 1.5
Cᵧ = 3.5 + 0 = 3.5
The magnitude of the resultant vector is found with Pythagorean theorem, and the direction is found with tangent.
C = √(Cₓ² + Cᵧ²) ≈ 3.9 m/s
θ = atan(Cᵧ / Cₓ) ≈ 67°
Answer:
I think golf does not demand high level of physical fitness. A golfer just need be skillful and able to play under pressure. In terms of physical exercise i think he/she just need to do some muscle stretches.
Explanation:
